##### Question 1:

If a line makes angles 90$\xb0$, 135$\xb0$, 45$\xb0$ with the x, y and z-axes respectively, find its direction-cosines.

##### Answer:

Direction–angles are 90$\xb0$, 135$\xb0$, 45$\xb0$.

$\therefore $ Direction–cosines are < cos 90$\xb0$, cos 135$\xb0$, cos 45$\xb0$ >

##### Question 2:

Find the direction-cosines of a line, which makes equal angles with the co-ordinate axes.

##### Answer:

Let each directional angle be ‘$\mathrm{\alpha}$’.

$\therefore $Direction–cosines are < cos $\mathrm{\alpha}$, cos $\mathrm{\alpha}$, cos $\mathrm{\alpha}$ >.

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

$\Rightarrow $cos^{2} $\mathrm{\alpha}$+ cos^{2} $\mathrm{\alpha}$+ cos^{2} $\mathrm{\alpha}$= 1 $\Rightarrow $3 cos^{2} $\mathrm{\alpha}$= 1

##### Question 3:

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

##### Answer:

##### Question 4:

Show that the points :

(2, 3, 4) ; (– 1, – 2, 1) ; (5, 8, 7) are collinear.

##### Answer:

Let A (2, 3, 4); B (– 1, – 2, 1) and C (5, 8, 7) be the given points.

$\therefore $ Direction-ratios of AB are :

< – 1 – 2, – 2 – 3, 1 – 4 >

i.e.< – 3, – 5, – 3 > i.e. < 3, 5, 3 >

Direction-ratios of BC are :

< 5 – (– 1), 8 – (– 2), 7 – 1 >

i.e. < 6, 10, 6 > i.e.< 3, 5, 3 >

$\Rightarrow $AB and BC have same direction-ratios

$\Rightarrow $ AB || BC.

But B is a common point.

Hence, 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:

Let A (3, 5, – 4), B (– 1, 1, 2) and C (– 5, – 5, – 2) be the vertices of ΔABC.

Now (I) Direction–ratios of AB are

< – 1 – 3, 1 – 5, 2 – (– 4) >

i.e. < – 4, – 4, 6 >

$\therefore $ Direction-cosines of AB are :

##### Question 6:

##### Answer:

##### Question 7:

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:

If A (1, – 1, 2) and B (3, 4, – 2) be the given points, then the direction–ratios of AB are :

< 3 – 1, 4 – (– 1), – 2 – 2 > i.e. < 2, 5, – 4 >.

Again if C (0, 3, 2) and D (3, 5, 6) be the given points,

then the direction ratios of CD are :

< 3 – 0, 5 – 3, 6 – 2 > i.e. < 3, 2, 4 >.

Now AB is perp. to CD if :

(2) (3) + (5) (2) + (– 4) (4) = 0

[a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0]

i.e. if 6 + 10 – 16 = 0, which is true.

Hence, AB is perpendicular to CD.

##### Question 8:

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:

If A (4, 7, 8) and B (2, 3, 4) be the given points, then the direction-ratios of AB are :

< 2 – 4, 3 – 7, 4 – 8 > i.e. < – 2, – 4, – 4 >

i.e. < 1, 2, 2 >.

If C (– 1, – 2, 1) and D (1, 2, 5) be the given points, then

the direction–ratios of CD are :

< 1 – (– 1), 2 – (– 2), 5 – 1 > i.e. < 2, 4, 4 >

i.e. < 1, 2, 2 >.

AB is parallel to CD if :

##### Question 9:

##### Answer:

##### Question 10:

##### Answer:

##### Question 11:

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

by :

##### Answer:

The given line is :

##### Question 12:

##### Answer:

##### Question 13:

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

##### Answer:

(i) The vector equations of the line is :

##### Question 14:

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

##### Answer:

##### Question 15:

Find the angle between the following pairs of lines :

##### Answer:

##### Question 16:

##### Answer:

##### Question 17:

Find the value of ‘p’ so that the lines :

##### Answer:

##### Question 18:

Show that the lines :

##### Answer:

##### Question 19:

##### Answer:

##### Question 20:

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##### Question 21:

Find the shortest distance between the lines whose vector equations are :

##### Answer:

##### Question 22:

Find the shortest distance between the following lines whose vector equations are :

##### Answer:

##### Question 23:

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

z = 2

##### Answer:

The given plane is z = 2 ....(1)

Direction-cosines of the normal to the plane are < 0, 0, 1 >.

And distance of (1) from the origin

##### Question 24:

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

x + y + z = 1

##### Answer:

The given plane is x + y + z = 1 ....(1)

Direction-ratios of the normal to the plane are : < 1, 1, 1 >

$\therefore $ Direction-cosines of the normal to the plane are

##### Question 25:

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

2x + 3y – z = 5

##### Answer:

The given plane is 2x + 3y – z = 5 ....(1)

Direction–ratios of the normal to the plane are < 2, 3, – 1 >.

$\therefore $ Direction–cosines of the normal to the plane are :

##### Question 26:

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

5y + 8 = 0.

##### Answer:

The given plane is 5y + 8 = 0 ....(1)

Direction–ratios of the normal to the plane are < 0, 5, 0 >

i.e. < 0, 1, 0 >

$\therefore $ Direction–cosines of the normal to the plane are < 0, 1, 0 > .

##### Question 27:

##### Answer:

##### Question 28:

Find the cartesian equation of the following planes :

##### Answer:

##### Question 29:

Find the cartesian equation of the following planes :

##### Answer:

##### Question 30:

Find the cartesian equation of the following planes :

##### Answer:

##### Question 31:

In the following cases, find the co-ordinates of the foot of the perpendicular drawn from the origin :

2x + 3y + 4z – 12 = 0

##### Answer:

The given plane is :

2x + 3y + 4z – 12 = 0 ...(1)

Direction-ratios of the normal to the plane are :

< 2, 3, 4 >

$\therefore $ The equations of the normal through (0, 0, 0) are :

##### Question 32:

In the following cases, find the co-ordinates of the foot of the perpendicular drawn from the origin :

3y + 4z – 6 = 0

##### Answer:

The given plane is 3y + 4z – 6 = 0 ....(1)

Direction-ratios of the normal are < 0, 3, 4 >.

$\therefore $ The equations of the normal though (0, 0, 0) are :

##### Question 33:

In the following cases, find the co-ordinates of the foot of the perpendicular drawn from the origin :

x + y + z = 1

##### Answer:

The given plane is :

x + y + z = 1 ....(1)

Direction-ratios of the normal to the plane are < 1, 1, 1 >.

##### Question 34:

In the following cases, find the co-ordinates of the foot of the perpendicular drawn from the origin :

5y + 8 = 0

##### Answer:

The given plane is :

5y + 8 = 0 ...(1)

Direction-ratios of the normal to the plane < 0, 5, 0 >

$\therefore $ The equations of the normal through (0, 0, 0) are :

##### Question 35:

Find the vector and cartesian equation of the plane :

##### Answer:

##### Question 36:

Find the vector and cartesian equation of the plane :

##### Answer:

##### Question 37:

Find the equation of the plane that passes through three points :

(1, 1, – 1), (6, 4 – 5), (– 4, – 2, 3)

##### Answer:

Any plane through (1, 1, – 1) is :

a (x – 1) + b (y – 1) + c (z + 1) = 0....(1)

Since the plane passes through the points

(6, 4, – 5) and (– 4, – 2, 3),

$\therefore $ a (6 – 1) + b (4 – 1) + c (– 5 + 1) = 0

and a (– 4 – 1) + b (– 2 – 1) + c (3 + 1) = 0

$\Rightarrow $ 5a + 3b – 4c = 0 ....(2)

and – 5a – 3b + 4c = 0 ....(3)

Solving (2) and (3),

$\Rightarrow $ a, b, c can’t be found.

Since the points are collinear,

$\therefore $ an infinite number of planes can be found through the given points.

##### Question 38:

Find the equation of the plane that passes through three points :

(1, 1, 0), (1, 2, 1), (– 2, 2, – 1).

##### Answer:

Any plane through (1, 1, 0) is :

a (x – 1) + b (y – 1) + cz = 0 ....(1)

Since the plane passes through the points

(1, 2, 1) and (– 2, 2, – 1),

$\therefore $ a (1 – 1) + b (2 – 1) + c (1) = 0

and a (– 2 – 1) + b (2 – 1) + c (– 1) = 0

$\Rightarrow $ 0 . a + b + c = 0 ....(2)

and – 3a + b – c = 0 ....(3)

Solving (2) and (3),

Putting these values of a, b, c in (1), we get :

– 2k (x – 1) – 3k (y – 1) + 3kz = 0

$\Rightarrow $ –2 (x – 1) – 3 (y – 1) + 3z = 0 [$\therefore $ k $\ne $ 0]

$\Rightarrow $ – 2x + 2 – 3y + 3 + 3z = 0

$\Rightarrow $ 2x + 3y – 3z = 5,

which is the reqd. equation.

##### Question 39:

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

##### Answer:

The given equation of the plane is :

2x + y – z = 5

##### Question 40:

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

##### Answer:

Clearly the plane passes through the point (0, 3, 0).

The direction-cosines of the normal to the plane are < 0, 1, 0 >

$\therefore $ The equation of the plane is :

0 . (x – 0) + 1 . (y – 3) + 0 . (z – 0) = 0

$\Rightarrow $ y – 3 = 0 $\Rightarrow $ y = 3.

##### Question 41:

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:

The given planes are :

3x – y + 2z – 4 = 0 ...(1)

and x + y + z – 2 = 0 ...(2)

Any plane through the intersection of (1) and (2) is :

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

Since it passes thro’ (2, 2, 1).

$\therefore $ (6 – 2 + 2 – 4) + k (2 + 2 + 1 – 2) = 0

$\Rightarrow $ 7x – 5y + 4z – 8 = 0,

which is the reqd. equation.

##### Question 42:

Find the vector equation of the plane passing through the intersection of the planes :

##### Answer:

The given planes are :

i.e. 2x + 2y – 3z – 7 = 0 ...(1)′

and 2x + 5y + 3z – 9 = 0 ...(2)′

Any plane through the intesection of (1)′ and (2)′ is :

(2x + 2y – 3z – 7) + k (2x + 5y + 3z – 9) = 9...(3)

Since it passes thro’ (2, 1, 3),

$\therefore $ (4 + 2 – 9 – 7) + k (4 + 5 + 9 – 9) = 0

##### Question 43:

Find the equation of the plane through the line of intersection of the planes :

x + y + z = 1 and 2x + 3y + 4z = 5,

which is perpendicular to the plane x – y + z = 0.

##### Answer:

Any plane through the line of intersection of
the planes :

x + y + z – 1 = 0 and 2x + 3y + 4z – 5 = 0 is :

(x + y + z – 1) + k (2x + 3y + 4z – 5) = 0

i.e.(1 + 2k) x + (1 + 3k) y + (1 + 4k) z

– (1 + 5k) = 0 ...(1)

Since it is perpendicular to the plane :

x – y + z = 0 ...(2)

$\therefore $ their normals are perpendicular

$\Rightarrow $ (1 + 2k) (1) + (1 + 3k) (– 1) + (1 + 4k) (1) = 0

$\Rightarrow $ 1 + 2k – 1 – 3k + 1 + 4k = 0

##### Question 44:

Find the angle between the planes whose vector equations are :

##### Answer:

##### Question 45:

In the following, determine whether the given planes are parallel or perpendicular and in case they are neither, find the angles between them :

7x + 5y + 6z + 30 = 0 and

3x – y – 10z + 4 = 0

##### Answer:

The given planes are :

7x + 5y + 6z + 30 = 0 ...(1)

and 3x – y – 10z + 4 = 0 ...(2)

##### Question 46:

In the following, determine whether the given planes are parallel or perpendicular and in case they are neither, find the angles between them :

2x + y + 3z – 2 = 0 and x – 2y + 5 = 0

##### Answer:

The given planes are :

2x + y + 3z – 2 = 0 ....(1)

and x – 2y + 5 = 0 ....(2)

Since (2) (1) + (1) (– 2) + (3) (0) = 0,

$\therefore $ planes are parallel.

##### Question 47:

In the following, determine whether the given planes are parallel or perpendicular and in case they are neither, find the angles between them :

2x – 2y + 4z + 5 = 0 and

3x – 3y + 6z – 1 = 0

##### Answer:

The given planes are :

2x – 2y + 4z + 5 = 0 ....(1)

and 3x – 3y + 6z – 1 = 0 . ...(2)

##### Question 48:

In the following, determine whether the given planes are parallel or perpendicular and in case they are neither, find the angles between them :

2x – y + 3z – 1 = 0 and

2x – y + 3z + 3 = 0

##### Answer:

The given planes are :

2x – y + 3z – 1 = 0 ....(1)

and 2x – y + 3z + 3 = 0 ....(2)

##### Question 49:

In the following, determine whether the given planes are parallel or perpendicular and in case they are neither, find the angles between them :

4x + 8y + z – 8 = 0 and y + z – 4 = 0.

##### Answer:

The given planes are :

4x + 8y + z – 8 = 0 ....(1)

and y + z – 4 = 0 ....(2)

##### Question 50:

In the following, find the distance of each of the given points from the corresponding given planes :

Point

(0, 0, 0)

Plane

3x – 4y + 12z = 3

##### Answer:

The given plane is :

3x – 4y + 12z – 3 = 0 ....(1)

$\therefore $ Distance, of (0, 0, 0) from plane (1)

##### Question 51:

In the following, find the distance of each of the given points from the corresponding given planes :

Point

(3, – 2, 1)

Plane

2x – y + 2z + 3 = 0

##### Answer:

The given plane is :

2x – y + 2z + 3 = 0 ....(1)

$\therefore $ Distance, (3, – 2, 1) from plane (1)

##### Question 52:

In the following, find the distance of each of the given points from the corresponding given planes :

Point

(2, 3, – 5)

Plane

x + 2y – 2z = 9

##### Answer:

The given plane is :

x + 2y – 2z – 9 = 0 ....(1)

$\therefore $ Distance of (2, 3, – 5) from plane (1)

##### Question 53:

In the following, find the distance of each of the given points from the corresponding given planes :

Point

(– 6, 0, 0)

Plane

2x – 3y + 6z – 2 = 0

##### Answer:

The given plane is 2x – 3y + 6z – 2 = 0 ....(1)

$\therefore $ Distance of (– 6, 0, 0) from place (1)

##### Question 54:

Distance between two planes :

2x + 3y + 4z = 5 and 4x + 6y + 8z = 12 is :

- 2 units
- 4 units
- 8 units

##### Answer:

##### Question 55:

The planes 2x – y + 4z = 3 and 5x – 2·5y + 10 z = 6 are :

- perpendicular
- parallel
- intersect along y-axis

##### Answer:

parallel