A plane passing through the point (3, 1, 1) contains two lines whose direction ratios are 1, –2, 2 and 2, 3, –1 respectively. If this plane also passes through the point ($\alpha$, –3, 5), then $\alpha$ is equal to:
Solution
As normal is perpendicular to both the lines so normal vector to the plane is<br><br>
$$\overrightarrow n = \left( {\widehat i - 2\widehat j + 2\widehat k} \right) \times \left( {2\widehat i + 3\widehat j - \widehat k} \right)$$<br><br>
$$\overrightarrow n = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & { - 2} & 2 \cr
2 & 3 & { - 1} \cr
} } \right|$$<br><br>
$$\overrightarrow n = \left( {2 - 6} \right)\widehat i - \left( { - 1 - 4} \right)\widehat j + \left( {3 + 4} \right)\widehat k$$<br><br>
$\overrightarrow n = - 4\widehat i + 5\widehat j + 7\widehat k$<br><br>
Now equation of plane passing through (3,1,1) is<br><br>
$\Rightarrow$ –4(x – 3) + 5(y – 1) + 7(z – 1) = 0<br><br>
$\Rightarrow$ –4x + 12 + 5y – 5 + 7z – 7 = 0<br><br>
$\Rightarrow$ –4x + 5y + 7z = 0 ...(1)<br><br>
Plane is also passing through ($\alpha$, –3, 5) so this point satisfies the equation of plane so put in equation (1)<br><br>
–4$\alpha$ + 5 × (–3) + 7 × (5) = 0<br><br>
$\Rightarrow$ –4$\alpha$ – 15 + 35 = 0<br><br>
$\Rightarrow\alpha$ = 5
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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