Question

A line with direction ratio proportional to $2,1,2$ meets each of the lines $x=y+a=z$ and $x+a=2y=2z$. The coordinates of each of the point of intersection are given by

• A. $(3a,3a,3a),(a,a,a)$
• B. $(3a,2a,3a),(a,a,a)$
• C. $(3a,3a,3a),(a,a,2a)$
• D. $(2a,3a,3a),(2a,a,a)$

Answer 1

The correct option is B $(3a,2a,3a),(a,a,a)$

Let the equation of line $AB$ be

$\frac{x-0}{1}=\frac{y+a}{1}=\frac{z-0}{1}=k$

$\therefore$ Coordinates of $E$ are $(k,k-a,k)$

Also, the equation of other line $CD$ is

$\frac{x+a}{2}=\frac{y-0}{1}=\frac{z-0}{1}=\lambda$

$\therefore$ Coordinates of $F$ are $(2\lambda -a,\lambda ,\lambda )$

Direction ratio of $FE$ are $\left\{\right(k-2\lambda +a),(k-\lambda -a),(k-\lambda \left)\right\}$

$\therefore \frac{k-2\lambda +a}{\left(i\right)}=\frac{k-\lambda -a}{\left(ii\right)}=\frac{k-\lambda }{\left(iii\right)}$

From 1st and 2nd term,

$k-2\lambda +a=2k-2\lambda -2a$

$k=3a$

and from 2nd and 3rd term,

$2k-2\lambda -2a=k-\lambda$

$\Rightarrow \lambda =k-2a=3a-2a$

$\therefore \lambda =a$

$\therefore$ Coordination of $E=(3a,2a,3a)$

and coordination of $F=(a,a,a)$