Let $${P_1}:\overrightarrow r \,.\,\left( {2\widehat i + \widehat j - 3\widehat k} \right) = 4$$ be a plane. Let P2 be another plane which passes through the points (2, $-$3, 2), (2, $-$2, $-$3) and (1, $-$4, 2). If the direction ratios of the line of intersection of P1 and P2 be 16, $\alpha$, $\beta$, then the value of $\alpha$ + $\beta$ is equal to ________________.
Answer (integer)
28
Solution
<p>Direction ratio of normal to ${P_1} \equiv < 2,1, - 3 >$</p>
<p>and that of $${P_2} \equiv \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
0 & 1 & { - 5} \cr
{ - 1} & { - 2} & 5 \cr
} } \right| = - 5\widehat i - \widehat j( - 5) + \widehat k(1)$$</p>
<p>i.e. $< - 5,5,1 >$</p>
<p>d.r's of line of intersection are along vector</p>
<p>$$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
2 & 1 & { - 3} \cr
{ - 5} & 5 & 1 \cr
} } \right| = \widehat i(16) - \widehat j( - 13) + \widehat k(15)$$</p>
<p>i.e. $< 16,13,15 >$</p>
<p>$\therefore$ $\alpha + \beta = 13 + 15 = 28$</p>
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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