Question

Solve previous year Question of iit jee
Chapter:- Vectors​

Answer 1

$\large\underline{\sf{Solution-}}$

The given equation of lines are,

$\rm :\longmapsto\:x = ay + b \: and \: z = cy + d - - - (1)$

and

$\rm :\longmapsto\:x = a'z + b' \: and \: y = c'z+ d' - - - (2)$

Now, Equation (1) can be rewritten as

$\rm :\longmapsto\:y = \dfrac{x - b}{a} \: \: and \: y = \dfrac{z - d}{c}$

So, it can further reduced to

$\rm :\longmapsto\:\dfrac{x - b}{a} \: = \:\: y = \dfrac{z - d}{c}$

$\rm :\longmapsto\:\dfrac{x - b}{a} \: = \:\: \dfrac{y - 0}{1} \: = \: \dfrac{z - d}{c}$

So, Direction ratios of this line is

$\rm :\longmapsto\:d. {r}'s \: = (a, \: 1, \: c)$

Now,

Equation (2) can be rewritten as

$\rm :\longmapsto\:x = a'z + b' \: and \: y = c'z+ d'$

$\rm :\longmapsto\:z = \dfrac{x - b'}{a'} \: \: and \: \: z = \dfrac{y - d'}{c'}$

So, it can further reduced to

$\rm :\longmapsto\: \dfrac{x - b'}{a'} \: = \: z \:= \dfrac{y - d'}{c'}$

$\rm :\longmapsto\: \dfrac{x - b'}{a'} \:= \: \dfrac{y - d'}{c'} \: = z$

$\rm :\longmapsto\: \dfrac{x - b'}{a'} \:= \: \dfrac{y - d'}{c'} \: = \dfrac{z - 0}{1}$

So, Direction ratios of this line is

$\rm :\longmapsto\:d. {r}'s \: = (a', \: c', \: 1)$

Since, it is given that Equation (1) and Equation (2) are perpendicular to each other.

Therefore, .

$\rm :\longmapsto\:a.a' + 1.c' + c.1 = 0$

$\bf :\longmapsto\:aa' + c' + c = 0$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underbrace{ \boxed{ \bf{ \: Option \: (c) \: is \: correct}}}$

Additional Information :-

Let us consider two lines

$\rm :\longmapsto\: \vec{r} \: = \: \vec{a} \: + \: \alpha \: \vec{b} \:$

and

$\rm :\longmapsto\: \vec{r} \: = \: \vec{c} \: + \: \beta \: \vec{d} \:$

Then,

1. Two lines are perpendicular iff

$\red{\rm :\longmapsto\:\vec{b} \:. \: \vec{d} \: = \: 0}$

2. Two lines are parallel iff

$\red{\rm :\longmapsto\:\vec{b} \: = k \: \vec{d} \:}$

or

$\red{\rm :\longmapsto\:\vec{b} \: \times \: \vec{d} \: = \: 0}$

3. Angle 'p' between 2 lines are given by

$\red{\rm :\longmapsto\:cosp \: = \: \dfrac{\vec{b} \:. \: \vec{d} \:}{ |\vec{b} \:| \: |\vec{d} \:| }}$