Let the points on the plane P be equidistant from the points ($-$4, 2, 1) and (2, $-$2, 3). Then the acute angle between the plane P and the plane 2x + y + 3z = 1 is :
Solution
<p>Let P(x, y, z) be any point on plane P<sub>1</sub></p>
<p>Then $${(x + 4)^2} + {(y - 2)^2} + {(z - 1)^2} = {(x - 2)^2} + {(y + 2)^2} + {(z - 3)^2}$$</p>
<p>$\Rightarrow 12x - 8y + 4z + 4 = 0$</p>
<p>$\Rightarrow 3x - 2y + z + 1 = 0$</p>
<p>And ${P_2}:2x + y + 3z = 0$</p>
<p>$\therefore$ angle between P<sub>1</sub> and P<sub>2</sub></p>
<p>$$\cos \theta =\left| {{{6 - 2 + 3} \over {14}}} \right| \Rightarrow \theta = {\pi \over 3}$$</p>
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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