Let the plane passing through the point ($-$1, 0, $-$2) and perpendicular to each of the planes 2x + y $-$ z = 2 and x $-$ y $-$ z = 3 be ax + by + cz + 8 = 0. Then the value of a + b + c is equal to :
Solution
Normal of required plane $$\left( {2\widehat i + \widehat j - \widehat k} \right) \times \left( {\widehat i - \widehat j - \widehat k} \right)$$<br><br>$= - 2\widehat i + \widehat j - 3\widehat k$<br><br>Equation of plane <br><br>$- 2(x + 1) + 1(y - 0) - 3(z + 2) = 0$<br><br>$- 2x + y - 3z - 8 = 0$<br><br>$2x - y + 3z + 8 = 0$<br><br>$a + b + c = 4$
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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