First slide

Planes in 3D

Question

A line having direction ratios $⟨ 1, 0, 3 ⟩$ cuts planes $x - y + 6 z = 6$ and $x - y + 6 z = 4$ at $P$ and $Q$ then $P Q .$

Moderate

A

$3 \sqrt{10}$
B

$\frac{2 \sqrt{10}}{19}$
C

$\sqrt{10}$
D

$\frac{3 \sqrt{10}}{19}$

Solution

Suppose that $θ$ be the acute angle between the line whose direction ratios are $⟨ 1, 0, 3 ⟩$ and the plane $x - y + 6 z = 6$
hence,
$\begin{matrix} \sin θ = \frac{a l + b m + c n}{\sqrt{a^{2} + b^{2} + c^{2}} \sqrt{l^{2} + m^{2} + n^{2}}} \\ = \frac{1 + 18}{\sqrt{1 + 9} \sqrt{1 + 1 + 36}} \\ = \frac{19}{\sqrt{10} \sqrt{38}} \end{matrix}$
and $\sin θ = \frac{P M}{P Q}$ where $P M$ is perpendicular distance from the point to the plane, or it is equal to the distance between two parallel planes.
hence,
$\begin{matrix} P M = \frac{c_{2} - c_{1}}{\sqrt{a^{2} + b^{2}}} \\ = \frac{2}{\sqrt{1 + 1 + 36}} \\ = \frac{2}{\sqrt{38}} \end{matrix}$
Therefore,
$\begin{matrix} P Q = \frac{P M}{\sin θ} \\ = \frac{2}{\sqrt{38}} \frac{\sqrt{10} \sqrt{38}}{19} \\ = \frac{2 \sqrt{10}}{19} \end{matrix}$

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