RBSE Solutions for Class 12 Maths Chapter 14 Three Dimensional Geometry Ex 14.2

Rajasthan Board RBSE Class 12 Maths Chapter 14 Three Dimensional Geometry Ex 14.2

Question 1.
Find the equation of the line which passes through the points (5, 7,9) and is parallel to the following axis :
(i) A-axis
(ii) F-axis
(iii) Z-axis
Solution:
Let position vector of point (5, 7, 9)

(i) The line parallel to X-axis passes through point (1,0,0) whose position vector

∴ Vector equation of required line.

Cartesian equation of line :

(ii) Let the line parallel to F-axis passes through point (0, 1,0).
∴ Position vector of point (0, 1, 0).

∴ required vector equation of line

(iii) The line parallel to Z-axis passes through (0, 0, 1).

∴ Required vector equation of line

Question 2.
Find the equation of the line in vector and in cartesian form that passes through the point with vector

and

is parallel to the vector
Solution:
Position vector of the given point is

Question 3.
Find the equation of the line parallel to the vector

and passes through the point (5,-2,4).
Solution :
Line passes through (5, -2, 4).
∴ Position vector of (5, – 2, 4)

Question 4.
Find the equation of the line passes through the point (2,-1,1) and is parallel to the line $$\frac { x-3 }{ 2 }$$ = $$\frac { y+1 }{ 7 }$$ = $$\frac { z-2 }{ -3 }$$
Solution:
The equation of line passing through point (2,-1, 1) and parallel to the given line $$\frac { x-3 }{ 2 }$$ = $$\frac { y+1 }{ 7 }$$ = $$\frac { z-2 }{ -3 }$$ because dc’s of parallel lines are same.

For vector equation of line passing through (2, – 1, 1) and parallel to $$\overrightarrow { m }$$,
The position vector of (2, -1, 1).

Question 5.
If cartesian equation of line is $$\frac { x-5 }{ 3 }$$ = $$\frac { y+4 }{ 7 }$$ = $$\frac { z-6 }{ 2 }$$, then find vector equation of line.
Solution:

Question 6.
Find the equation of a line in cartesian form which passes through (1, 2, 3) and is parallel to $$\frac { -x-2 }{ 1 }$$ = $$\frac { y+3 }{ 7 }$$ = $$\frac { 2z-6 }{ 3 }$$
Solution:
Let a, b, c are dc’s of line passing through (x1,y1,z1) then equation of the line

Here line passes through point (1, 2, 3) and parallel to line

Question 7.
Coordinates of three vertices of parallelogram ABCD are A(4, 5, 10), B(2, 3, 4) and C(1, 2, – 1). Find vector and cartesian equation of AB and AC. Also Find coordinates of D.
Solution:
Let O is origin.
∴ Position vector of points A, B and C are respectively.

(iii) For coordinates of point D.
Let (x1, y1, z1,) are the coordinates of point D.
∵ ABCD is a parallelogram whose diagonals AC and BD bisect each other at point D. Therefore mid points of AC and BD will coincide. So the coordinate of mid point of AC.

Question 8.
Cartesian equation of a line is 3x + 1 = 6y – 2 = 1 -z. Find the point from where it passes and also find its direction ratios and vector equation.
Solution:
The equation of given line

Question 9.
Find an equation passes through (1,2,3) and is parallel to the vector

Solution:

Question 10.
Find the equation of line in vector and cartesian form that passes through the point with position vector

and is in the direction

Solution:

Question 11.
Find the cartesian equation of a line passes through the point (- 2, 4, -5) and is parallel to

Solution:
Let line passes through point (x1,y1,z1) whose dc’s are a,b,c then equation of line is

Question 12.
The cartesian equation of a line is

Write its vector form.
Solution:
The cartesian equation of line is

Question 13.
Find the equation of line in vector and cartesian form that passes through origin and (5, -2, 3).
Solution:
(i) Position vector of origin O(0,0,0) is $$\overrightarrow { a }$$ = $$\overrightarrow { 0 }$$ and position vector equation of (5, – 2, 3) is

(ii) Line passing through origin O(0, 0, 0) whose dc’s are 5,-2, 3.

Question 14.
Find the equation of line in vector and cartesian form that passes through the points (3, -2,-5) and (3,-2,6).
Solution:
Let line passes through point (3,-2,-5) and (3,-2,6).

(ii) Line passes through A(3, -2,-5) and 5(3, – 2, 6). Cartesian equation of line AB