First slide

Straight lines in 3D

Question

A line with direction ratios $〈2, 1, 2〉$ intersect each of the lines $x = y + a = z$ and $x + a = 2 y = 2 z$ the co-ordinates of each of the points of intersection are

Moderate

A

$(3 a, 3 a, 3 a), (a, a, a)$
B

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

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

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

Solution

The given lines are $\frac{x}{1} = \frac{y + a}{1} = \frac{z}{1}, \frac{x + a}{2} = \frac{y}{1} = \frac{z}{1}$
Suppose that the line $L$ intersect the above two lines at $P, Q$
The direction rations of $P Q$ are $〈2, 1, 2〉$
The points $P (s, s - a, s)$ and $Q (2 t - a, t, t)$ , these are general points on the given two lines.
Direction ratios of $P Q$ are $〈2 t - a - s, t - s + a, t - s〉$ , these are proportional to $〈2, 1, 2〉$
It implies that $\frac{2 t - a - s}{2} = \frac{t - s + a}{1} = \frac{t - s}{2}$
Hence, $s = 3 a, t = a$
Therefore, the points of intersections are $P (3 a, 2 a, 3 a), Q (a, a, a)$

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