##### Class 12^{th} Mathematics Part Ii CBSE Solution

**Exercise 11.1**- If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively,…
- Find the direction cosines of a line which makes equal angles with the…
- If a line has the direction ratios -18, 12, -4, then what are its direction…
- Show that the points (2, 3, 4), (-1, -2, 1), (5, 8, 7) are collinear.…
- Find the direction cosines of the sides of the triangle whose vertices are (3,…

**Exercise 11.2**- Show that the three lines with direction cosines 12/13 , -3/13 , -4/13 4/13 ,…
- Show that the line through the points (1, -1, 2), (3, 4, -2) is perpendicular…
- Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the…
- Find the equation of the line which passes through the point (1, 2, 3) and is…
- Find the equation of the line in vector and in cartesian form that passes…
- Find the cartesian equation of the line which passes through the point (-2, 4,…
- The Cartesian equation of a line is x-5/3 = y+4/7 = z-6/2 . Write its vector…
- Find the vector and the cartesian equations of the lines that passes through…
- . Find the vector and the cartesian equations of the line that passes through…
- Find the angle between the following pairs of lines:
- . Find the angle between the following pair of lines: x-2/2 = y-1/5 - z+3/-3…
- Find the values of p so that the lines 1-x/3 = 7y-14/2p = z-3/2 7-7x/3p =…
- Show that the lines x-5/7 = y+2/-5 = z/1 x/1 = y/2 = z/3 are perpendicular to…
- vector r = (i+2 j + k) + lambda (i - j + k) vector r = 2 i - j - k + μ (2 i +…
- x+1/7 = y+1/-6 = z+1/1 x-3/1 = y-5/-2 = z-7/1 Find the shortest distance…
- Find the shortest distance between the lines whose vector equations are…
- vector r = (1-t) i + (t-2) j + (3-2t) k vector r = (s+1) i + (2s-1) j - (2s+1)…

**Exercise 11.3**- z = 2 In each of the following cases, determine the direction cosines of the…
- x + y + z = 1 In each of the following cases, determine the direction cosines…
- 2x + 3y - z = 5 In each of the following cases, determine the direction…
- 5y + 8 = 0 In each of the following cases, determine the direction cosines of…
- Find the vector equation of a plane which is at a distance of 7 units from the…
- vector t (i + j - k) = 2 Find the Cartesian equation of the following planes:…
- Find the Cartesian equation of the following planes:
- Find the Cartesian equation of the following planes:
- 2x + 3y + 4z - 12 = 0 In the following cases, find the coordinates of the foot…
- 2x + 3y + 4z - 12 = 0 In the following cases, find the coordinates of the foot…
- 3y + 4z - 6 = 0 In the following cases, find the coordinates of the foot of…
- x + y + z = 1 In the following cases, find the coordinates of the foot of the…
- that passes through the point (1, 0, -2) and the normal to the plane is i + j…
- that passes through the point (1,4, 6) and the normal vector to the plane is…
- (1, 1, -1), (6, 4, -5), (-4, -2, 3) Find the equations of the planes that…
- (1, 1, 0), (1, 2, 1), (-2, 2, -1) Find the equations of the planes that passes…
- Find the intercepts cut off by the plane 2x + y - z = 5.
- Find the equation of the plane with intercept 3 on the y-axis and parallel to…
- Find the equation of the plane through the intersection of the planes 3x - y +…
- Find the vector equation of the plane passing through the intersection of the…
- Find the equation of the plane through the line of intersection of the planes…
- Find the angle between the planes whose vector equations are
- 7x + 5y + 6z + 30 = 0 and 3x - y - 10z + 4 = 0 In the following cases,…
- 2x + y + 3z - 2 = 0 and x - 2y + 5 = 0 In the following cases, determine…
- 2x - 2y + 4z + 5 = 0 and 3x - 3y + 6z - 1 = 0 In the following cases,…
- 2x - 2y + 4z + 5 = 0 and 3x - 3y + 6z - 1 = 0 In the following cases,…
- 4x + 8y + z - 8 = 0 and y + z - 4 = 0 In the following cases, determine…
- Point Plane (0, 0, 0) 3x - 4y + 12 z = 3 In the following cases, find the…
- Point Plane (3, - 2, 1) 2x - y + 2z + 3 = 0 In the following cases, find the…
- Point Plane (2, 3, - 5) x + 2y - 2z = 9 In the following cases, find the…
- Point Plane (-6, 0, 0) 2x - 3y + 6z - 2 = 0 In the following cases, find the…

**Miscellaneous Exercise**- Show that the line joining the origin to the point (2, 1, 1) is perpendicular…
- If l1, m1, n1 and l2, m2, n2 are the direction cosines of two mutually…
- Find the angle between the lines whose direction ratios are a, b, c and b - c,…
- Find the equation of a line parallel to x - axis and passing through the…
- If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (-4, 3,…
- If the lines x-1/3k = y-2/1 = z-3/-5 and x-1/3k = y-2/1 = z-3/-5 are…
- Find the vector equation of the line passing through (1, 2, 3) and…
- Find the equation of the plane passing through (a, b, c) and parallel to the…
- Find the shortest distance between lines vector r = (6 i+2 j+2 k) + lambda (1…
- Find the coordinates of the point where the line through (5, 1, 6) and (3,…
- Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,…
- Find the coordinates of the point where the line through (3, -4, -5) and (2,…
- Find the equation of the plane passing through the point (-1, 3, 2) and…
- If the points (1, 1, p) and (-3, 0, 1) be equidistant from the plane vector r…
- Find the equation of the plane passing through the line of intersection of the…
- If O be the origin and the coordinates of P be (1, 2, -3), then find the…
- Find the equation of the plane which contains the line of intersection of the…
- Find the distance of the point (-1, -5, -10) from the point of intersection of…
- Find the vector equation of the line passing through (1, 2, 3) and parallel to…
- Find the vector equation of the line passing through the point (1, 2, - 4) and…
- Prove that if a plane has the intercepts a, b, c and is at a distance of p…
- Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 isA. 2…
- The planes: 2x - y + 4z = 5 and 5x - 2.5y + 10z = 6 areA. Perpendicular B.…

**Exercise 11.1**

- If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively,…
- Find the direction cosines of a line which makes equal angles with the…
- If a line has the direction ratios -18, 12, -4, then what are its direction…
- Show that the points (2, 3, 4), (-1, -2, 1), (5, 8, 7) are collinear.…
- Find the direction cosines of the sides of the triangle whose vertices are (3,…

**Exercise 11.2**

- Show that the three lines with direction cosines 12/13 , -3/13 , -4/13 4/13 ,…
- Show that the line through the points (1, -1, 2), (3, 4, -2) is perpendicular…
- Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the…
- Find the equation of the line which passes through the point (1, 2, 3) and is…
- Find the equation of the line in vector and in cartesian form that passes…
- Find the cartesian equation of the line which passes through the point (-2, 4,…
- The Cartesian equation of a line is x-5/3 = y+4/7 = z-6/2 . Write its vector…
- Find the vector and the cartesian equations of the lines that passes through…
- . Find the vector and the cartesian equations of the line that passes through…
- Find the angle between the following pairs of lines:
- . Find the angle between the following pair of lines: x-2/2 = y-1/5 - z+3/-3…
- Find the values of p so that the lines 1-x/3 = 7y-14/2p = z-3/2 7-7x/3p =…
- Show that the lines x-5/7 = y+2/-5 = z/1 x/1 = y/2 = z/3 are perpendicular to…
- vector r = (i+2 j + k) + lambda (i - j + k) vector r = 2 i - j - k + μ (2 i +…
- x+1/7 = y+1/-6 = z+1/1 x-3/1 = y-5/-2 = z-7/1 Find the shortest distance…
- Find the shortest distance between the lines whose vector equations are…
- vector r = (1-t) i + (t-2) j + (3-2t) k vector r = (s+1) i + (2s-1) j - (2s+1)…

**Exercise 11.3**

- z = 2 In each of the following cases, determine the direction cosines of the…
- x + y + z = 1 In each of the following cases, determine the direction cosines…
- 2x + 3y - z = 5 In each of the following cases, determine the direction…
- 5y + 8 = 0 In each of the following cases, determine the direction cosines of…
- Find the vector equation of a plane which is at a distance of 7 units from the…
- vector t (i + j - k) = 2 Find the Cartesian equation of the following planes:…
- Find the Cartesian equation of the following planes:
- Find the Cartesian equation of the following planes:
- 2x + 3y + 4z - 12 = 0 In the following cases, find the coordinates of the foot…
- 2x + 3y + 4z - 12 = 0 In the following cases, find the coordinates of the foot…
- 3y + 4z - 6 = 0 In the following cases, find the coordinates of the foot of…
- x + y + z = 1 In the following cases, find the coordinates of the foot of the…
- that passes through the point (1, 0, -2) and the normal to the plane is i + j…
- that passes through the point (1,4, 6) and the normal vector to the plane is…
- (1, 1, -1), (6, 4, -5), (-4, -2, 3) Find the equations of the planes that…
- (1, 1, 0), (1, 2, 1), (-2, 2, -1) Find the equations of the planes that passes…
- Find the intercepts cut off by the plane 2x + y - z = 5.
- Find the equation of the plane with intercept 3 on the y-axis and parallel to…
- Find the equation of the plane through the intersection of the planes 3x - y +…
- Find the vector equation of the plane passing through the intersection of the…
- Find the equation of the plane through the line of intersection of the planes…
- Find the angle between the planes whose vector equations are
- 7x + 5y + 6z + 30 = 0 and 3x - y - 10z + 4 = 0 In the following cases,…
- 2x + y + 3z - 2 = 0 and x - 2y + 5 = 0 In the following cases, determine…
- 2x - 2y + 4z + 5 = 0 and 3x - 3y + 6z - 1 = 0 In the following cases,…
- 2x - 2y + 4z + 5 = 0 and 3x - 3y + 6z - 1 = 0 In the following cases,…
- 4x + 8y + z - 8 = 0 and y + z - 4 = 0 In the following cases, determine…
- Point Plane (0, 0, 0) 3x - 4y + 12 z = 3 In the following cases, find the…
- Point Plane (3, - 2, 1) 2x - y + 2z + 3 = 0 In the following cases, find the…
- Point Plane (2, 3, - 5) x + 2y - 2z = 9 In the following cases, find the…
- Point Plane (-6, 0, 0) 2x - 3y + 6z - 2 = 0 In the following cases, find the…

**Miscellaneous Exercise**

- Show that the line joining the origin to the point (2, 1, 1) is perpendicular…
- If l1, m1, n1 and l2, m2, n2 are the direction cosines of two mutually…
- Find the angle between the lines whose direction ratios are a, b, c and b - c,…
- Find the equation of a line parallel to x - axis and passing through the…
- If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (-4, 3,…
- If the lines x-1/3k = y-2/1 = z-3/-5 and x-1/3k = y-2/1 = z-3/-5 are…
- Find the vector equation of the line passing through (1, 2, 3) and…
- Find the equation of the plane passing through (a, b, c) and parallel to the…
- Find the shortest distance between lines vector r = (6 i+2 j+2 k) + lambda (1…
- Find the coordinates of the point where the line through (5, 1, 6) and (3,…
- Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,…
- Find the coordinates of the point where the line through (3, -4, -5) and (2,…
- Find the equation of the plane passing through the point (-1, 3, 2) and…
- If the points (1, 1, p) and (-3, 0, 1) be equidistant from the plane vector r…
- Find the equation of the plane passing through the line of intersection of the…
- If O be the origin and the coordinates of P be (1, 2, -3), then find the…
- Find the equation of the plane which contains the line of intersection of the…
- Find the distance of the point (-1, -5, -10) from the point of intersection of…
- Find the vector equation of the line passing through (1, 2, 3) and parallel to…
- Find the vector equation of the line passing through the point (1, 2, - 4) and…
- Prove that if a plane has the intercepts a, b, c and is at a distance of p…
- Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 isA. 2…
- The planes: 2x - y + 4z = 5 and 5x - 2.5y + 10z = 6 areA. Perpendicular B.…

###### Exercise 11.1

**Question 1.**If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its direction cosines.

**Answer:**Let the direction cosines of the line making ∠ α with x-axis, β – with y axis and γ- with z axis are l, m and n

⇒ l = cos α, m = cos β and n = cos γ

Here α = 90°, β = 135° and γ = 45°

So direction cosines are

l = cos 90° = 0

m = cos 135°= cos (180° - 45°) = -cos 45° =

n = cos 45° =

⇒ Direction cosines of the line

**Question 2.**Find the direction cosines of a line which makes equal angles with the coordinate axes.

**Answer:**Let the direction cosines of the line making ∠ α with x-axis, β – with y axis and γ- with z axis are l, m and n

⇒ l = cos α, m = cos β and n = cos γ

Here given α = β = γ (line makes equal angles with the coordinate axes) ……….1

Direction Cosines are

⇒ l = cos α, m = cos β and n = cos γ

We have

l^{2} + m ^{2} + n^{2} = 1

cos^{2} α + cos^{2}β + cos^{2}γ = 1

From 1 we have

cos^{2} α + cos^{2} α + cos^{2} α = 1

3 cos^{2} α = 1

The direction cosines are

**Question 3.**If a line has the direction ratios –18, 12, –4, then what are its direction cosines?

**Answer:**If the direction ratios of the line are a, b and c

Then the direction cosines are

Given direction ratios are – 18, 12 and – 4

⇒ a= -18, b = 12 and c= -4

=

=

Direction cosines are

**Question 4.**Show that the points (2, 3, 4), (–1, –2, 1), (5, 8, 7) are collinear.

**Answer:**If the direction ratios of two lines segments are proportional, then the lines are collinear.

Given A(2, 3, 4), B(−1, −2, 1), C(5, 8, 7)

Direction ratio of line joining A (2,3,4) and B (−1, −2, 1), are

(−1−2), (−2−3), (1−4)

= (−3, −5, −3)

So a_{1} = -3, b_{1} = -5, c_{1} = -3

Direction ratio of line joining B (−1, −2, 1) and C (5, 8, 7) are

(5− (−1)), (8−(−2)), (7−1)

= (6, 10, 6)

So a_{2} = 6, b_{2} = 10 and c_{2} =6

It is clear that the direction ratios of AB and BC are of same proportions

As

and

Therefore A, B, C are collinear.

**Question 5.**Find the direction cosines of the sides of the triangle whose vertices are (3, 5, –4), (-1, 1, 2) and (–5, –5, –2).

**Answer:**

The direction cosines of the two points passing through A(x_{1}, y_{1}, z_{1}) and B(x_{2}, y_{2}, z_{2}) is given by

(x_{2} – x_{1}), (y_{2}-y_{1}), (z_{2}-z_{1})

And the direction cosines of the line AB is

Where AB =

**Question 1.**

If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its direction cosines.

**Answer:**

Let the direction cosines of the line making ∠ α with x-axis, β – with y axis and γ- with z axis are l, m and n

⇒ l = cos α, m = cos β and n = cos γ

Here α = 90°, β = 135° and γ = 45°

So direction cosines are

l = cos 90° = 0

m = cos 135°= cos (180° - 45°) = -cos 45° =

n = cos 45° =

⇒ Direction cosines of the line

**Question 2.**

Find the direction cosines of a line which makes equal angles with the coordinate axes.

**Answer:**

Let the direction cosines of the line making ∠ α with x-axis, β – with y axis and γ- with z axis are l, m and n

⇒ l = cos α, m = cos β and n = cos γ

Here given α = β = γ (line makes equal angles with the coordinate axes) ……….1

Direction Cosines are

⇒ l = cos α, m = cos β and n = cos γ

We have

l^{2} + m ^{2} + n^{2} = 1

cos^{2} α + cos^{2}β + cos^{2}γ = 1

From 1 we have

cos^{2} α + cos^{2} α + cos^{2} α = 1

3 cos^{2} α = 1

The direction cosines are

**Question 3.**

If a line has the direction ratios –18, 12, –4, then what are its direction cosines?

**Answer:**

If the direction ratios of the line are a, b and c

Then the direction cosines are

Given direction ratios are – 18, 12 and – 4

⇒ a= -18, b = 12 and c= -4

=

=

Direction cosines are

**Question 4.**

Show that the points (2, 3, 4), (–1, –2, 1), (5, 8, 7) are collinear.

**Answer:**

If the direction ratios of two lines segments are proportional, then the lines are collinear.

Given A(2, 3, 4), B(−1, −2, 1), C(5, 8, 7)

Direction ratio of line joining A (2,3,4) and B (−1, −2, 1), are

(−1−2), (−2−3), (1−4)

= (−3, −5, −3)

So a_{1} = -3, b_{1} = -5, c_{1} = -3

Direction ratio of line joining B (−1, −2, 1) and C (5, 8, 7) are

(5− (−1)), (8−(−2)), (7−1)

= (6, 10, 6)

So a_{2} = 6, b_{2} = 10 and c_{2} =6

It is clear that the direction ratios of AB and BC are of same proportions

As

and

Therefore A, B, C are collinear.

**Question 5.**

Find the direction cosines of the sides of the triangle whose vertices are (3, 5, –4), (-1, 1, 2) and (–5, –5, –2).

**Answer:**

The direction cosines of the two points passing through A(x_{1}, y_{1}, z_{1}) and B(x_{2}, y_{2}, z_{2}) is given by

(x_{2} – x_{1}), (y_{2}-y_{1}), (z_{2}-z_{1})

And the direction cosines of the line AB is

Where AB =

###### Exercise 11.2

**Question 1.**Show that the three lines with direction cosines are mutually perpendicular.

**Answer:**We know that

If l_{1}, m_{1}, n_{1} and l_{2}, m_{2}, n_{2} are the direction cosines of two lines; and θ is the acute angle between the two lines; then cos θ = |l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2}|

If two lines are perpendicular, then the angle between the two is θ = 90°

⇒ For perpendicular lines, | l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} | = cos 90° = 0, i.e.

| l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} | = 0

So, in order to check if the three lines are mutually perpendicular, we compute | l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} | for all the pairs of the three lines.

Now let the direction cosines of L_{1}, L_{2} and L_{3} be l_{1}, m_{1}, n_{1}; l_{2}, m_{2}, n_{2} and l_{3}, m_{3}, n_{3}.

First, consider

⇒

⇒ L_{1}⊥ L_{2} ……(i)

Next, consider

⇒

⇒

⇒ L_{2}⊥ L_{3} …(ii)

Now, consider

⇒

⇒

⇒ L_{1}⊥ L_{3} …(iii)

∴ By (i), (ii) and (iii), we have

L_{1}, L_{2} and L_{3} are mutually perpendicular.

**Question 2.**Show that the line through the points (1, –1, 2), (3, 4, –2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

**Answer:**We know that

Two lines with direction ratios a_{1}, b_{1}, c_{1} and a_{2}, b_{2}, c_{2} are perpendicular if the angle between them is θ = 90°, i.e. a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

Also, we know that the direction ratios of the line segment joining (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) is taken as x_{2} – x_{1}, y_{2} – y_{1}, z_{2} – z_{1} (or x_{1} – x_{2}, y_{1} – y_{2}, z_{1} – z_{2}).

⇒ The direction ratios of the line through the points (1, –1, 2) and (3, 4, –2) is:

a_{1} = 3 – 1 = 2, b_{1} = 4 – (-1) = 4 + 1 = 5, c_{1} = -2 –2 = -4

and the direction ratios of the line through the points (0, 3, 2) and (3, 5, 6) is:

a_{2} = 3 – 0 = 3, b_{2} = 5 – 3 = 2, c_{2} = 6 – 2 = 4

Now, consider

a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 2 × 3 + 5 × 2 + (-4) × 4 = 6 + 10 + (-16) = 16 + (-16) = 0

⇒ The line through the points (1, –1, 2), (3, 4, –2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

**Question 3.**Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (–1, –2, 1), (1, 2, 5).

**Answer:**We know that

Two lines with direction ratios a_{1}, b_{1}, c_{1} and a_{2}, b_{2}, c_{2} are parallel if the angle between them is θ = 0°, i. e.

⇒

Also, we know that the direction ratios of the line segment joining (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) is taken as x_{2} – x_{1}, y_{2} – y_{1}, z_{2} – z_{1} (or x_{1} – x_{2}, y_{1} – y_{2}, z_{1} – z_{2}).

⇒ The direction ratios of the line through the points (4, 7, 8) and (2, 3, 4) is:

a_{1} = 2 – 4 = -2, b_{1} = 3 – 7 = -4, c_{1} = 4 – 8 = -4

And the direction ratios of the line through the points (– 1, – 2, 1) and (1, 2, 5) is:

a_{2} = 1 – (-1) = 1 + 1 = 2, b_{2} = 2 – (-2) = 2 + 2 = 4, c_{2} = 5 – 1 = 4

Consider

⇒

∴ The line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (–1, –2, 1), (1, 2, 5).

**Question 4.**Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector

**Answer:**We know that

Vector equation of a line that passes through a given point whose position vector is and parallel to a given vector is .

So, here the position vector of the point (1, 2, 3) is given by and the parallel vector is .

∴ The vector equation of the required line is:

, where is a real number.

**Question 5.**Find the equation of the line in vector and in cartesian form that passes through the point with position vector and is in the direction

**Answer:**We know that

Vector equation of a line that passes through a given point whose position vector is and parallel to a given vector is .

Here, and

⇒ The vector equation of the required line is:

⇒

Also, we know that

The Cartesian equation of a line through a point (x_{1}, y_{1}, z_{1}) and having direction cosines l, m, n is .

Also, we know that if the direction ratios of the line are a, b, c, then

⇒

⇒ The Cartesian equation of a line through a point (x_{1}, y_{1}, z_{1}) and having direction ratios a, b, c is .

Here, x_{1} = 2, y_{1} = -1, z_{1} = 4 and a = 1, b = 2, c = -1

⇒ The Cartesian equation of the required line is:

⇒

**Question 6.**Find the cartesian equation of the line which passes through the point (–2, 4, –5) and parallel to the line given by

**Answer:**We know that

The Cartesian equation of a line through a point (x_{1}, y_{1}, z_{1}) and having direction ratios a, b, c is .

Here, The point (x_{1}, y_{1}, z_{1}) is (-2, 4, -5) and the direction ratios are:

a = 3, b = 5, c = 6

⇒ The Cartesian equation of the required line is:

⇒

**Question 7.**The Cartesian equation of a line is . Write its vector form.

**Answer:**We know that

The Cartesian equation of a line through a point (x_{1}, y_{1}, z_{1}) and having direction cosines l, m, n is .

Comparing this standard form with the given equation, we get

x_{1} = 5, y_{1} = -4, z_{1} = 6 and l = 3, m = 7, n = 2

⇒ The point through which the line passes has the position vector and the vector parallel to the line is given by .

Now, ∵ Vector equation of a line that passes through a given point whose position vector is and parallel to a given vector is .

∴ The vector equation of the required line is:

⇒

**Question 8.**Find the vector and the cartesian equations of the lines that passes through the origin and (5, –2, 3).

**Answer:**We know that

The vector equation of as line which passes through two points whose position vectors are and is .

Here, the position vectors of the two points (0, 0, 0) and (5, -2, 3) are and , respectively.

So, The vector equation of the required line is:

⇒

⇒

Now, we know that

Cartesian equation of a line that passes through two points (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) is

So, the Cartesian equation of the line that passes through the origin (0, 0, 0) and (5, – 2, 3) is

**Question 9.**. Find the vector and the cartesian equations of the line that passes through the points (3, –2, –5), (3, –2, 6).

**Answer:**We know that

The vector equation of as line which passes through two points whose position vectors are and is .

Here, the position vectors of the two points (3, –2, –5) and (3, –2, 6) are and , respectively.

So, the vector equation of the required line is:

⇒

⇒

⇒

Now, we also know that

Cartesian equation of a line that passes through two points (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) is

So, the Cartesian equation of the line that passes through the origin (3, -2, -5) and (3, -2, 6) is

**Question 10.**Find the angle between the following pairs of lines:

**Answer:**We know that

If θ is the acute angle between and , then

……(i)

(i) and

Here and

So, from (i), we have

……(ii)

⇒

⇒

And

Now,

⇒

⇒ By (ii), we have

⇒

⇒

(ii) and

Here, and

So, from (i), we have

…(iii)

⇒

⇒

And

Now,

⇒

⇒ By (iii), we have

⇒

⇒

**Question 11.**. Find the angle between the following pair of lines:

**Answer:**We know that

If and are the equations of two lines, then the acute angle between the two lines is given by

cos θ = | l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} | ……(i)

(i) and

Here, a_{1} = 2, b_{1} = 5, c_{1} = -3 and a_{2} = -1, b_{2} = 8, c_{2} = 4

Now, ……(ii)

Here,

And

So, from (ii), we have

⇒

And

∴ From (i), we have

⇒

⇒

(ii) and

Here, a_{1} = 2, b_{1} = 2, c_{1} = 1 and a_{2} = 4, b_{2} = 1, c_{2} = 8

Here,

And

So, from (ii), we have

⇒

And

∴ From (i), we have

⇒

⇒

**Question 12.**Find the values of p so that the lines are at right angles.

**Answer:**For any two lines to be at right angles, the angle between them should be θ = 90°.

⇒ a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0, where a_{1}, b_{1}, c_{1} and a_{2}, b_{2}, c_{2} are the direction ratios of two lines.

The standard form of a pair of Cartesian lines is:

and …(i)

Now, first we rewrite the given equations according to the standard form, i.e.

and , i.e.

and …(ii)

Now, comparing (i) and (ii), we get

and

Now, as both the lines are at right angles,

so a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

⇒

⇒

⇒

⇒

⇒ 11p = 70

⇒

**Question 13.**Show that the lines are perpendicular to each other.

**Answer:**We know that

Two lines with direction ratios a_{1}, b_{1}, c_{1} and a_{2}, b_{2}, c_{2} are perpendicular if the angle between them is θ = 90°, i.e. a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

Also, here the direction ratios are:

a_{1} = 7, b_{1} = -5, c_{1} = 1 and a_{2} = 1, b_{2} = 2, c_{2} = 3

Now, Consider

a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 7 × 1 + (-5) × 2 + 1 × 3 = 7 -10 + 3 = - 3 + 3 = 0

∴ The two lines are perpendicular to each other.

**Question 14.**Find the shortest distance between the lines

**Answer:**We know that

Shortest distance between two lines and is

…(i)

Here, , and

,

Now,

…(ii)

Now,

⇒

……….(iii)

……….(iv)

Now,

……….(v)

Now, using (i), we have

The shortest distance between the two lines, d [From (iv) and (v)]

⇒

Rationalizing the fraction by multiplying the numerator and denominator by √2,

⇒

**Question 15.**Find the shortest distance between the lines

**Answer:**We know that

_{Shortest distance between the lines:}

and is

…(i)

The standard form of a pair of Cartesian lines is:

and

And the given equations are: and

_{Comparing the given equations with the standard form, we get}

_{x}_{1} = -1, y_{1} = -1, z_{1} = -1; x_{2} = 3, y_{2} = 5, z_{2} = 7

_{a}_{1} = 7, b_{1} = -6, c_{1} = 1; a_{2} = 1, b_{2} = -2, c_{2} = 1

_{Now, consider}

_{⇒}

⇒

⇒

⇒

⇒

⇒

Next, consider

⇒

⇒

⇒

⇒

_{⇒}_{From (i), we have}

⇒

**Question 16.**Find the shortest distance between the lines whose vector equations are

**Answer:**We know that

Shortest distance between two lines and is

……….(i)

Here, , and

,

Now,

……….(ii)

Now,

⇒

……….(iii)

……….(iv)

Now,

……….(v)

Now, using (i), we have

The shortest distance between the two lines,

**Question 17.**Find the shortest distance between the lines whose vector equations are

**Answer:**Firstly, consider

⇒

⇒

⇒

⇒

⇒

⇒

So, we need to find the shortest distance between and .

Now, We know that

Shortest distance between two lines and is

…(i)

Here, and

⇒

Now,

…(ii)

Now,

⇒

……….(iii)

……….(iv)

Now,

……….(v)

Now, using (i), we have

The shortest distance between the two lines,

**Question 1.**

Show that the three lines with direction cosines are mutually perpendicular.

**Answer:**

We know that

If l_{1}, m_{1}, n_{1} and l_{2}, m_{2}, n_{2} are the direction cosines of two lines; and θ is the acute angle between the two lines; then cos θ = |l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2}|

If two lines are perpendicular, then the angle between the two is θ = 90°

⇒ For perpendicular lines, | l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} | = cos 90° = 0, i.e.

| l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} | = 0

So, in order to check if the three lines are mutually perpendicular, we compute | l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} | for all the pairs of the three lines.

Now let the direction cosines of L_{1}, L_{2} and L_{3} be l_{1}, m_{1}, n_{1}; l_{2}, m_{2}, n_{2} and l_{3}, m_{3}, n_{3}.

First, consider

⇒

⇒ L_{1}⊥ L_{2} ……(i)

Next, consider

⇒

⇒

⇒ L_{2}⊥ L_{3} …(ii)

Now, consider

⇒

⇒

⇒ L_{1}⊥ L_{3} …(iii)

∴ By (i), (ii) and (iii), we have

L_{1}, L_{2} and L_{3} are mutually perpendicular.

**Question 2.**

Show that the line through the points (1, –1, 2), (3, 4, –2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

**Answer:**

We know that

Two lines with direction ratios a_{1}, b_{1}, c_{1} and a_{2}, b_{2}, c_{2} are perpendicular if the angle between them is θ = 90°, i.e. a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

Also, we know that the direction ratios of the line segment joining (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) is taken as x_{2} – x_{1}, y_{2} – y_{1}, z_{2} – z_{1} (or x_{1} – x_{2}, y_{1} – y_{2}, z_{1} – z_{2}).

⇒ The direction ratios of the line through the points (1, –1, 2) and (3, 4, –2) is:

a_{1} = 3 – 1 = 2, b_{1} = 4 – (-1) = 4 + 1 = 5, c_{1} = -2 –2 = -4

and the direction ratios of the line through the points (0, 3, 2) and (3, 5, 6) is:

a_{2} = 3 – 0 = 3, b_{2} = 5 – 3 = 2, c_{2} = 6 – 2 = 4

Now, consider

a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 2 × 3 + 5 × 2 + (-4) × 4 = 6 + 10 + (-16) = 16 + (-16) = 0

⇒ The line through the points (1, –1, 2), (3, 4, –2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

**Question 3.**

Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (–1, –2, 1), (1, 2, 5).

**Answer:**

We know that

Two lines with direction ratios a_{1}, b_{1}, c_{1} and a_{2}, b_{2}, c_{2} are parallel if the angle between them is θ = 0°, i. e.

⇒

Also, we know that the direction ratios of the line segment joining (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) is taken as x_{2} – x_{1}, y_{2} – y_{1}, z_{2} – z_{1} (or x_{1} – x_{2}, y_{1} – y_{2}, z_{1} – z_{2}).

⇒ The direction ratios of the line through the points (4, 7, 8) and (2, 3, 4) is:

a_{1} = 2 – 4 = -2, b_{1} = 3 – 7 = -4, c_{1} = 4 – 8 = -4

And the direction ratios of the line through the points (– 1, – 2, 1) and (1, 2, 5) is:

a_{2} = 1 – (-1) = 1 + 1 = 2, b_{2} = 2 – (-2) = 2 + 2 = 4, c_{2} = 5 – 1 = 4

Consider

⇒

∴ The line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (–1, –2, 1), (1, 2, 5).

**Question 4.**

Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector

**Answer:**

We know that

Vector equation of a line that passes through a given point whose position vector is and parallel to a given vector is .

So, here the position vector of the point (1, 2, 3) is given by and the parallel vector is .

∴ The vector equation of the required line is:

, where is a real number.

**Question 5.**

Find the equation of the line in vector and in cartesian form that passes through the point with position vector and is in the direction

**Answer:**

We know that

Vector equation of a line that passes through a given point whose position vector is and parallel to a given vector is .

Here, and

⇒ The vector equation of the required line is:

⇒

Also, we know that

The Cartesian equation of a line through a point (x_{1}, y_{1}, z_{1}) and having direction cosines l, m, n is .

Also, we know that if the direction ratios of the line are a, b, c, then

⇒

⇒ The Cartesian equation of a line through a point (x_{1}, y_{1}, z_{1}) and having direction ratios a, b, c is .

Here, x_{1} = 2, y_{1} = -1, z_{1} = 4 and a = 1, b = 2, c = -1

⇒ The Cartesian equation of the required line is:

⇒

**Question 6.**

Find the cartesian equation of the line which passes through the point (–2, 4, –5) and parallel to the line given by

**Answer:**

We know that

The Cartesian equation of a line through a point (x_{1}, y_{1}, z_{1}) and having direction ratios a, b, c is .

Here, The point (x_{1}, y_{1}, z_{1}) is (-2, 4, -5) and the direction ratios are:

a = 3, b = 5, c = 6

⇒ The Cartesian equation of the required line is:

⇒

**Question 7.**

The Cartesian equation of a line is . Write its vector form.

**Answer:**

We know that

The Cartesian equation of a line through a point (x_{1}, y_{1}, z_{1}) and having direction cosines l, m, n is .

Comparing this standard form with the given equation, we get

x_{1} = 5, y_{1} = -4, z_{1} = 6 and l = 3, m = 7, n = 2

⇒ The point through which the line passes has the position vector and the vector parallel to the line is given by .

Now, ∵ Vector equation of a line that passes through a given point whose position vector is and parallel to a given vector is .

∴ The vector equation of the required line is:

⇒

**Question 8.**

Find the vector and the cartesian equations of the lines that passes through the origin and (5, –2, 3).

**Answer:**

We know that

The vector equation of as line which passes through two points whose position vectors are and is .

Here, the position vectors of the two points (0, 0, 0) and (5, -2, 3) are and , respectively.

So, The vector equation of the required line is:

⇒

⇒

Now, we know that

Cartesian equation of a line that passes through two points (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) is

So, the Cartesian equation of the line that passes through the origin (0, 0, 0) and (5, – 2, 3) is

**Question 9.**

. Find the vector and the cartesian equations of the line that passes through the points (3, –2, –5), (3, –2, 6).

**Answer:**

We know that

The vector equation of as line which passes through two points whose position vectors are and is .

Here, the position vectors of the two points (3, –2, –5) and (3, –2, 6) are and , respectively.

So, the vector equation of the required line is:

⇒

⇒

⇒

Now, we also know that

Cartesian equation of a line that passes through two points (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) is

So, the Cartesian equation of the line that passes through the origin (3, -2, -5) and (3, -2, 6) is

**Question 10.**

Find the angle between the following pairs of lines:

**Answer:**

We know that

If θ is the acute angle between and , then

……(i)

(i) and

Here and

So, from (i), we have

……(ii)

⇒

⇒

And

Now,

⇒

⇒ By (ii), we have

⇒

⇒

(ii) and

Here, and

So, from (i), we have

…(iii)

⇒

⇒

And

Now,

⇒

⇒ By (iii), we have

⇒

⇒

**Question 11.**

. Find the angle between the following pair of lines:

**Answer:**

We know that

If and are the equations of two lines, then the acute angle between the two lines is given by

cos θ = | l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} | ……(i)

(i) and

Here, a_{1} = 2, b_{1} = 5, c_{1} = -3 and a_{2} = -1, b_{2} = 8, c_{2} = 4

Now, ……(ii)

Here,

And

So, from (ii), we have

⇒

And

∴ From (i), we have

⇒

⇒

(ii) and

Here, a_{1} = 2, b_{1} = 2, c_{1} = 1 and a_{2} = 4, b_{2} = 1, c_{2} = 8

Here,

And

So, from (ii), we have

⇒

And

∴ From (i), we have

⇒

⇒

**Question 12.**

Find the values of p so that the lines are at right angles.

**Answer:**

For any two lines to be at right angles, the angle between them should be θ = 90°.

⇒ a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0, where a_{1}, b_{1}, c_{1} and a_{2}, b_{2}, c_{2} are the direction ratios of two lines.

The standard form of a pair of Cartesian lines is:

and …(i)

Now, first we rewrite the given equations according to the standard form, i.e.

and , i.e.

and …(ii)

Now, comparing (i) and (ii), we get

and

Now, as both the lines are at right angles,

so a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

⇒

⇒

⇒

⇒

⇒ 11p = 70

⇒

**Question 13.**

Show that the lines are perpendicular to each other.

**Answer:**

We know that

Two lines with direction ratios a_{1}, b_{1}, c_{1} and a_{2}, b_{2}, c_{2} are perpendicular if the angle between them is θ = 90°, i.e. a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

Also, here the direction ratios are:

a_{1} = 7, b_{1} = -5, c_{1} = 1 and a_{2} = 1, b_{2} = 2, c_{2} = 3

Now, Consider

a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 7 × 1 + (-5) × 2 + 1 × 3 = 7 -10 + 3 = - 3 + 3 = 0

∴ The two lines are perpendicular to each other.

**Question 14.**

Find the shortest distance between the lines

**Answer:**

We know that

Shortest distance between two lines and is

…(i)

Here, , and

,

Now,

…(ii)

Now,

⇒

……….(iii)

……….(iv)

Now,

……….(v)

Now, using (i), we have

The shortest distance between the two lines, d [From (iv) and (v)]

⇒

Rationalizing the fraction by multiplying the numerator and denominator by √2,

⇒

**Question 15.**

Find the shortest distance between the lines

**Answer:**

We know that

_{Shortest distance between the lines:}

and is

…(i)

The standard form of a pair of Cartesian lines is:

and

And the given equations are: and

_{Comparing the given equations with the standard form, we get}

_{x}_{1} = -1, y_{1} = -1, z_{1} = -1; x_{2} = 3, y_{2} = 5, z_{2} = 7

_{a}_{1} = 7, b_{1} = -6, c_{1} = 1; a_{2} = 1, b_{2} = -2, c_{2} = 1

_{Now, consider}

_{⇒}

⇒

⇒

⇒

⇒

⇒

Next, consider

⇒

⇒

⇒

⇒

_{⇒}_{From (i), we have}

⇒

**Question 16.**

Find the shortest distance between the lines whose vector equations are

**Answer:**

We know that

Shortest distance between two lines and is

……….(i)

Here, , and

,

Now,

……….(ii)

Now,

⇒

……….(iii)

……….(iv)

Now,

……….(v)

Now, using (i), we have

The shortest distance between the two lines,

**Question 17.**

Find the shortest distance between the lines whose vector equations are

**Answer:**

Firstly, consider

⇒

⇒

⇒

⇒

⇒

⇒

So, we need to find the shortest distance between and .

Now, We know that

Shortest distance between two lines and is

…(i)

Here, and

⇒

Now,

…(ii)

Now,

⇒

……….(iii)

……….(iv)

Now,

……….(v)

Now, using (i), we have

The shortest distance between the two lines,

###### Exercise 11.3

**Question 1.**In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

z = 2

**Answer:**The eq. of the plane

z = 2

Direction ratio of the normal (0,0,1)

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

Direction cosines = 0,0,1

Distance(d) = 2

**Question 2.**In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

x + y + z = 1

**Answer:**The eq. of the plane

x + y + z = 1

Direction ratio of the normal (1,1,1)

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

**Question 3.**In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

2x + 3y – z = 5

**Answer:**The eq. of the plane

2x + 3y-z = 5

Direction ratio of the normal (2, 3, -1)

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

**Question 4.**In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

5y + 8 = 0

**Answer:**The eq. of the plane

0x-5y + 0z = 8

Direction ratio of the normal (0, -5, 0)

= 5

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

Direction cosine = 0, -1, 0

**Question 5.**Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector

**Answer:**Vector eq. of the plane with position vector is

**Question 6.**Find the Cartesian equation of the following planes:

**Answer:**(Letbe the position vector of P(x,y,z)

Hence,

So, Cartesian eq. is

x + y - z = 2

**Question 7.**Find the Cartesian equation of the following planes:

**Answer:**(Letbe the position vector of P(x,y,z)

.

.

So, Cartesian eq. is

2x + 3y - 4z = 1

**Question 8.**Find the Cartesian equation of the following planes:

**Answer:**Letbe the position vector of P(x,y,z)

Hence,

So, Cartesian eq. is

(s-2t)x + (3-t)y + (2s + t)z=15

**Question 9.**In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

2x + 3y + 4z – 12 = 0

**Answer:**Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x,y,z).

2x + 3y + 4z = 12

Direction ratio (2,3,4)

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

**Question 10.**In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

2x + 3y + 4z – 12 = 0

**Answer:**Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x,y,z).

0x + 3y + 4z = 6

Direction ratio (0,3,4)

.

= 5

his is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

Coordinate of the foot (ld,md,nd)

**Question 11.**In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

3y + 4z – 6 = 0

**Answer:**Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x,y,z).

x + y + z = 1

Direction ratio (1,1,1)

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

**Question 12.**In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

x + y + z = 1

**Answer:**Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x,y,z).

0x – 5y + 0z = 8

Direction ratio (0,-5,0)

= 5

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

Coordinate of the foot (ld,md,nd)

**Question 13.**Find the vector and cartesian equations of the planes

that passes through the point (1, 0, –2) and the normal to the plane is

**Answer:**Let the position vector of the point

Normal⊥to the plane

Vector eq. of the plane,

x – 1 + y – z – 2 = 0

x + y – z – 3 = 0

Required Cartesian eq. of the plane

x + y – z = 3

**Question 14.**Find the vector and cartesian equations of the planes

that passes through the point (1,4, 6) and the normal vector to the plane is

**Answer:**Let the position vector of the point

Vector eq. of the plane,

x – 1 – 2y + 8 + z – 6 = 0

x – 2y + z + 1 = 0

Required Cartesian eq. of the plane

x – 2y + z = - 1

**Question 15.**Find the equations of the planes that passes through three points.

(1, 1, –1), (6, 4, –5), (–4, –2, 3)

**Answer:**The given points are (1, 1, -1), (6, 4, -5), (-4, -2, 3).

Let,

= 1(12 - 10) – 1(18 - 20) -1 (-12 + 16)

= 2 + 2 – 4

= 0

Since, the value of determinant is 0.

Therefore, these points are collinear as there will be infinite planes passing through the given 3 points.

**Question 16.**Find the equations of the planes that passes through three points.

(1, 1, 0), (1, 2, 1), (–2, 2, –1)

**Answer:**The given points are (1, 1, 0), (1, 2, 1), (-2, 2, -1).

Let,

= 1(-2 - 2) – 1(-1 + 2)

= -4 – 1

= -5 ≠ 0

There passes a unique plane from the given 3 points.

Equation of the plane passes through the points, (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) and (x_{3}, y_{3}, z_{3}), i.e.,

⇒ (x - 1)(-2) – (y - 1)(3) + 3z = 0

⇒ -2x + 2 – 3y + 3 + 3z = 0

⇒ 2x + 3y – 3z = 5

This is the required eq. of the plane.

**Question 17.**Find the intercepts cut off by the plane 2x + y – z = 5.

**Answer:**We know that, the eq. of the plane in intercept form

where a, b, c are the intercepts cut-off by the plane at x, y and z axes respectively.

⇒ 2x + y – z = 5 (i)

Dividing both side of (i)eq. by 5, we get

Thus, the intercepts cut-off by the plane are 5/2, 5 and -5.

**Question 18.**Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.

**Answer:**We know that the eq. of the plane ZOX is

y = 0

Eq. of plane parallel to it is of the form, y = a

Hence, the required eq. of the plane is

y = 3

**Question 19.**Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1).

**Answer:**Eq. of the plane passes through the intersection of the plane is given by

(3x – y + 2z – 4) + λ(x + y + z – 2) = 0

∵ Plane passes through the points (2,2,1)

(3 × 2 – 2 + 2 × 1 – 4) + λ(2 + 2 + 1 – 2) = 0

2 + 3λ = 0

3λ = -2

(i)

Hence, the required eq. of the plane

7x – 5y + 4z – 8 = 0

This is the required eq. of the plane.

**Question 20.**Find the vector equation of the plane passing through the intersection of the planes and through the point (2, 1, 3).

**Answer:**Let the vector eq. of the plane passing through the intersection of the planes

Here,

**Question 1.**

In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

z = 2

**Answer:**

The eq. of the plane

z = 2

Direction ratio of the normal (0,0,1)

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

Direction cosines = 0,0,1

Distance(d) = 2

**Question 2.**

In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

x + y + z = 1

**Answer:**

The eq. of the plane

x + y + z = 1

Direction ratio of the normal (1,1,1)

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

**Question 3.**

In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

2x + 3y – z = 5

**Answer:**

The eq. of the plane

2x + 3y-z = 5

Direction ratio of the normal (2, 3, -1)

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

**Question 4.**

In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

5y + 8 = 0

**Answer:**

The eq. of the plane

0x-5y + 0z = 8

Direction ratio of the normal (0, -5, 0)

= 5

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

Direction cosine = 0, -1, 0

**Question 5.**

Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector

**Answer:**

Vector eq. of the plane with position vector is

**Question 6.**

Find the Cartesian equation of the following planes:

**Answer:**

(Letbe the position vector of P(x,y,z)

Hence,

So, Cartesian eq. is

x + y - z = 2

**Question 7.**

Find the Cartesian equation of the following planes:

**Answer:**

(Letbe the position vector of P(x,y,z)

.

.

So, Cartesian eq. is

2x + 3y - 4z = 1

**Question 8.**

Find the Cartesian equation of the following planes:

**Answer:**

Letbe the position vector of P(x,y,z)

Hence,

So, Cartesian eq. is

(s-2t)x + (3-t)y + (2s + t)z=15

**Question 9.**

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

2x + 3y + 4z – 12 = 0

**Answer:**

Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x,y,z).

2x + 3y + 4z = 12

Direction ratio (2,3,4)

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

**Question 10.**

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

2x + 3y + 4z – 12 = 0

**Answer:**

Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x,y,z).

0x + 3y + 4z = 6

Direction ratio (0,3,4)

.

= 5

his is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

Coordinate of the foot (ld,md,nd)

**Question 11.**

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

3y + 4z – 6 = 0

**Answer:**

Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x,y,z).

x + y + z = 1

Direction ratio (1,1,1)

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

**Question 12.**

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

x + y + z = 1

**Answer:**

Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x,y,z).

0x – 5y + 0z = 8

Direction ratio (0,-5,0)

= 5

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

Coordinate of the foot (ld,md,nd)

**Question 13.**

Find the vector and cartesian equations of the planes

that passes through the point (1, 0, –2) and the normal to the plane is

**Answer:**

Let the position vector of the point

Normal⊥to the plane

Vector eq. of the plane,

x – 1 + y – z – 2 = 0

x + y – z – 3 = 0

Required Cartesian eq. of the plane

x + y – z = 3

**Question 14.**

Find the vector and cartesian equations of the planes

that passes through the point (1,4, 6) and the normal vector to the plane is

**Answer:**

Let the position vector of the point

Vector eq. of the plane,

x – 1 – 2y + 8 + z – 6 = 0

x – 2y + z + 1 = 0

Required Cartesian eq. of the plane

x – 2y + z = - 1

**Question 15.**

Find the equations of the planes that passes through three points.

(1, 1, –1), (6, 4, –5), (–4, –2, 3)

**Answer:**

The given points are (1, 1, -1), (6, 4, -5), (-4, -2, 3).

Let,

= 1(12 - 10) – 1(18 - 20) -1 (-12 + 16)

= 2 + 2 – 4

= 0

Since, the value of determinant is 0.

Therefore, these points are collinear as there will be infinite planes passing through the given 3 points.

**Question 16.**

Find the equations of the planes that passes through three points.

(1, 1, 0), (1, 2, 1), (–2, 2, –1)

**Answer:**

The given points are (1, 1, 0), (1, 2, 1), (-2, 2, -1).

Let,

= 1(-2 - 2) – 1(-1 + 2)

= -4 – 1

= -5 ≠ 0

There passes a unique plane from the given 3 points.

Equation of the plane passes through the points, (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) and (x_{3}, y_{3}, z_{3}), i.e.,

⇒ (x - 1)(-2) – (y - 1)(3) + 3z = 0

⇒ -2x + 2 – 3y + 3 + 3z = 0

⇒ 2x + 3y – 3z = 5

This is the required eq. of the plane.

**Question 17.**

Find the intercepts cut off by the plane 2x + y – z = 5.

**Answer:**

We know that, the eq. of the plane in intercept form

where a, b, c are the intercepts cut-off by the plane at x, y and z axes respectively.

⇒ 2x + y – z = 5 (i)

Dividing both side of (i)eq. by 5, we get

Thus, the intercepts cut-off by the plane are 5/2, 5 and -5.

**Question 18.**

Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.

**Answer:**

We know that the eq. of the plane ZOX is

y = 0

Eq. of plane parallel to it is of the form, y = a

Hence, the required eq. of the plane is

y = 3

**Question 19.**

Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1).

**Answer:**

Eq. of the plane passes through the intersection of the plane is given by

(3x – y + 2z – 4) + λ(x + y + z – 2) = 0

∵ Plane passes through the points (2,2,1)

(3 × 2 – 2 + 2 × 1 – 4) + λ(2 + 2 + 1 – 2) = 0

2 + 3λ = 0

3λ = -2

(i)

Hence, the required eq. of the plane

7x – 5y + 4z – 8 = 0

This is the required eq. of the plane.

**Question 20.**

Find the vector equation of the plane passing through the intersection of the planes and through the point (2, 1, 3).

**Answer:**

Let the vector eq. of the plane passing through the intersection of the planes

Here,