Question

Answer this Question of iit jee . Chapter → Three dimensional Geometry​

Answer 1

Answer:

option A) (3a, 3a, 3a,) ;(a, a, a)

Answer 2

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

Given equation of line

$\rm :\longmapsto\:x = y + a = z$

can be rewritten as

$\rm :\longmapsto\:\dfrac{x}{1} = \dfrac{y + a}{1} = \dfrac{z}{1} = k$

So, any point on this line, say P is given as

Coordinates of P = ( k, k - a, k ) -------(1)

Now, second equation of line is

$\rm :\longmapsto\:x + a = 2y = 2z$

can be rewritten as

$\rm :\longmapsto\:\dfrac{x + a}{2} = \dfrac{y}{1} = \dfrac{z}{1} = p$

So, any point on this line, say Q is given by

Coordinates of Q = ( 2p - a, p, p ) ------(2)

So,

Direction ratios of line joining PQ is evaluated as

= ( 2p - a - k, p - k + a, p - k )

Now, it is given that the direction ratios of PQ are proportional with ( 2, 1, 2 ).

So,

$\rm :\longmapsto\:\dfrac{2p - a - k}{2} = \dfrac{p - k + a}{1} = \dfrac{p - k}{2}$

Taking first and second member, we get

$\rm :\longmapsto\:\dfrac{2p - a - k}{2} = \dfrac{p - k + a}{1}$

$\rm :\longmapsto\:2p - a - k= 2p - 2k + 2a$

$\bf\implies \:k = 3a - - - (3)$

Now, taking second and third member, we get

$\rm :\longmapsto\:\dfrac{p - k + a}{1} = \dfrac{p - k}{2}$

$\rm :\longmapsto\:2p - 2k + 2a = p - k$

$\bf :\implies\:p = k - 2a = 3a - 2a = a$

Now,

Substituting the value of k = 3a in equation (1), we get

Coordinates of P = ( k, k - a, k ) = (3a, 2a, 3a)

and

Substituting the value of p = a in equation (2), we get

Coordinates of Q = ( 2p - a, p, p ) = (a, a, a)

Hence,

• Option (b) is correct