# 11- 3D GEOMETRY

Here we are going through the ncert solutions for class 12 maths chapter 11– 3D geometry . so before going through the ncert solutions make sure to go through the textbook that helps you to understand the solutions more easily.

# EXERCISE 11.1

#### Question 1:

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

Let direction cosines of the line be lm, 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.

Let the direction cosines of the line make an angle α with each of the coordinate axes.

l = cos αm = cos αn = 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?

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.

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, (x1y1z1) and (x2y2z2), are given by, x2 − x1y2 − 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)

The vertices of ΔABC 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

−217√−317√−217√-217, -317, -217

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

Therefore the direction cosines​​ of CA are

8(8)2 + (10)2 + (−2)210(8)2 + (10)2 + (−2)2−2(8)2 + (10)2 + (−2)28242√10242√−2242√442√542√−142√882 + 102 + -22, 1082 + 102 + -22, -282 + 102 + -228242, 10242, -2242442, 542, -142

#### Question 1:

Show that the three lines with direction cosines

are mutually perpendicular.

Two lines with direction cosines, l1m1n1 and l2m2n2, 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).

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, a1b1c1, of AB are (3​​ − 1), (4 − (−1)), and (−2 − 2) i.e., 2, 5, and −4.

The direction ratios, a2b2c2, 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 + b1b2c1c2 = 0

a1a2 + b1b2c1c2 = 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).

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, a1b1c1, of AB are (2 − 4), (3 − 7), and (4 − 8) i.e., −2, −4, and −4.

The direction ratios, a2b2c2, 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.

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 .

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

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 (x1y1z1) and with direction​​ ratios, abc, is given by

Therefore the equation of the required line is

#### Question 7:

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

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).

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 (x1y1z1) and direction ratios abc 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).

Let the line passing through the points, P (3, −2, −5) and Q (3, −2, 6), be PQ.

Since PQ passes through P (3, −2, −5), its​​ position vector is given by,

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

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

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

The given lines are parallel to the​​ vectors, and , respectively.

(ii) The given lines are parallel to the vectors, and , respectively.

#### Question 11:

Find the angle between the following pairs of lines:

(i)

(ii)

• Let and  be the vectors parallel to the pair of​​ lines, respectively.

and

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.

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, a1b1c1 and a2b2c2, 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.

The equations of the given lines areand

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

Two lines with direction ratios, a1b1c1 and a2b2c2, 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

The equations of the given lines are

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

d = ∣∣∣∣(b1→×b2).(a2→−a1)∣∣∣b1→×b2→∣∣∣∣∣∣∣d = b1→×b2→.a2→-a1→b1→×b2→

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

The given lines are  and

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

Comparing the given equations, we obtain

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

The given lines are and

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

Comparing the given equations with  and , we obtain

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

The given lines are

r→=(s+1)iˆ+(2s−1)jˆ−(2s+1)kˆ⇒r→=(iˆ−jˆ−kˆ)+s(iˆ+2jˆ−2kˆ)             ...(2)r→=(s+1)i^+(2s-1)j^-(2s+1)k^r→=(i^-j^-k^)+s(i^+2j^-2k^)             ...(2)

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

For the given equations,

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

Therefore, the shortest​​ distance between the lines isunits.

#### 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)