Let the foot of the perpendicular from the point (1, 2, 4) on the line ${{x + 2} \over 4} = {{y - 1} \over 2} = {{z + 1} \over 3}$ be P. Then the distance of P from the plane $3x + 4y + 12z + 23 = 0$ is :
Solution
<p>$L:{{x + 2} \over 4} = {{y - 1} \over 2} = {{z + 1} \over 3} = t$</p>
<p>Let P = (4t $-$ 2, 2t + 1, 3t $-$ 1)</p>
<p>$\because$ P is the foot of perpendicular of (1, 2, 4)</p>
<p>$\therefore$ $4(4t - 3) + 2(2t - 1) + 3(3t - 5) = 0$</p>
<p>$\Rightarrow 29t = 29 \Rightarrow t = 1$</p>
<p>$\therefore$ P = (2, 3, 2)</p>
<p>Now, distance of P from the plane</p>
<p>$3x + 4y + 12z + 23 = 0$, is</p>
<p>$$\left| {{{6 + 12 + 24 + 23} \over {\sqrt {9 + 16 + 144} }}} \right| = {{65} \over {13}} = 5$$</p>
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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