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 $\left(a, a, a\right).$