First slide

Planes in 3D

Question

$Angle between the two planes of which one plane is 4 x + y + 3 z = 0 and another plane$
$containing the lines \frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 1}{3}, \frac{x - 2}{3} = \frac{y - 3}{2} = \frac{z - 2}{3} is$

Moderate

A

$\frac{π}{3}$
B

$\frac{π}{2}$
C

$\frac{π}{6}$
D

$\frac{2 π}{3}$

Solution

$Angle between the planes \cos θ = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{\sum a_{1}^{2}} \sqrt{\sum a_{2}^{2}}}$
$\begin{array}{l} E q u a t i o n o f t h e p l a n e c o n t a i n i n g t h e l i n e s \\ \frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 1}{3}, \frac{x - 2}{3} = \frac{y - 3}{2} = \frac{z - 2}{3} i s \\ |\begin{matrix} x - 3 & y - 2 & z - 1 \\ 2 & 3 & 3 \\ 3 & 2 & 3 \end{matrix}| = 0 \\ \Rightarrow 3 (x - 3) + 3 (y - 2) - 5 (z - 1) = 0 \\ \Rightarrow 3 x + 3 y - 5 z - 10 = 0 \\ ∴ \cos θ = \frac{12 + 3 - 15}{\sqrt{22} \sqrt{22}} = 0 \\ \Rightarrow θ = \frac{π}{2} \end{array}$

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