If the mirror image of the point (1, 3, 5) with respect to the plane
4x $-$ 5y + 2z = 8 is ($\alpha$, $\beta$, $\gamma$), then 5($\alpha$ + $\beta$ + $\gamma$) equals :
Solution
Image of (1, 3, 5) in the plane 4x $-$ 5y + 2z = 8 is ($\alpha$, $\beta$, $\gamma$)<br><br>$$ \Rightarrow {{\alpha - 1} \over 4} = {{\beta - 3} \over { - 5}} = {{\gamma - 5} \over 2} = - {{(4(1) - 5(3) + 2(5) - 8)} \over {{4^2} + {5^2} + {2^2}}} = {2 \over 5}$$<br><br>$\therefore$ $\alpha = 1 + 4\left( {{2 \over 5}} \right) = {{13} \over 5}$<br><br>$\beta = 3 - 5\left( {{2 \over 5}} \right) = 1 = {5 \over 5}$<br><br>$\gamma = 5 + 2\left( {{2 \over 5}} \right) = {{29} \over 5}$<br><br>Thus, $$5(\alpha + \beta + \gamma ) = 5\left( {{{13} \over 5} + {5 \over 5} + {{29} \over 5}} \right) = 47$$
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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