First slide

Planes in 3D

Question

A line having direction ratios $⟨ 3, 4, 5 ⟩$ cuts two planes $2 x - 3 y + 6 z - 12 = 0$ and $2 x - 3 y + 6 z + 2 = 0$ at point $P$ and at point $Q$ then the length of PQ

Difficult

A

$\frac{35 \sqrt{2}}{12}$
B

$\frac{35 \sqrt{2}}{24}$
C

$\frac{35 \sqrt{2}}{6}$
D

$\frac{35 \sqrt{2}}{8}$

Solution

If $θ$ is acute angle between the line having direction ratios $⟨ 3, 4, 5 ⟩$ and the planes given, and the perpendicular distance $P M$ from $P$ to the plane $2 x - 3 y + 6 z - 12 = 0$ then $\sin θ = \frac{P M}{P Q}$
Here
it implies that $\sin θ = \frac{a l + b m + c n}{\sqrt{a^{2} + b^{2} + c^{2}} \sqrt{l^{2} + m^{2} + n^{2}}}$ and $P M = \frac{a x_{1} + b y_{1} + c z_{1} + d}{\sqrt{a^{2} + b^{2} + c^{2}}}$
$\begin{matrix} \frac{3 (2) + 4 (- 3) + 5 (6)}{\sqrt{9 + 16 + 25} \cdot \sqrt{4 + 9 + 36}} = \frac{\frac{12 + 2}{\sqrt{4 + 9 + 36}}}{P Q} \\ \frac{24}{\sqrt{50} \sqrt{49}} = \frac{2}{P Q} \\ P Q = \frac{35 \sqrt{2}}{12} \end{matrix}$

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