Three Dimensional Geometry Class 12th Mathematics Part Ii CBSE Solution

Class 12th Mathematics Part Ii CBSE Solution
Exercise 11.1
  1. If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively,…
  2. Find the direction cosines of a line which makes equal angles with the…
  3. If a line has the direction ratios -18, 12, -4, then what are its direction…
  4. Show that the points (2, 3, 4), (-1, -2, 1), (5, 8, 7) are collinear.…
  5. Find the direction cosines of the sides of the triangle whose vertices are (3,…
Exercise 11.2
  1. Show that the three lines with direction cosines 12/13 , -3/13 , -4/13 4/13 ,…
  2. Show that the line through the points (1, -1, 2), (3, 4, -2) is perpendicular…
  3. Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the…
  4. Find the equation of the line which passes through the point (1, 2, 3) and is…
  5. Find the equation of the line in vector and in cartesian form that passes…
  6. Find the cartesian equation of the line which passes through the point (-2, 4,…
  7. The Cartesian equation of a line is x-5/3 = y+4/7 = z-6/2 . Write its vector…
  8. Find the vector and the cartesian equations of the lines that passes through…
  9. . Find the vector and the cartesian equations of the line that passes through…
  10. Find the angle between the following pairs of lines:
  11. . Find the angle between the following pair of lines: x-2/2 = y-1/5 - z+3/-3…
  12. Find the values of p so that the lines 1-x/3 = 7y-14/2p = z-3/2 7-7x/3p =…
  13. Show that the lines x-5/7 = y+2/-5 = z/1 x/1 = y/2 = z/3 are perpendicular to…
  14. vector r = (i+2 j + k) + lambda (i - j + k) vector r = 2 i - j - k + μ (2 i +…
  15. x+1/7 = y+1/-6 = z+1/1 x-3/1 = y-5/-2 = z-7/1 Find the shortest distance…
  16. Find the shortest distance between the lines whose vector equations are…
  17. vector r = (1-t) i + (t-2) j + (3-2t) k vector r = (s+1) i + (2s-1) j - (2s+1)…
Exercise 11.3
  1. z = 2 In each of the following cases, determine the direction cosines of the…
  2. x + y + z = 1 In each of the following cases, determine the direction cosines…
  3. 2x + 3y - z = 5 In each of the following cases, determine the direction…
  4. 5y + 8 = 0 In each of the following cases, determine the direction cosines of…
  5. Find the vector equation of a plane which is at a distance of 7 units from the…
  6. vector t (i + j - k) = 2 Find the Cartesian equation of the following planes:…
  7. Find the Cartesian equation of the following planes:
  8. Find the Cartesian equation of the following planes:
  9. 2x + 3y + 4z - 12 = 0 In the following cases, find the coordinates of the foot…
  10. 2x + 3y + 4z - 12 = 0 In the following cases, find the coordinates of the foot…
  11. 3y + 4z - 6 = 0 In the following cases, find the coordinates of the foot of…
  12. x + y + z = 1 In the following cases, find the coordinates of the foot of the…
  13. that passes through the point (1, 0, -2) and the normal to the plane is i + j…
  14. that passes through the point (1,4, 6) and the normal vector to the plane is…
  15. (1, 1, -1), (6, 4, -5), (-4, -2, 3) Find the equations of the planes that…
  16. (1, 1, 0), (1, 2, 1), (-2, 2, -1) Find the equations of the planes that passes…
  17. Find the intercepts cut off by the plane 2x + y - z = 5.
  18. Find the equation of the plane with intercept 3 on the y-axis and parallel to…
  19. Find the equation of the plane through the intersection of the planes 3x - y +…
  20. Find the vector equation of the plane passing through the intersection of the…
  21. Find the equation of the plane through the line of intersection of the planes…
  22. Find the angle between the planes whose vector equations are
  23. 7x + 5y + 6z + 30 = 0 and 3x - y - 10z + 4 = 0 In the following cases,…
  24. 2x + y + 3z - 2 = 0 and x - 2y + 5 = 0 In the following cases, determine…
  25. 2x - 2y + 4z + 5 = 0 and 3x - 3y + 6z - 1 = 0 In the following cases,…
  26. 2x - 2y + 4z + 5 = 0 and 3x - 3y + 6z - 1 = 0 In the following cases,…
  27. 4x + 8y + z - 8 = 0 and y + z - 4 = 0 In the following cases, determine…
  28. Point Plane (0, 0, 0) 3x - 4y + 12 z = 3 In the following cases, find the…
  29. Point Plane (3, - 2, 1) 2x - y + 2z + 3 = 0 In the following cases, find the…
  30. Point Plane (2, 3, - 5) x + 2y - 2z = 9 In the following cases, find the…
  31. Point Plane (-6, 0, 0) 2x - 3y + 6z - 2 = 0 In the following cases, find the…
Miscellaneous Exercise
  1. Show that the line joining the origin to the point (2, 1, 1) is perpendicular…
  2. If l1, m1, n1 and l2, m2, n2 are the direction cosines of two mutually…
  3. Find the angle between the lines whose direction ratios are a, b, c and b - c,…
  4. Find the equation of a line parallel to x - axis and passing through the…
  5. If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (-4, 3,…
  6. If the lines x-1/3k = y-2/1 = z-3/-5 and x-1/3k = y-2/1 = z-3/-5 are…
  7. Find the vector equation of the line passing through (1, 2, 3) and…
  8. Find the equation of the plane passing through (a, b, c) and parallel to the…
  9. Find the shortest distance between lines vector r = (6 i+2 j+2 k) + lambda (1…
  10. Find the coordinates of the point where the line through (5, 1, 6) and (3,…
  11. Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,…
  12. Find the coordinates of the point where the line through (3, -4, -5) and (2,…
  13. Find the equation of the plane passing through the point (-1, 3, 2) and…
  14. If the points (1, 1, p) and (-3, 0, 1) be equidistant from the plane vector r…
  15. Find the equation of the plane passing through the line of intersection of the…
  16. If O be the origin and the coordinates of P be (1, 2, -3), then find the…
  17. Find the equation of the plane which contains the line of intersection of the…
  18. Find the distance of the point (-1, -5, -10) from the point of intersection of…
  19. Find the vector equation of the line passing through (1, 2, 3) and parallel to…
  20. Find the vector equation of the line passing through the point (1, 2, - 4) and…
  21. Prove that if a plane has the intercepts a, b, c and is at a distance of p…
  22. Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 isA. 2…
  23. 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


l2 + m 2 + n2 = 1


cos2 α + cos2β + cos2γ = 1


From 1 we have


cos2 α + cos2 α + cos2 α = 1


3 cos2 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 a1 = -3, b1 = -5, c1 = -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 a2 = 6, b2 = 10 and c2 =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(x1, y1, z1) and B(x2, y2, z2) is given by


(x2 – x1), (y2-y1), (z2-z1)


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 l1, m1, n1 and l2, m2, n2 are the direction cosines of two lines; and θ is the acute angle between the two lines; then cos θ = |l1l2 + m1m2 + n1n2|


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


⇒ For perpendicular lines, | l1l2 + m1m2 + n1n2 | = cos 90° = 0, i.e.


| l1l2 + m1m2 + n1n2 | = 0


So, in order to check if the three lines are mutually perpendicular, we compute | l1l2 + m1m2 + n1n2 | for all the pairs of the three lines.


Now let the direction cosines of L1, L2 and L3 be l1, m1, n1; l2, m2, n2 and l3, m3, n3.


First, consider


⇒ 


⇒ 


⇒ L1⊥ L2 ……(i)


Next, consider


⇒ 


⇒ 


⇒ L2⊥ L3 …(ii)


Now, consider


⇒ 


⇒ 


⇒ L1⊥ L3 …(iii)


∴ By (i), (ii) and (iii), we have


L1, L2 and L3 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 a1, b1, c1 and a2, b2, c2 are perpendicular if the angle between them is θ = 90°, i.e. a1a2 + b1b2 + c1c2 = 0


Also, we know that the direction ratios of the line segment joining (x1, y1, z1) and (x2, y2, z2) is taken as x2 – x1, y2 – y1, z2 – z1 (or x1 – x2, y1 – y2, z1 – z2).


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


a1 = 3 – 1 = 2, b1 = 4 – (-1) = 4 + 1 = 5, c1 = -2 –2 = -4


and the direction ratios of the line through the points (0, 3, 2) and (3, 5, 6) is:
a2 = 3 – 0 = 3, b2 = 5 – 3 = 2, c2 = 6 – 2 = 4


Now, consider


a1a2 + b1b2 + c1c2 = 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 a1, b1, c1 and a2, b2, c2 are parallel if the angle between them is θ = 0°, i. e.


⇒ 


Also, we know that the direction ratios of the line segment joining (x1, y1, z1) and (x2, y2, z2) is taken as x2 – x1, y2 – y1, z2 – z1 (or x1 – x2, y1 – y2, z1 – z2).


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


a1 = 2 – 4 = -2, b1 = 3 – 7 = -4, c1 = 4 – 8 = -4


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


a2 = 1 – (-1) = 1 + 1 = 2, b2 = 2 – (-2) = 2 + 2 = 4, c2 = 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 (x1, y1, z1) 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 (x1, y1, z1) and having direction ratios a, b, c is .


Here, x1 = 2, y1 = -1, z1 = 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 (x1, y1, z1) and having direction ratios a, b, c is .


Here, The point (x1, y1, z1) 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 (x1, y1, z1) and having direction cosines l, m, n is .


Comparing this standard form with the given equation, we get


x1 = 5, y1 = -4, z1 = 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 (x1, y1, z1) and (x2, y2, z2) 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 (x1, y1, z1) and (x2, y2, z2) 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 = | l1l2 + m1m2 + n1n2 | ……(i)


(i)  and 


Here, a1 = 2, b1 = 5, c1 = -3 and a2 = -1, b2 = 8, c2 = 4


Now,  ……(ii)


Here, 


And 


So, from (ii), we have


⇒ 


And 


∴ From (i), we have


⇒ 


⇒ 


(ii)  and 


Here, a1 = 2, b1 = 2, c1 = 1 and a2 = 4, b2 = 1, c2 = 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°.

⇒ a1a2 + b1b2 + c1c2 = 0, where a1, b1, c1 and a2, b2, c2 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 a1a2 + b1b2 + c1c2 = 0


⇒ 


⇒ 


⇒ 


⇒ 


⇒ 11p = 70


⇒ 



Question 13.

Show that the lines  are perpendicular to each other.


Answer:

We know that

Two lines with direction ratios a1, b1, c1 and a2, b2, c2 are perpendicular if the angle between them is θ = 90°, i.e. a1a2 + b1b2 + c1c2 = 0


Also, here the direction ratios are:


a1 = 7, b1 = -5, c1 = 1 and a2 = 1, b2 = 2, c2 = 3


Now, Consider


a1a2 + b1b2 + c1c2 = 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


x1 = -1, y1 = -1, z1 = -1; x2 = 3, y2 = 5, z2 = 7


a1 = 7, b1 = -6, c1 = 1; a2 = 1, b2 = -2, c2 = 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, (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3), 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,