Question

`vec"AB" = 3hat"i" - hat"j" + hat"k"` and `vec"CD" = -3hat"i" + 2hat"j" + 4hat"k"` are two vectors. The position vectors of the points A and C are `6hat"i" + 7hat"j" + 4hat"k"` and `-9hat"j" + 2hat"k"`, respectively. Find the position vector of a point P on the line AB and a point Q on the line Cd such that `vec"PQ"` is perpendicular to `vec"AB"` and `vec"CD"` both.

Answer 1

Position vector of A is `6hat"i" + 7hat"j" + 4hat"k"` and `vec"AB" = 3hat"i" - hat"j" + hat"k"`

So, equation of any line passing through A and parallel to `vec"AB"`

`vec"r" = (6hat"i" + 7hat"j" + 4hat"k") + lambda(3hat"i" - hat"j" + hat"k")`  .....(i)

Now any point P on `vec"AB" = (6 + 3lambda, 7 - lambda, 4 + lambda)`

Similarly, position vector of C is `-9hat"j" + 2hat"k"`

And `vec"CD" = -3hat"i" + 2hat"j" + 4hat"k"`

So, equation of any line passing through C and parallel to CD is

`vec"r" = (-9hat"j" + 2hat"k") + mu(-3hat"i" + 2hat"j" + 4hat"k")`  .....(ii)

Any point Q on `vec"CD" = (-3mu, -9 + 2mu, 2 + 4mu)`

d’ratios of PQ are `(6 + 3lambda + 3mu, 7 - lambda + 9 - 2mu, 4 + lambda - 2 - 4mu)`

⇒ `(6 + 3lambda + 3mu), (16 - lambda - 2mu), (2 + lambda - 4mu)`

Now `vec"PQ"` is ⊥ to equation (i), then

3(6 + 3λ + 3m) – 1(16 – λ – 2m) + 1(2 + λ – 4m) = 0

⇒ 18 + 9λ + 9m – 16 + λ + 2m + 2 + λ – 4m = 0

⇒ 11λ + 7m + 4 = 0  .....(iii)

`vec"PQ"` is also ⊥ to equation (ii), then

`-3(6 + 3lambda + 3mu) + 2(16 - lambda - 2mu) + 4(2 + lambda - 4mu)` = 0

⇒ – 18 – 9λ – 9m + 32 – 2λ – 4m + 8 + 4λ – 16m = 0

⇒ – 7λ – 29m + 22 = 0

⇒ 7λ + 29m – 22 = 0  ......(iv)

Solving equation (iii) and (iv) we get

77λ +   49μ +  28 = 0
77λ + 319μ – 242 = 0
(–)       (–)    (+)            
       – 270μ  + 270 = 0

∴ μ = 1

Now using μ = 1 in equation (iv) we get

7λ + 29 – 22 = 0

⇒ λ = – 1

∴ Position vector of P = [6 + 3(– 1), 7 + 1, 4 – 1]

= (3, 8, 3)

And position vector of Q = [– 3(1), –9 + 2(1), 2 + 4(1)]

= (– 3, –7, 6)

Hence, the position vectors of P = `3hat"i" + 8hat"j" + 3hat"k"` and Q = `-3hat"i" - 7hat"j" + 6hat"k"`