Let the equation of the plane, that passes through the point (1, 4, $-$3) and contains the line of intersection of the
planes 3x $-$ 2y + 4z $-$ 7 = 0
and x + 5y $-$ 2z + 9 = 0, be
$\alpha$x + $\beta$y + $\gamma$z + 3 = 0, then $\alpha$ + $\beta$ + $\gamma$ is equal to :
Solution
3x $-$ 2y + 4z $-$ 7 + $\lambda$(x + 5y $-$ 2z + 9) = 0<br><br>(3 + $\lambda$)x + (5$\lambda$ $-$ 2)y + (4 $-$ 2$\lambda$)z + 9$\lambda$ $-$ 7 = 0<br><br>passing through (1, 4, $-$3)<br><br>$\Rightarrow$ 3 + $\lambda$ + 20$\lambda$ $-$ 8 $-$ 12 + 6$\lambda$ + 9$\lambda$ $-$ 7 = 0<br><br>$\Rightarrow$ $\lambda$ = ${2 \over 3}$<br><br>$\Rightarrow$ equation of plane is<br><br>$-$11x $-$ 4y $-$ 8z + 3 = 0<br><br>$\Rightarrow$ $\alpha$ + $\beta$ + $\gamma$ = $-$23
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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