Question

A line with direction cosines proportional to $2, 1, 2$ meets each of the lines $x = y + a = z$ and $x + a = 2y = 2z$. The co-ordinates of each of the points of intersection are given by :

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

Answer 1

Answer: (B)

Let the equation of line $AB$ be
$\frac{x-0}{1}=\frac{y+a}{1}=\frac{z-0}{1}=k$ (say)
$\therefore $ co-ordinate of $E$ is $(k, k - a, k)$. Also the equation of other line $CD$ is
$\frac{x+a}{2}=\frac{y-0}{1}-\frac{z-0}{1}=\lambda$ (say)
$\therefore $ co-ordinate of $F$ is $\left(2\lambda-a, \lambda, \lambda\right)$ Direction Ratio of $EF$ are $\left(k-2\lambda+a\right)$, $\left(k-\lambda-a\right)$, $\left(k-\lambda\right)$
$\therefore \frac{k-2\lambda+a}{2}=\frac{k-\lambda-a}{1}=\frac{k-\lambda}{2}$
On solving first and second fraction
$\frac{k-2\lambda+a}{2}=\frac{k-\lambda-a}{1}$
$k-2\lambda+a=2k-2\lambda-2a$
$\Rightarrow k=3a$
On solving second and third fraction
$\frac{k-\lambda-a}{1}=\frac{k-\lambda}{2}$
$2k-2\lambda-2a=k-\lambda$
$k-\lambda=2a$
$\lambda=k-2a=3a-2a$
$\Rightarrow \lambda=a$
$\therefore $ co-ordinate of $E \left(3a, 2a, 3a\right)$ and co-ordinate of $F=\left(a, a, a\right)$