Suppose, the line ${{x - 2} \over \alpha } = {{y - 2} \over { - 5}} = {{z + 2} \over 2}$ lies on the plane $x + 3y - 2z + \beta = 0$. Then $(\alpha + \beta )$ is equal to _______.
Answer (integer)
7
Solution
<p>Given equation of line</p>
<p>${{x - 2} \over \alpha } = {{y - 2} \over { - 5}} = {{z + 2} \over 2}$ ...... (i)</p>
<p>and plane x + 3y $-$ 2z + $\beta$ = 0 ...... (ii)</p>
<p>Line (i) passes through (2, 2, $-$2)</p>
<p>which lies on plane (ii).</p>
<p>$\therefore$ 2 + 6 + 4 + $\beta$ = 0 $\Rightarrow$ $\beta$ = $-$ 12</p>
<p>Also, given line is perpendicular to normal of the plane</p>
<p>$\alpha$(1) $-$ 5(3) + 2($-$2) = 0 $\Rightarrow$ $\alpha$ = 19</p>
<p>$\therefore$ $\alpha$ + $\beta$ = 19 + (-12) = 19 - 12 = 7</p>
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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