Question

A line with direction cosines proportional to $2,1,2$ meets each of the line $x=y+a=z$ 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)

$(3a,2a,3a),(a,a,a)$

Let the equation of line $AB$ is $\frac{x-0}{1}=\frac{y+a}{1}=\frac{z-0}{1}=k$ (let)

Therefore coordinate of $E$ is $(k,k-a,k)$

Also the equation of other line $CD$ is $\frac{x+a}{2}=\frac{y-0}{1}=\frac{z-0}{1}=\lambda$ (let)

Therefore coordinate of $F$ is $(2\lambda -a,\lambda ,\lambda )$.

Direction ratio of $EF$ are $(k-2\lambda +a),(k-\lambda -a),(k-\lambda )$

$\therefore \frac{k-2\lambda +a}{2}=\frac{k-\lambda -a}{1}=\frac{k-\lambda }{2}$

On solving first and second dfraction, we get

$\frac{k-2\lambda +a}{2}=\frac{k-\lambda -a}{1}$

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

On solving second and third dfraction, we get

$\frac{k-\lambda -a}{1}=\frac{k-\lambda }{2}2k-2\lambda -2a=k-\lambda k-\lambda =2a\lambda =k-2a=3a-2a\lambda =a$

Therefore coordinate of $E=(3a,2a,3a)$ and $F=(a,a,a)$