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 point 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)

Any point on the line $\frac{x}{1}=\frac{y+a}{1}=\frac{z}{1}=t_1$ (say) is $\left(t_1,t_1-a,t_1\right)$ and any point on the line
$\frac{x+a}{2}=\frac{y}{1}=\frac{z}{1}=t_2$ (say) is $\left(2t_2-a,t_2,t_2\right)$.
Now direction cosine of the lines intersecting the above lines is proportional to
$\left(2t_2-a-t_1,t_2-t_1+a,t_2-t_1\right)$.
Hence $2t_2-a-t_1=2k,t_2-t_1+a=k$ and $t_2-t_1=2k$
On solving these, we get $t_1=3a,t_2=a$.
Hence points are $\left(3a,2a,3a\right)$ and $\left(a,a,a\right).$