# Class 12 Maths NCERT Solutions for Chapter 11 Three Dimensional Geometry Exercise 11.2

### Three Dimensional Geometry Exercise 11.2 Solutions

**1. Show that the three lines with direction cosines 12/13, -3/13, -4/13; 4/13, 12/13, 3/13; 3/13, -4/13, 12/13 are mutually perpendicular. **

**Solution**

Two lines with direction cosines, *l*_{1}, *m*_{1}, *n*_{1} and *l*_{2}, *m*_{2}, *n*_{2}, are perpendicular to each other, if *l*_{1}*l*_{2} + *m*_{1}*m*_{2} + *n*_{1}*n*_{2} = 0

(i) For the lines with direction cosines, 12/3, -3/13, -4/13 and 4/13, 12/13, 3/13, we obtain

Therefore, the lines are perpendicular.

(ii) For the lines with direction cosines, 4/13, 12/13, 3/13 and 3/13, -4/13, 12/13, we obtain

Therefore, the lines are perpendicular.

(iii) For the lines with direction cosines, 3/13, -4/13, 12/13 and 12/13, -3/13, -4/13, we obtain

Therefore, the lines are perpendicular.

Thus, all the lines are mutually perpendicular.

**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).**

**Solution**

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, *a*_{1}, *b*_{1}, *c*_{1}, of AB are (3 −1), (4 −(−1)), and (−2− 2) i.e., 2, 5, and −4.

The direction ratios, *a*_{2}, *b*_{2}, *c*_{2}, 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 *a*_{1}*a*_{2} + *b*_{1}*b*_{2}+ *c*_{1}*c*_{2} = 0*a*_{1}*a*_{2} + *b*_{1}*b*_{2}+ *c*_{1}*c*_{2} = 2 × 3 + 5 × 2 + (− 4) × 4

= 6 + 10 − 16

= 0

Therefore, AB and CD are perpendicular to each other.

**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).**

**Solution**

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, *a*_{1}, *b*_{1}, *c*_{1}, of AB are (2 −4), (3 −7), and (4 −8) i.e., −2, −4, and −4.

The direction ratios, *a*_{2}, *b*_{2}, *c*_{2}, of CD are (1 −(−1)), (2 −(−2)), and (5 − 1) i.e., 2, 4, and 4.

AB will be parallel to CD, if *a*_{1}/*a*_{2}, *b*_{1}/*b*_{2} , *c*_{1}/*c*_{2},

Thus, AB is parallel to CD.

**4. Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector 3i ^+ 2j ^- 2k ^.**

**Solution**

It is given that the line passes through the point A(1, 2, 3). Therefore, the position vector through A is a⃗ = i ^ + 2j ^ - 3k ^ , b⃗ = 3i ^ + 2j ^ - 2k ^

It is known that the line which passes through point A and parallel to b⃗ is given by r⃗ = a⃗ + Î»b⃗ , where Î» is a constant.

This is the required equation of the line.

**5. Find the equation of the line in vector and in Cartesian form that passes through the point with position vector 2i ^- j ^+ 4k ^ and is in the direction i ^ + 2j ^ - k ^ .**

**Solution**

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

a⃗ = i ^ + 2j ^ - 3k ^ **...(1)**

b⃗ = 3i ^ + 2j ^ - 2k ^ **...(2)**

It is known that a line through a point with position vector a⃗ and parallel to b⃗ is given by the equation, r⃗ = a⃗ + Î»b⃗

⇒ r⃗ = 2i ^ - j ^ + 4k ^ + Î»( i ^ + 2j ^ - k ^ )

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

Eliminating Î», we obtain the Cartesian form equation as

(x -2)/1 = (y + 1)/2 = (z- 4)/-1

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

**6. Find the Cartesian equation of the line which passes through the point (-2, 4, -5) and parallel to the line given by (x + 3)/3 = (y - 4)/5 = (z + 8)/6 **

**Solution**

It is given that the line passes through the point (-2, 4, -5) and is parallel to (x + 3)/3 = (y - 4)/5 = (z + 8)/6

The direction ratios of the line, (x + 3)/3 = (y - 4)/5 = (z + 8)/6 , are 3, 5 and 6.

The required line is parallel to (x + 3)/3 = (y - 4)/5 = (z + 8)/6

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

It is known that the equation of the line through the point ( *x*_{1}, *y*_{1}, *z*_{1}) and with direction ratios, a, b, c, is given by

Therefore the equation of the required line is

**7. The Cartesian equation of a line is (x - 5)/3 = (y + 4)/7 = (z - 6)/2 . Write its vector form. **

**Solution**

The cartesian equation of the line is (x - 5)/3 = (y + 4)/7 = (z - 6)/2 **...(1)**

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

a⃗ = 5i ^ - 4j ^ +6k ^

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

This means that the line is in the direction of vector, b⃗ = 3i ^ + 7j ^ + 2k ^

It is known that the line through position vector a⃗ and in the direction of the vector b⃗ is given by the equation, r⃗ = a⃗ + Î»b⃗ , Î» ∊ R

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

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

**Solution**

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

a⃗ = 0⃗ **...(1) **

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 b⃗ = 5i ^ - 2j ^ + 3k ^

The equation of the line in vector form through a point with position vector a⃗ and parallel to b⃗ is, r⃗ = a⃗ + Î»b⃗ , Î» ∊ R

The equation of the line through the point (*x*_{1}, *y*_{1}, *z*_{1}) and direction ratios *a*, *b*, *c* is given by,

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

(x - 0)/5 = (y - 0)/-2 = (z - 0)/3

⇒ x/5 = y/-2 = z/3

**9. Find the vector and the Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).**

**Solution**

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,

a⃗ = 3i ^ - 2j ^ - 5k ^

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

b⃗ = 0.i ^ - 0.j ^ + 11k ^ = 11k ^

The equation of PQ in Cartesian form is

**10. Find the angle between the following pairs of lines: (i) r⃗ = 2i ^- 5j ^ + k ^+ Î» (3i ^ - 2j ^ + 6k ^) and r⃗ = 7i ^ - 6k ^ + Î¼ (i ^ + 2j ^ + 2k ^) (ii) r⃗ = 3i ^ + j ^- 2k ^ + Î» (i ^ - j ^- 2k ^) and r⃗ = 2i ^ - j ^- 56k ^ + Î¼ (3i ^ - 5j ^ - 4k ^) **

**Solution**

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

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

The given lines are parallel to the vectors, respectively.

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

**11. Find the angle between the following pairs of lines : (i) (x - 2)/2 = (y - 1)/5 = (z + 3)/-3 and (x +2)/-1 = (y- 4)/8 = (z - 5)/4 (ii) x/2 = y/2 = z/1 and (x -5)/4 = (y - 2)/1 = (z - 3)/8 **

**Solution**

(i) Let be the vectors parallel to the pair of lines, (x-2)/2 = (y-1)/5 = (z+3)/-3 and (x+2)/-1 = (y- 4)/8 = (z-5)/4, respectively.

= 2(-1) + 5×8 + (-3)×4

= -2 + 40 - 12

= 26

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, x/2 = y/2 = z/1 and (x-5)/4 = (y-5)/1 = (z-3)/8, respectively.

= 2×4 + 2×1 + 1×8

= 8 + 2 + 8

= 18

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

⇒ cos Q = 18/(3 × 9) = 2/3

⇒ Q = cos^{-1} (2/3)

**12. Find the values of p so the line (1-x)/3 = (7y-14)/2p = (z-3)/2 and (7-7x)/3p = (y-5)/1 = (6-z)/5 are right angles. **

**Solution**

The given equations can be written in the standard form as

The direction ratios of the lines are -3, 2p/7, 2 and -3p/7, 1, -5 respectively.

Two lines with direction ratios, a_{1}, b_{1}, c_{1} and a_{2}, b_{2}, c_{2} are perpendicular to each other, if a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

Thus, the values of p is 70/11.

**13. Show the the lines (x - 5)/7 = (y + 2)/-5 = z/1 and x/1 = y/2 = z/3 are perpendicular to each other. **

**Solution**

The equations of the given lines are (x - 5)/7 = (y + 2)/-5 = z/1 and x/1 = y/2 = z/3

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

Two lines with direction ratios, a_{1} , b_{1}, c_{1} and a_{2} , b_{2}, c_{2} are perpendicular to each other, if a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

∴ 7×1 + (-5)×2 + 1×3

= 7 - 10 + 3

= 0

Therefore, the given lines are perpendicular to each other.

**14. Find the shortest distance between the lines r⃗ = (i ^- 2j ^+ k ^) + Î» (i ^- j ^+ k ^) and r⃗ = 2i ^- j ^- k ^ + Î¼ (2i ^+ j ^+ 2k ^) **

**Solution**

The equations of the given lines are

r⃗ = (i ^ - 2j ^ + k ^) + Î» (i ^ - j ^ + k ^)

r⃗ = 2i ^ - j ^ - k ^ + Î¼ (2i ^ + j ^ + 2k ^)

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

Comparing the given equations, we obtain

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

Therefore, the shortest distance between the two lines is 3√2/2 units.

**15. Find the shortest distance between the lines (x+1)/7 = (y + 1)/-6 = (z + 1)/1 and (x - 3)/1 = (y - 5)/-2 = (z - 7)/1 .**

**Solution**

The given lines are (x+1)/7 = (y + 1)/-6 = (z + 1)/1 and (x - 3)/1 = (y - 5)/-2 = (z - 7)/1 .

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

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

d = -116/2√29 = -58/√29 = (-2 × 29)/√29 = -2√29

Since distance is always non-negative, the distance between the given lines is 2√29 units.

**16. Find the shortest distance between the lines whose vector equations are r⃗ = (i ^+ 2j ^+ 3k ^) + Î» (i ^- 3j ^+ 2k ^) and r⃗ = 4i ^+ 5j ^+ 6k ^ + Î¼ (2i ^+ 3j ^+ k ^) **

**Solution**

The given lines are r⃗ = (i ^ + 2j ^ + 3k ^) + Î» (i ^ - 3j ^ + 2k ^) and r⃗ = 4i ^ + 5j ^ + 6k ^ + Î¼ (2i ^ + 3j ^ + k ^)

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

Comparing the given equations with

= -9×3 + 3×3 + 9×3

= 9

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

Therefore, the shortest distance between the two given lines is 3/√19 units.

**17. Find the shortest distance between the lines whose vector equations are r⃗ = (1 -t)i ^ + (t-2)j ^ + (3-2t) k ^ and r⃗ = (s + 1)i ^ + (2s- 1)j ^ + (2s+1)k ^ **

**Solution**

The given lines are

r⃗ = (1 -t)i ^ + (t - 2)j ^ + (3 - 2t) k ^

⇒ r⃗ = (i ^ - 2j ^ + 3k ^) + t(-i ^ + j ^ - 2k ^) **...(1)**

r⃗ = (s + 1)i ^ + (2s - 1)j ^ + (2s + 1)k ^

⇒ r⃗ = (i ^ - j ^ - k ^) + s(i ^ +2j ^ - 2k ^) **...(2)**

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

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

Therefore, the shortest distance between the line is 8/√29 units.