Consider the planes 3x - 6y + 2z + 5 = 0 and 4x - 12y + 3z = 3. The plane 67x - 162y + 47z + 44 = 0 bisects the angle between the planes which.

contains origin

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is right angle

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is obtuse

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acute angle

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Solution

The correct option is D acute angle
For $3 x - 6 y + 2 z + 5 = 0$ and $- 4 x + 12 y - 3 z + 3 = 0$ bisector are
$\frac{3 x - 6 y + 2 z + 5}{\sqrt{9 + 36 + 4}} = \pm \frac{- 4 x + 12 y - 3 z + 3}{\sqrt{16 + 144 + 9}}$
The plane which bisects the angle between the plane that contains the origin
$13 (3 x - 6 y + 2 z + 5) = 7 (- 4 x + 12 y - 3 z + 3)$
$\Rightarrow 39 x - 78 y + 26 z + 65 + 28 x - 84 y + 21 z - 21 = 0$
$\Rightarrow 67 x - 162 y + 47 z + 44 = 0$
Further $3 \times (- 4) + (- 6) \times 12 + 2 \times (- 3) < 0$
Hence, the origin lies in the acute angle.

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3 x - 6 y + 2 z + 5 = 0

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