STUDYORB

EXERCISE 5 (a)
1. A line passes through A (2 iˆ − ˆj + 4 kˆ) and is in the direction of iˆ + ˆj − 2 kˆ . Obtain its equation in the
vector and in Cartersian form. (Pb 1991 C)
x − 3 y +1 z − 3
2. The cartesian equations of a line are = = . Find a vector equation for the line.
2 −2 5
3. (i) Find the equation of the line, both vector and cartesian which is parallel to the vector 2 iˆ − ˆj + 3 kˆ
and which passes through the point (5, −2, 4).
(ii) Find the vector equation of the line through the point A (1, 2, 3) and parallel to the vector
iˆ − 2 ˆj + 3 kˆ . Deduce the Cartesian form of the equation. (CBSE 1990)
(iii) Write the equation of the straight line passing through (−1, 2, 3) and equally inclined to the axes.
(H.B. 1997)
4. (i) Find the vector equation of the line through the point A (3, 4, − 7) and B (1, −1, 6)
(ii) Find the vector form as well as in Cartesian form, the equation of the line passing through the
points A (1, 2, −1) and B (2, 1, 1).
(iii) A (−1, 4, −2) and B (0, 2, −1). (H.B. 1999 C)
5. Find the vector equation of a line through (2, −1, 1) and parallel to the line whose equations are
x − 3 y +1 z − 2
= = .
2 7 −3
6. Find the Cartesian equations of a line passing through (1, −1, 2) and parallel to the line whose
equations are x − 3 = y − 1 = z + 1 . Also reduce the equations obtained to the vector form.
1 2 −2
7. ABCD is a parallelogram. The position vectors of the points A, B and C are respectively,
4 iˆ + 5 ˆj − 10 kˆ, 2 iˆ − 3 ˆj + 4 kˆ and −iˆ + 2 ˆj + kˆ . Find the vector equation of the line BD. Also reduce it to
Cartesian form.
x−2 2y − 5
8. (i) Find the direction cosines of the line = , z = − 1. Also find the vector equation of the
2 −3
line.
(ii) The Cartesian equations of a line are 3x + 1 = 6y − 2 = 1 − z. Find its d.r.’s and write it in vector form.
[CBSE (SP) 2000]
(iii) The Cartesian equations of a line are x = ay + b, z = cy + d. Find the its directions ratios and reduce
it to vector form.
x−b y−0 z−d
[Hint. x = ay + b, z = cy + d ⇒ = = . d.r’s are a, 1, c and the vector equation is
a 1 c
→
r = (b iˆ + 0 ˆj + d kˆ) + λ (a iˆ + b ˆj + c kˆ) ].
9. Find the vector equation of the line passing through the point A (2, −1, 1) and parallel to the line joining
the points B (−1, 4, 1) and C (1, 2, 2). Also find the cartesian equation of the line.
[Same as solved Ex. 2] (CBSE 2003)
10. Find the coordinates of the point where the line joining (2, −3, 1) and (3, 4, −5) meets the xy-plane.
11. Find the points on the line through the points A (3, 2, 1), B (9, 0, 4) at a distance of 21 units from the mid-
points of AB.
→ → → →
12. Show that the points a , b , 3 a − 2 b are collinear.

→ → → → → → → →
13. Show that the points a − 2 b + 3 c , 2 a + 3 b − 4 c , − 7 b + 10 c are collinear.

14. Show that the points whose position vectors are given by 5 iˆ + 5 kˆ, 2 iˆ + ˆj + 3 kˆ and −4 iˆ + 3 ˆj − kˆ are
collinear.
[Hint. Type solved Ex. 14]
15. Prove that the points (1, 2, 3), (4, 0, 4), (−2, 4, 2), (7, −2, 5) are collinear.
[Hint. Let A, B, C, D be the given points. Find equations of AB and CD. Show that coordinates of C
and D satisfy the equation of AB.]

16. Show that the lines x + 3 = y + 5 = z − 7 ; x + 1 = y + 1 = z + 1 are intersecting.

2 3 −3 4 5 −1
x −1 y − 2 z − 3 x − 4 y −1
17. Find the point of intersection of the lines = = and = =z (CBSE)
2 3 4 5 2
 STUDYORB

→ →
18. Show that the lines r = iˆ + ˆj − kˆ + λ (3 iˆ − ˆj ) and r = 4iˆ − µ (2 iˆ + ˆj ) intersect. Find the point of
intersection P of the lines.
x−5 y−7 z+3 x−8 y −4 z −5
19. Show that the lines = = and = = intersect. Find their point of
4 4 −5 7 1 3
intersection. (CBSE 2002)
20. Prove that the line through A (0, −1, −1) and B (4, 5, 1) intersects the line through C (3, 9, 4) and D (− 4,
4, 4). Also find their point of intersection.
21. Show that the lines x − 1 = y + 1 = z − 1 and x − 2 = y − 1 = z + 1 do not intersect. (CBSE 2002)
3 2 5 4 3 −2
→ → → → → → → →
22. Show that the straight lines r = a + t ( b + c ) and r = b + s ( c + a ) intersect and find also the point
of intersection.
→ → → → → → → → → → →
23. Show that the lines →
r = − 3 a + 6 b + t (− 4 a + 3 b + 2c) and r = − 2 a + 7 c + s (− 4 a + b + c ) do not
intersect.

ANSWERS
→ x − 2 y +1 z − 4 →
1. r = 2 iˆ − ˆj + 4 kˆ + λ (iˆ + ˆj − 2 kˆ), = = 2. r = 3 iˆ − ˆj + 3 kˆ + λ (2iˆ − 2 ˆj + 5 kˆ)
1 1 −2
3. (i) r = 5 iˆ − 2 ˆj + 4 kˆ + λ ( 2iˆ − 2 ˆj + 3 kˆ), x − 5 = y + 2 = z − 4
→
2 −1 3
→
(ii) r = i + 2 j + 3 k + λ (i − 2 j + 3 k ), x − 1 y − 2 z − 3
ˆ ˆ ˆ ˆ ˆ ˆ = =
1 −2 3
(iii) x + 1 = y − 2 = z − 3
→
4. (i) r = 3iˆ + 4 ˆj − 7 kˆ + λ (−2 iˆ − 5 ˆj + 13 kˆ)
→ x −1 y − 2 z −1
(ii) r = (iˆ + 2 ˆj − kˆ) + λ ( iˆ − ˆj + 2 kˆ ), = =
1 −1 2
→ y−4
(iii) r = (− iˆ + 4 ˆj − 2 kˆ) + λ (iˆ − 2 ˆj + kˆ), x + 1 = =z+2
−2
→
5. r = 2 iˆ − ˆj + kˆ + λ (2 iˆ + 7 ˆj − 3 kˆ)
x −1 y +1 z − 2 → ˆ ˆ
6. = = , r = (i − j + 2 kˆ) + λ (iˆ + 2 ˆj − 2 kˆ)
1 2 −2
→ x−2 y +3 z −4
7. r = 2 iˆ − 3 ˆj + 4 kˆ + λ (iˆ − 13 ˆj + 17 kˆ ), = =
1 −13 17

8. (i) ( 54 , −53 , 0), r = (2iˆ + 52 ˆj − kˆ ) + λ (2iˆ − 32 ˆj + 0 . kˆ )

→

→ 1 1
(ii) 2, 1, - 6, r = − iˆ + ˆj + kˆ + λ (2 iˆ + ˆj − 6 kˆ)
3 3
10. (132 , −211, 0)
11. (24, − 5, 232 ), (−12, 7, 132 ) 17. (− 1, − 1, − 1) 18. (4, 0, − 1)
→ →
19. (1, 3, 2) 20. (10, 14, 4) 22. →
a+b+c
 STUDYORB

EXERCISE 5 (b)
1. Find the angle betweeen the following pairs of lines :
x + 4 y −1 z + 3 x +1 y − 4 z − 5 5 − x y + 3 1− z x 1− y z + 5
(i) = = ; = = (ii) = = ; = =
3 5 4 1 1 2 −2 1 3 3 −2 −1
x−2 y+3 x + 1 2y − 3 z − 5
(iii) = = z − 5; = = [Hint. Type solved Ex. 18 (v)]
3 −2 1 3 2
(iv) x= y= z ;x= y=z
1 0 −1 3 4 5
2. Find the angle between the following pairs of lines :
→ →
(i) r = 4 iˆ − ˆj + λ (iˆ + 2 ˆj − 2 kˆ); r = iˆ − ˆj + 2 kˆ − µ (2 iˆ + 4 ˆj − 4 kˆ)
→
[Hint. The first line is in the direction of b1 = iˆ + 2 ˆj − 2 kˆ and the second line is in the direction of
→
b2 = 2 iˆ + 4 ˆj − 4 kˆ .
→ →
Note. that we can treat − µ as a single parameter. Since b2 = 2 b1 , it follows that the given lines are
parallel to each other.]
→ →
(ii) r = 3iˆ + 2 ˆj − 4 kˆ + λ (iˆ + 2 ˆj + 2 kˆ); r = 5 ˆj − 2 kˆ + µ (3iˆ + 2 ˆj + 6 kˆ)
→ →
(iii) r = λ (iˆ + ˆj + 2 kˆ); r = 2 ˆj + µ [( 3 − 1) iˆ − ( 3 + 1) ˆj + 4 kˆ]
3. Find the angle between the pairs of lines with direction ratios :
(i) 2, 2, 1 and 4, 1, 8 (ii) 1, 2, − 2 and − 2, 2, 1.
4. If P, Q are (2, 3, − 6), (3, − 4, 5) then find the actue angle that OP makes with OQ.
[Hint. Direction ratios of OP and OQ are 2, − 0, 3 − 0, 6 − 0 and 3 − 0, 4 − 0, 5 − 0, i.e., 2, 3, − 6 and 3, −
4, 5 respectively.]
5. A, B, C are the points (1, 4, 2), (−2, 1, 2), (2, − 3, 4). Find the angles of the triangle ABC.
6. The coordinates of the angular points A, B, C, D of a tetrahedra are (− 2, 1, 3), (3, −1, 2), (2, 4, −1) and
(1, 2, 3) respectively. Find the angle between the edges AC and BD.
x−3 y−3 z
7. Find the equations of the two lines through the origin and which intersect the line = =
2 1 1
π
at angles of 3 .
[Hint. Any pt. on the given line is P (3 + 2t, 3 + t, t). D.r.’s of OP are 3 + 2t − 0, 3 + t − 0, t − 0, i.e., 3 +
2t, 3 + t, t.
π O (0, 0, 0)
Since OP makes angle of 3 with the given line, we have

cos π = 2 (3 + 2t) + 1 (3 + t ) + 1. t ⇒ t = − 1, − 2 P π/3 π/3 x−3 y−3 z

= =
3 2
6 (3 + 2t ) + (3 + t ) + t
2 2 2 1 1

Hence, the two possible positions of P are (1, 2, −1) and (−1, 1, −2).
x= y=z x y z
8. Prove that the two lines and = = are at right angles.
1 2 1 1 −1 1
9. P, Q, R, S are the points (−2, 3, 4), (−4, 4, 6), (4, 3, 5) and (0, 1, 2). Show that PQ is perpendicular to RS.
10. (i) Determine the equations of k so that the line joining the points A (k, −1, − 1), B (2, 0, 2k) is
perpendicular to the line joining the points C (4, 2k, 1) and D (2, 3, 2).
(ii) For what value of p will the line through (4, 1, 2) and (5, p, 0) be perpendicular to the line through
(2, 1, 1) and (3, 3, −1) ?
11. Find the values of of p and q so that the line joining the points (7, p, 2) and (q, −2, 5) may be parallel to
the line joining the points (2, −3, 5) and (−6, −15, 11).
12. Find the equations of the line passing through the point (2, 1, 3) and perpendicular to the lines
x −1 y − 2 z − 3 x = y=z
= = and .
1 2 3 −3 2 5
13. Find the equations of the line passing through the point A (−1, 3, −2) and perpendicular to the lines
→ →
r = λ (iˆ + ˆj + kˆ) and r = − 2 iˆ + ˆj − 5 kˆ + µ (− 3 iˆ + 2 ˆj + 5 kˆ).

14. Find the equation of the line passing through the point iˆ + ˆj − 3 kˆ and perpendicular to the lines
→ →
r = iˆ + λ (2 iˆ − ˆj − 3 kˆ) and r = (2 iˆ + ˆj − kˆ) + µ (iˆ + ˆj + kˆ) .
15. Find the equations of the line passing through the point (1, −1, 1) and perpendicular to the lines
joining the points (4, 3, 2), (1, −1, 0) and (1, 2, −1), (2, 1, 1).
 STUDYORB

16. Find the coordinates of the foot of the perpendicular from (1, 1, 1) on the line joining (5, 4, 4) and
(1, 4, 6).
x + 2 y −1 z − 3
17. Find the foot of the perpendicular from (0, 2, 7) on the line = = . (CBSE 1999)
−1 3 −2
18. A (0, 6, −9), B (−3, −6, 3) and C (7, 4, −1) are three points. Find the equation of the line AB. If D is the
foot of perpendicular drawn for C to the line AB, find coordinates of the point D. (CBSE 1998 C)
19. Find the distance of (−2, 1, 5) from the line through (2, 3, 5) whose d.c.’s are propotional to 2, −3, 6.
20. Find the equations and length of the perpendicular drawn from the point (2, 4, −1) to the line
y+3 z−6
x+5= = . (H.B. 1996, CBSE)
4 −9
4 − x y 1− z
21. Find the perp. distance of the pt. (2, 3, 4) from the line = = .
2 6 3
Also find the coordinates of the foot of the perpendicular. (CBSE 1995, Pb)
x − 1 y + 1 z + 10
22. Find the perp. distance of the pt. (1, 0, 0) from the line = = .
2 −3 8
Also find the coordinates of the foot of the perpendicular. (H.B. 1998, CBSE)
x−2 y−3 z−4
23. From the point P (1, 2, 3), PN is drawn perpendicular to the straight line = = . Find the
3 4 5
distance PN, the equations to PN and the coordinates of N.
24. Find the distance of A (1, −2, 3) from the line, PQ through P, (2, −3, 5), which makes equal angles with
the axes.
→
25. Find the length of the perpendicular drawn from the point (5, 4, −1) to the line r = iˆ + λ (2 iˆ + 9 ˆj + 5 kˆ) .
26. Find the (a) length, (b) equation and (c) foot of the perpendicular from the point A having position
→
vector iˆ + 6 ˆj + 3 kˆ to the line r = ( ˆj + 2 kˆ) + λ (iˆ + 2 ˆj + 3 kˆ) .
27. Find the image of the point
(i) (5, 9, 3) in the line x − 1 = y − 2 = z − 3 .
2
3 4
− −
x y 1 = z 2.
(ii) (3, 5, 3) in the line = (CBSE 2002 C)
1 2 3
→
28. Find the image of the point (2, −1, 5) in the line r = (11iˆ − 2 ˆj − 8 kˆ) + λ (10 iˆ − 4 ˆj − 11kˆ) .
29. (i) Find the coordinates of the foot of the perpendicular from the point (1, 8, 4) to the line joining
B (0, −1, 3) and C (2, −3, −1). (AICBSE 1995)
(ii) Find the point in which the line joining the points (−9, 4, 5) and (11, 0, − 1) is met by the perpendicular
from the origin. (Pb 1997 C)
30. Find the angle between the two lines whose direction cosines are given by the equations.
(a) l+m+n = 0 ...(i)
l +m −n = 0
2 2 2
...(ii)
[Hint. From (i), l = −m − n. Putting this value in (ii), and solving we get l = − n, 0, m = 0, − n]
∴ Direction numbers of the two lines are −n, 0, n and 0, −n, n, i.e., −1, 0, 1 and 0, −1, 1. If θ be the actue
angle between the lines, then
| − 1.0 + 0. (−1) + 1.1|
cos θ = = 1 ⇒ θ= π]
2 2
(−1) + 0 + 1
2 2
0 + (−1) + 1
2 2 2 3

(b) 2l − m + 2n = 0 (c) 3l + m + 5n = 0
mn + nl + lm = 0 6mn − 2nl + 5 lm = 0
31. Find the direction cosines of the four diagonals and the angle between two daigonals of a cube.
 STUDYORB

ANSWERS
−1 8 3 π
1. (a) cos −1 8 3 (ii) cos (iii) (iv) cos −1 1
15 15 2 5
−1 19 π π
2. (i) parallel (ii) cos (iii) 3. (i) cos −1 2 (ii) 2
21 3 3
4. (i) cos −1 18 2 5. 90º, cos−1  1  , cos −1  2  6. cos−1  3 
35  3  3  574 
p = −3
y
x= = z , x = = z y
7. 10. (i) k = 0 (ii)
1 2 −1 −1 1 −2 2
x − 2 y −1 z − 3
11. p = 4, q = 3 12. = =
2 −7 4
→ x + 1 y − 3 z + 2
13. r = − iˆ + 3 ˆj − 2 kˆ + λ (2 iˆ − 7 ˆj + 4 kˆ) , = =
2 −7 4
→ x − 1 y + 1 z −1
14. r = iˆ + ˆj + 3 kˆ + λ (4 iˆ − 5 ˆj + kˆ) 15. = = 16. (3, 4, 5)
10 −4 −7
17. ( − 3, − 1, 4
2 2 ) 18. x =
1
y−6 z+9
4
=
−4
; (−1, 2, −5) 19. 14
3

20. (−1, 1, −3) ;7,

x − 2 y − 4 z +1
6
=
3
=
2 7 (
21. 3 101 , 120 , 78 , 10
49 49 49 ) 22. 2 6 , (3, −4, −2)

5 7 1 −5 (
23. 1 3, x − 1 = y − 2 = z − 3 , 32 , 51 , 14
25 25 5 ) 24.
14
3
25.
2109
110
26. (a) 13 (b) iˆ + 3 ˆj + 5 kˆ (c) 6 iˆ + 3 k
+ λ (−3 ˆj + 2 kˆ)
27. (i) (1, 1, 11) (ii) (− 1, 1, 7) 28. (0, 5, 1) (3 3 3)
29. (i) −5 , 2 , 19

(ii) (1, 2, 2) 30. (a) π (b) 90º (c) cos ( 1 )

−1
3 6
31. 1 , 1 , 1 ; 1 , 1 , − 1 ; − 1 , 1 , 1 ; 1 , −1 , 1 ; cos −1 1
3 3 3 3 3 3 3 3 3 3 3 3 3 ()
 STUDYORB

EXERCISE 5 (c)
Find the shortest distance between the following pairs of lines whose equations are
→
1. r = iˆ + ˆj + λ (2 iˆ − ˆj + kˆ) and 2 iˆ + ˆj − kˆ + µ (3iˆ − 5 ˆj + 2 kˆ) (CBSE 1991, 90 C)
→ →
2. r = iˆ + 2 ˆj + kˆ + λ (iˆ − ˆj + kˆ) and r = 2 iˆ − ˆj + kˆ + µ (2 iˆ + ˆj + 2 kˆ) (CBSE 1994 C)
→ →
3. r = (4 iˆ − ˆj) + λ ( iˆ + 2 ˆj − 3 kˆ) and r = ( iˆ − ˆj + 2 kˆ) + µ (2 iˆ + 4 ˆj − 5 kˆ) (CBSE 1992)
→ →
4. r = (λ − 1) iˆ + (λ + 1) ĵ − (1 + λ) k̂ and r = (1 − µ) iˆ + (2µ − 1) ĵ + (µ + 2) k̂ . (CBSE 1997 C)
→ →
5. r = iˆ + 2 ˆj + 3 kˆ + λ (2 iˆ + 3 ˆj + 4 kˆ) and r = (2 iˆ + 4 ˆj + 5 kˆ) + µ (3 iˆ + 4 ˆj + 5 kˆ) (CBSE 1992)
x−2 y−4 z −5
6. x − 1 = y − 2 = z − 3 and = = (Pb 1991, NMOC)
2 3 4 3 4 5
x −1 y +1 x +1 y − 2
7. = = z and = ;z=2
2 3 3 1
[Hint. The second equations of the two given lines are
x −1 y +1 z − 0 x +1 y − 2 z − 2
= = ...(i) and = = ...(ii)]
2 3 1 3 1 0
x−5 y−4 z−4 x −1 y + 2 z + 4
8. = = and = =
1 −2 1 7 −6 1
9. Find the shortest distance between the lines through the points A (6, 2, 2) and A′ (−4, 0, −1) in the
directions 1, −2, 2 and 3, −2, −2 respectively. (J & K 1995 C)
→ →
10. Show that the lines r = (iˆ − ˆj) + λ (2 iˆ + kˆ) and r = (2 iˆ − ˆj) + µ (iˆ + ˆj − kˆ) do not intersect.
(CBSE 1994, H.B. 1992)
[Hint. Show that S.D. ≠ 0]
11. Show that the lines intersect and find their point of intersection.
y −2 z +3 − − −
(i) x = = and x 2 = y 6 = z 3
2 3 2 3 4
x−5 y−7 z+3 x −8 y −4 z −5
(ii) = = and = = (AICBSE 2002)
4 4 −5 7 1 3
12. By computing the shortest distance determine whether the following pairs of lines intersect or not:
→ →
(i) r = (iˆ + ˆj − kˆ) + λ (3 iˆ − ˆj) and r = (4 iˆ − kˆ) + µ (2 iˆ + 3 kˆ)

(ii) x − 5 = y − 7 = z + 3 , x − 8 = y − 7 = z − 5 (H.B. 1993 C)

4 −5 −5 7 1 3
→ →
(iii) r = (3 − t ) iˆ + (4 + 2t ) ˆj + (t − 2) kˆ r = (1 + s) iˆ + (3s − 7) ˆj + (2s − 2) kˆ (Pb 1998 C)
− + − x + 2 y −1 z +1
(iv) x 1 = y 1 = z 1 and = = (AICBSE 2002)
3 2 5 4 3 −2
13. Find the coordinates of the points where the line of shortest distance between the lines
x − 12 y − 1 z − 5 x − 23 9 y − 19 z − 25
= = and = = meets them.
−9 4 2 −6 −4 3
14. Find the shortest distance between the following pairs of parallel lines.
→ →
(i) r = (iˆ + 2 ˆj + 3 kˆ) + λ (iˆ − ˆj + kˆ) and r = (2 iˆ − ˆj − kˆ) + µ (− iˆ + ˆj − kˆ).
→ →
(ii) r = (iˆ + ˆj) + λ (2 iˆ − ˆj + kˆ) and r = (2 iˆ + ˆj − kˆ) + µ (4 iˆ − 2 ˆj + 2 kˆ).
15. Define the line of shortest distance between two skew lines. Find the shortest distance and the vector
equation of the line of shortest distance between the lines given by : [Hint. Type solved ex. 38]
→ →
(i) r = (8 + 3 λ) iˆ − (9 + 16 λ) ˆj + (10 + 7 λ) kˆ and r = 15 iˆ + 29 ˆj + 5 kˆ + µ (3 iˆ + 8 ˆj − 5 kˆ) (CBSE 1996)
→ →
(ii) r = (3 iˆ + 8 ˆj + 3 kˆ) + λ (3iˆ − ˆj + kˆ) and r = (−3 µ − 3) iˆ + (2 µ − 7) ˆj + (4 µ + 6) kˆ (CBSE 1999)
→ →
(iii) r = (2 ˆj − 3 kˆ) + λ (2 iˆ − ˆj) and r = (4 iˆ + 3 kˆ) + µ (3 iˆ + ˆj + kˆ) (CBSE 1998)
→ →
(iv) r = (− iˆ + ˆj + 9 kˆ) + λ (2 iˆ + ˆj − 3 kˆ) and r = (3 iˆ − 15 ˆj + 9 kˆ) + µ (2 iˆ − 7 ˆj + 5 kˆ) (CBSE 1999 C)
→ →
(v) r = (1 − λ) iˆ + (λ − 2) ˆj + (3 − 2 λ) kˆ and r = (µ + 1) iˆ + (2 µ − 1) ˆj − (2 µ + 1) kˆ
 STUDYORB

16. Find the length and the equation of the line of shortest distance between the lines
x−3 y −8 z −3 x+3 y+7 z−6
(i) = = and = = . Also find the points where the line of shortest
3 −1 1 −3 2 4
distance meets the given lines. [Hint. See solved Ex. 36]
x − 8 y + 9 z − 10 x − 15 y − 29 5 − z
(ii) = = and = = (Pb 1995)
3 −16 7 3 8 5

ANSWERS

1. 10 2. 1 3 2 3. 6 units 4. 5 5. 1
59 2 5 2 6

3
6. 1 7. 8. 2 29 9. 9 units 10. No., S.D. = 1
6 59 14
11. (i) (2, 6, 3) (ii) (1, 3, 2) 12. (i) yes, (ii) yes (iii) No. S.D. = 35 (iv) No

13. (3, 5, 7), (11, 11, 31) 14. (i) 78 (ii) 10

3 6
→
15. (i) S.D. = 14 units, r = (5 iˆ + 7 ˆj + 3 kˆ) + t ′ (2 iˆ + 3 ˆj + 6 kˆ)
→
(ii) 3 30 units; r = 3 iˆ + 8 ˆj + 3 kˆ + t (2 iˆ + 5 ˆj − kˆ)
→
(iii) 30 units ; r = 2 iˆ + ˆj − 3 kˆ + t (− iˆ − 2 ˆj + 5 kˆ)