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 $F=\left(a,a,a\right)$