Let P be a plane containing the line ${{x - 1} \over 3} = {{y + 6} \over 4} = {{z + 5} \over 2}$ and parallel to the line ${{x - 1} \over 4} = {{y - 2} \over { - 3}} = {{z + 5} \over 7}$. If the point (1, $-$1, $\alpha$) lies on the plane P, then the value of |5$\alpha$| is equal to ____________.
Answer (integer)
38
Solution
<p>Equation of required plane is $$\left| {\matrix{
{x - 1} & {y + 6} & {z + 5} \cr
3 & 4 & 2 \cr
4 & { - 3} & 7 \cr
} } \right| = 0$$</p>
<p>Since, (1, $-$1, $\alpha$) lies on it,</p>
<p>So, replace x by 1, y by ($-$1) and z and $\alpha$.</p>
<p>$$\left| {\matrix{
0 & 5 & {\alpha + 5} \cr
3 & 4 & 2 \cr
4 & { - 3} & 7 \cr
} } \right| = 0$$</p>
<p>$\Rightarrow 5\alpha + 38 = 0 \Rightarrow 5\alpha = - 38$</p>
<p>$\therefore$ $\left| {5\alpha } \right| = \left| { - 38} \right| = 38$</p>
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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