Question

Show that the line joining the pair of points 6a - 4b + 4c, -4c and the line joining the
pair of points -a - 2b - 3c, a + 2b + 5c intersect at the point -4c when a, b, c are
non-coplanar vectors. (Mar-19)







Can anyone solve and give the solution in a proper way

Answer 1

Answer:

i wont cheat promise.

Step-by-step explanation:

B

C

D

A

Answer :

A

Solution :

Coordinate of points A and B are (6,-4,4) and (0,0,-4) and coordinate of points C and D are (-1,-2,-3) and (1,2,-5)

Now, equation of line passing through (0,0,-4) and (6,-4,4) is

x−06=y−0−4=z+44+4=k [say]

⇒x=k,=−4kandz=8k−4 . . .(i)

Again, equation of line passing through

(-1,-2,-3) and (1,2,-5) is

x+11+1=y+22+2=z+3−5+3

⇒ x+12=y+24=3+3−2 . . . (ii)

Since, two lines are intersect, therefore point

(6k,-4km8k-4) satisfy Eq. (ii), we get

⇒ 6k+1=−2k+1=−(8k−1)

∴ 6k+1=−2k+1

⇒ 8k=0⇒k=0

∴ x=6×0,y=−4×0andz=8×0−4

⇒x=0,y=0andz=−4, which is equal to the B coordinate. long work my dude

Answer 2

Question :

Show that the line joining the pair of points 6a - 4b + 4c, -4c and the line joining the pair of points -a - 2b - 3c, a + 2b - 5c intersect at the point -4c when a, b, c are non-coplanar vectors.

The Vector equation of Line passing through two points a, b is given by,

⇒ r = (1 - s) a + sb where s is a scalar.

The line joining the pair of points 6a - 4b + 4c, -4c is,

⇒r = (1 - s)( 6a - 4b + 4c) + s ( - 4c)

⇒ r = - s ( 6a - 4b + 8c) + 6a - 4b + 4c

⇒ r = (-6s + 6 ) a + (4s - 4 ) b + (- 8s + 4)c

The line joining the pair of points -a - 2b - 3c, a + 2b - 5c is,

⇒ r' = (1-t)(-a - 2b - 3c) + t (a + 2b - 5c)

⇒ r' = t ( 2a +4b - 2c) - ( a + 2b + 3c)

⇒r' = (2t - 1)a + (4t - 2)b + (-2t - 3)c

For the point of intersection,

⇒ r = r'

⇒(-6s + 6 ) a + (4s - 4 ) b + (- 8s + 4)c = (2t - 1)a + (4t - 2)b + (-2t-3)c

So,

⇒- 6s + 6 = 2t - 1 __(1)

⇒4s - 4 = 4t - 2 ___(2)

⇒- 8s + 4 = - 2t - 3 ___(3)

Equations :

⇒ 2t + 6s = 7

⇒4t - 4s = - 2

⇒8s - 2t = 7

Solving (1) & (2)

⇒4t + 12s = 14

⇒4t - 4s = - 2

Subtracting them gives,

⇒16s = 16

⇒s = 1

Value of t

⇒4t - 4 = - 2

⇒4t = 2

⇒t = 1/2

So the point of intersection is when s = 1 or t = 1/2

The point of intersection vector is,

⇒ (-6 + 6 ) a + (4 - 4 ) b + (- 8 + 4)c

⇒ 0a + 0b - 4c

⇒ - 4c

Therefore, The lines joining the above points intersect at - 4c