First slide

Straight lines in 3D

Question

If lines x =y= z and $x = \frac{y}{2} = \frac{z}{3}$ and third through passing through (1, 1,1) form a triangle of area $\sqrt{6}$ uni1s, then the point of intersection of third line with the second line will be

Moderate

A

(1,2,3)
B

(2,4,6)
C

$(\frac{4}{3}, \frac{8}{3}, \frac{12}{3})$
D

none of these

Solution

Let any point on the second line be $(λ, 2 λ, 3 λ)$
$\begin{array}{l} \cos θ = \frac{6}{\sqrt{42}}, \sin θ = \frac{\sqrt{6}}{\sqrt{42}} \\ Δ_{OAB} = \frac{1}{2} (OA) OBsin θ \\ \Rightarrow \frac{1}{2} \sqrt{3} λ \sqrt{14} \times \frac{\sqrt{6}}{\sqrt{42}} = \sqrt{6} \\ or λ = 2 \end{array}$
So B is (2, 4,6)

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