Exercise 11.1

Question 1:

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

Answer

Let direction cosines of the line be l, m, and n.

Therefore, the direction cosines of the line are

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 make an angle α with each of the coordinate axes.

∴ l = cos α, m = cos α, n = cos α

Thus, the direction cosines of the line, which is equally inclined to the coordinate axes,

are

Question 3:

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

If a line has direction ratios of −18, 12, and −4, then its direction cosines are

Thus, the direction cosines are .

Question 4:

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

The given points are A (2, 3, 4), B (− 1, − 2, 1), and C (5, 8, 7).

It is known that the direction ratios of line joining the points, (x₁, y₁, z₁) and (x₂, y₂, z₂), are given by, x₂ − x₁, y₂ − y₁, and z₂ − z₁.

The direction ratios of AB are (−1 − 2), (−2 − 3), and (1 − 4) i.e., −3, −5, and −3.

The direction ratios of BC are (5 − (− 1)), (8 − (− 2)), and (7 − 1) i.e., 6, 10, and 6. It can be seen that the direction ratios of BC are −2 times that of AB i.e., they are proportional.

Therefore, AB is parallel to BC. Since point B is common to both AB and BC, points A, B, and 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 vertices of ∆ABC are A (3, 5, −4), B (−1, 1, 2), and C (−5, −5, −2).

The direction ratios of side AB are (−1 − 3), (1 − 5), and (2 − (−4)) i.e., −4, −4, and 6.

Therefore, the direction cosines of AB are

The direction ratios of BC are (−5 − (−1)), (−5 − 1), and (−2 − 2) i.e., −4, −6, and −4. Therefore, the direction cosines of BC are

The direction ratios of CA are (−5 − 3), (−5 − 5), and (−2 − (−4)) i.e., −8, −10, and 2. Therefore, the direction cosines of AC are

Exercise 11.2

Question 1:

Show that the three lines with direction cosines

are mutually perpendicular.

Answer

Two lines with direction cosines, l₁, m₁, n₁ and l₂, m₂, n₂, are perpendicular to each other, if l₁l₂ + m₁m₂ + n₁n₂ = 0

(i)  For the lines with direction cosines, and   , we obtain

Therefore, the lines are perpendicular.

(ii)  For the lines with direction cosines,  and  , we obtain

Therefore, the lines are perpendicular.

(iii)  For the lines with direction cosines,                  and                   , we obtain

Therefore, the lines are perpendicular.

Thus, all the lines 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

Let AB be the line joining the points, (1, −1, 2) and (3, 4, − 2), and CD be the line joining the points, (0, 3, 2) and (3, 5, 6).

The direction ratios, a₁, b₁, c₁, of AB are (3 − 1), (4 − (−1)), and (−2 − 2) i.e., 2, 5, and

−4.

The direction ratios, a₂, b₂, c₂, of CD are (3 − 0), (5 − 3), and (6 −2) i.e., 3, 2, and 4.

AB and CD will be perpendicular to each other, if a₁a₂ + b₁b₂+ c₁c₂ = 0

a₁a₂ + b₁b₂+ c₁c₂ = 2 × 3 + 5 × 2 + (− 4) × 4

= 6 + 10 − 16

= 0

Therefore, AB and CD are perpendicular to each other.

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

Let AB be the line through the points, (4, 7, 8) and (2, 3, 4), and CD be the line through the points, (−1, −2, 1) and (1, 2, 5).

The directions ratios, a₁, b₁, c₁, of AB are (2 − 4), (3 − 7), and (4 − 8) i.e., −2, −4, and

−4.

The direction ratios, a₂, b₂, c₂, of CD are (1 − (−1)), (2 − (−2)), and (5 − 1) i.e., 2, 4,

and 4.

AB will be parallel to CD, if

Thus, AB is parallel to CD.

Question 4:

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

the vector .

Answer

It is given that the line passes through the point A (1, 2, 3). Therefore, the position

vector through A is

It is known that the line which passes through point A and parallel to  is given by   is a constant.

This is the required equation of the line.

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

It is given that the line passes through the point with position vector

It is known that a line through a point with position vector  and parallel to   is given by the equation,

This is the required equation of the line in vector form.

Eliminating λ, we obtain the Cartesian form equation as

This is the required equation of the given line in Cartesian form.

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

It is given that the line passes through the point (−2, 4, −5) and is parallel to

The direction ratios of the line,   , are 3, 5, and 6. The required line is parallel to

Therefore, its direction ratios are 3k, 5k, and 6k, where k ≠ 0

It is known that the equation of the line through the point (x₁, y₁, z₁) and with direction

ratios, a, b, c, is given by

Therefore the equation of the required line is

Question 7:

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

The Cartesian equation of the line is

The given line passes through the point (5, −4, 6). The position vector of this point is

Also, the direction ratios of the given line are 3, 7, and 2.

This means that the line is in the direction of vector,

It is known that the line through position vector     and in the direction of the vector  is given by the equation,

This is the required equation of the given line in vector form.

Question 8:

Find the vector and the Cartesian equations of the lines that pass through the origin and (5, −2, 3).

Answer

The required line passes through the origin. Therefore, its position vector is given by,

The direction ratios of the line through origin and (5, −2, 3) are (5 − 0) = 5, (−2 − 0) = −2, (3 − 0) = 3

The line is parallel to the vector given by the equation,

The equation of the line in vector form through a point with position vector  and parallel

to  is,

The equation of the line through the point (x₁, y₁, z₁) and direction ratios a, b, c is given by,

Therefore, the equation of the required line in the Cartesian form 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

The direction ratios of PQ are given by, (3 − 3) = 0, (−2 + 2) = 0, (6 + 5) = 11

The equation of the vector in the direction of PQ is

The equation of PQ in vector form is given by,

The equation of PQ in Cartesian form is

i.e.,

Question 10:

Find the angle between the following pairs of lines:

(i)

(ii)  and

Answer

(i)  Let Q be the angle between the given lines.

The angle between the given pairs of lines is given by,

(ii) 

and

Question 11:

Find the angle between the following pairs of lines:

(i)

(ii)

Answer

Let    and  be the vectors parallel to the pair of lines,

The angle, Q, between the given pair of lines is given by the relation,

(ii) Let   be the vectors parallel to the given pair of lines,  and

, respectively.

If Q is the angle between the given pair of lines, then

Question 12:

Find the values of p so the line   and

  are at right angles.

Answer

The given equations can be written in the standard form as

and

The direction ratios of the lines are −3,     , 2 and    respectively.

Two lines with direction ratios, a₁, b₁, c₁ and a₂, b₂, c₂, are perpendicular to each other, if

a₁a₂ + b₁ b₂ + c₁c₂ = 0

Thus, the value of p is       .

Question 13:

Show that the lines  and  are perpendicular to each other. Answer

The equations of the given lines are  and

The direction ratios of the given lines are 7, −5, 1 and 1, 2, 3 respectively.

Two lines with direction ratios, a₁, b₁, c₁ and a₂, b₂, c₂, are perpendicular to each other, if

a₁a₂ + b₁ b₂ + c₁c₂ = 0

∴ 7 × 1 + (−5) × 2 + 1 × 3

= 7 − 10 + 3

= 0

Therefore, the given lines are perpendicular to each other.

Question 14:

Find the shortest distance between the lines

Answer

The equations of the given lines are

Comparing the given equations, we obtain

Substituting all the values in equation (1), we obtain

Therefore, the shortest distance between the two lines is          units.

Question 15:

Find the shortest distance between the lines  and  Answer

The given lines are  and  It is known that the shortest distance between the two lines,

Comparing the given equations, we obtain

, is given by,

Substituting all the values in equation (1), we obtain

Since distance is always non-negative, the distance between the given lines is  units.

Question 16:

Find the shortest distance between the lines whose vector equations are

Answer

The given lines are  and

It is known that the shortest distance between the lines,    and  , is given by,

Substituting all the values in equation (1), we obtain

Therefore, the shortest distance between the two given lines is  units.

Question 17:

Find the shortest distance between the lines whose vector equations are

Answer

The given lines are

For the given equations,

Substituting all the values in equation (3), we obtain

Therefore, the shortest distance between the lines is         units.

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.

(a)z = 2 (b)

(c)         (d)5y + 8 = 0

Answer

(a) The equation of the plane is z = 2 or 0x + 0y + z = 2 … (1) The direction ratios of normal are 0, 0, and 1.

∴

Dividing both sides of equation (1) by 1, we obtain

This is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the origin is 2 units.

(b) x + y + z = 1 … (1)

The direction ratios of normal are 1, 1, and 1.

∴

Dividing both sides of equation (1) by  , we obtain

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal are and the distance of normal from the origin is units.

(c) 2x + 3y − z = 5 … (1)

The direction ratios of normal are 2, 3, and −1.

Dividing both sides of equation (1) by   , we obtain

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal to the plane are  and the distance of normal from the origin is  units.

(d) 5y + 8 = 0

⇒ 0x − 5y + 0z = 8 … (1)

The direction ratios of normal are 0, −5, and 0.

Dividing both sides of equation (1) by 5, we obtain

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal to the plane are 0, −1, and 0 and the

distance of normal from the origin is units.

Question 2:

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

normal to the vector . Answer

The normal vector is,

It is known that the equation of the plane with position vector  is given by,

This is the vector equation of the required plane.

Question 3:

Find the Cartesian equation of the following planes:

(a)   (b)

(c)

Answer

(a) It is given that equation of the plane is

For any arbitrary point P (x, y, z) on the plane, position vector   is given by,

Substituting the value of  in equation (1), we obtain

This is the Cartesian equation of the plane.

(b)

For any arbitrary point P (x, y, z) on the plane, position vector  is given by,

Substituting the value of    in equation (1), we obtain

This is the Cartesian equation of the plane.

(c)

For any arbitrary point P (x, y, z) on the plane, position vector  is given by,

Substituting the value of    in equation (1), we obtain

This is the Cartesian equation of the given plane.

Question 4:

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

(a)    (b)

(c)               (d)

Answer

(a)  Let the coordinates of the foot of perpendicular P from the origin to the plane be (x₁, y₁, z₁).

2x + 3y + 4z − 12 = 0

⇒ 2x + 3y + 4z = 12 … (1)

The direction ratios of normal are 2, 3, and 4.

Dividing both sides of equation (1) by    , we obtain

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by (ld, md, nd).

Therefore, the coordinates of the foot of the perpendicular are

(b)  Let the coordinates of the foot of perpendicular P from the origin to the plane be (x₁,

y₁, z₁).

⇒  … (1)

The direction ratios of the normal are 0, 3, and 4.

Dividing both sides of equation (1) by 5, we obtain

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by (ld, md, nd).

Therefore, the coordinates of the foot of the perpendicular are

(c)   Let the coordinates of the foot of perpendicular P from the origin to the plane be (x₁,

y₁, z₁).

… (1)

The direction ratios of the normal are 1, 1, and 1.

Dividing both sides of equation (1) by     , we obtain

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by (ld, md, nd).

Therefore, the coordinates of the foot of the perpendicular are

(d)  Let the coordinates of the foot of perpendicular P from the origin to the plane be (x₁,

y₁, z₁).

⇒ 0x − 5y + 0z = 8 … (1)

The direction ratios of the normal are 0, −5, and 0.

Dividing both sides of equation (1) by 5, we obtain

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by (ld, md, nd).

Therefore, the coordinates of the foot of the perpendicular are

Question 5:

Find the vector and Cartesian equation of the planes

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

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

.

Answer

(a)   The position vector of point (1, 0, −2) is

The normal vector   perpendicular to the plane is

The vector equation of the plane is given by,

 is the position vector of any point P (x, y, z) in the plane.

Therefore, equation (1) becomes

This is the Cartesian equation of the required plane.

(b)   The position vector of the point (1, 4, 6) is   The normal vector    perpendicular to the plane is

 is the position vector of any point P (x, y, z) in the plane.

Therefore, equation (1) becomes

This is the Cartesian equation of the required plane.

Question 6:

Find the equations of the planes that passes through three points. (a) (1, 1, −1), (6, 4, −5), (−4, −2, 3)

(b) (1, 1, 0), (1, 2, 1), (−2, 2, −1)

Answer

(a) The given points are A (1, 1, −1), B (6, 4, −5), and C (−4, −2, 3).

Since A, B, C are collinear points, there will be infinite number of planes passing through the given points.

(b) The given points are A (1, 1, 0), B (1, 2, 1), and C (−2, 2, −1).

Therefore, a plane will pass through the points A, B, and C.

It is known that the equation of the plane through the points,       , and  , is

This is the Cartesian equation of the required plane.

Question 7:

Dividing both sides of equation (1) by 5, we obtain

It is known that the equation of a plane in intercept form is    , where a, b, c

Thus, the intercepts cut off by the plane are .

Question 8:

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

The equation of the plane ZOX is

y = 0

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

Since the y-intercept of the plane is 3,

∴ a = 3

Thus, the equation of the required plane is y = 3

Question 9:

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

and                          and the point (2, 2, 1)

Answer

The plane passes through the point (2, 2, 1). Therefore, this point will satisfy equation (1).

This is the required equation of the plane.

Question 10:

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

 and through the point (2, 1, 3) Answer

The equations of the planes are

The equation of any plane through the intersection of the planes given in equations (1) and (2) is given by,

, where

The plane passes through the point (2, 1, 3). Therefore, its position vector is given by,

Substituting in equation (3), we obtain

Substituting   in equation (3), we obtain

This is the vector equation of the required plane.

Question 11:

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

 and                          which is perpendicular to the plane

Answer

The direction ratios, a₁, b₁, c₁, of this plane are (2λ + 1), (3λ + 1), and (4λ + 1). The plane in equation (1) is perpendicular to

in equation (1), we obtain

This is the required equation of the plane.

Question 12:

Find the angle between the planes whose vector equations are

and

.

Answer

The equations of the given planes are                                 and

Here,

Question 13:

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

(a)

(b)

(c)

(d)

(e) Answer

The direction ratios of normal to the plane,    , are a₁, b₁, c₁ and

.

The angle between L₁ and L₂ is given by,

(a)  The equations of the planes are 7x + 5y + 6z + 30 = 0 and 3x − y − 10z + 4 = 0

Here, a₁ = 7, b₁ =5, c₁ = 6

Therefore, the given planes are not perpendicular.

It can be seen that,

(b)  The equations of the planes are   and  Here,  and

Thus, the given planes are perpendicular to each other.

(c) 

and

Here,

Thus, the given planes are not perpendicular to each other.

∴

Thus, the given planes are parallel to each other.

(d)  The equations of the planes are  and  Here,         and

∴

Thus, the given lines are parallel to each other.

(e) 

and

Here,

and

Therefore, the given lines are not perpendicular to each other.

∴

Question 14:

In the following cases, find the distance of each of the given points from the corresponding given plane.

Point Plane

(a) (0, 0, 0)       

(b) (3, −2, 1)    

(c) (2, 3, −5)      

(d) (−6, 0, 0)    

Answer

It is known that the distance between a point, p(x₁, y₁, z₁), and a plane, Ax + By + Cz =

D, is given by,

(a)  The given point is (0, 0, 0) and the plane is

(b)  The given point is (3, − 2, 1) and the plane is

∴

(c)  The given point is (2, 3, −5) and the plane is

(d)  The given point is (−6, 0, 0) and the plane is

Miscellaneous Solutions

Question 1:

Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, −1), (4, 3, −1).

Answer

Let OA be the line joining the origin, O (0, 0, 0), and the point, A (2, 1, 1).

Also, let BC be the line joining the points, B (3, 5, −1) and C (4, 3, −1).

The direction ratios of OA are 2, 1, and 1 and of BC are (4 − 3) = 1, (3 − 5) = −2, and

(−1 + 1) = 0

OA is perpendicular to BC, if a₁a₂ + b₁b₂ + c₁c₂ = 0

∴ a₁a₂ + b₁b₂ + c₁c₂ = 2 × 1 + 1 (−2) + 1 ×0 = 2 − 2 = 0

Thus, OA is perpendicular to BC.

Question 2:

If l₁, m₁, n₁ and l₂, m₂, n₂ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are m₁n₂ − m₂n₁, n₁l₂ − n₂l₁, l₁m₂ − l₂m₁.

Answer

Let l, m, n be the direction cosines of the line which is perpendicular to the line with direction cosines l₁, m₁, n₁ and l₂, m₂, n₂.

l, m, n are the direction cosines of the line.

∴ l² + m² + n² = 1 … (5)

It is known that,

∴

Substituting the values from equations (5) and (6) in equation (4), we obtain

Thus, the direction cosines of the required line are

Question 3:

Find the angle between the lines whose direction ratios are a, b, c and b − c, c − a, a − b.

Answer

Thus, the angle between the lines is 90°.

Question 4:

Find the equation of a line parallel to x-axis and passing through the origin. Answer

The line parallel to x-axis and passing through the origin is x-axis itself.

Let A be a point on x-axis. Therefore, the coordinates of A are given by (a, 0, 0), where

a ∈ R.

Thus, the equation of line parallel to x-axis and passing through origin is

Question 5:

If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

Answer

The coordinates of A, B, C, and D are (1, 2, 3), (4, 5, 7), (−4, 3, −6), and

(2, 9, 2) respectively.

The direction ratios of AB are (4 − 1) = 3, (5 − 2) = 3, and (7 − 3) = 4

The direction ratios of CD are (2 −(− 4)) = 6, (9 − 3) = 6, and (2 −(−6)) = 8

It can be seen that,   Therefore, AB is parallel to CD.

Thus, the angle between AB and CD is either 0° or 180°.

sQuestion 6:

If the lines  and   are perpendicular, find the value of k.

Answer

The direction of ratios of the lines,  and  , are −3, 2k, 2 and 3k, 1, −5 respectively.

Therefore, for , the given lines are perpendicular to each other.

Question 7:

Find the vector equation of the plane passing through (1, 2, 3) and perpendicular to the

plane  Answer

The position vector of the point (1, 2, 3) is

The direction ratios of the normal to the plane,  , are 1, 2, and −5 and the normal vector is

The equation of a line passing through a point and perpendicular to the given plane is

Question 8:

Find the equation of the plane passing through (a, b, c) and parallel to the plane

Answer

Any plane parallel to the plane,    , is of the form

The plane passes through the point (a, b, c). Therefore, the position vector   of this point is