Let $\alpha x+\beta y+\gamma z=1$ be the equation of a plane passing through the point $(3,-2,5)$ and perpendicular to the line joining the points $(1,2,3)$ and $(-2,3,5)$. Then the value of $\alpha \beta y$ is equal to _____________.
Answer (integer)
6
Solution
Plane :
<br/><br/>$a(x-3)+b(y+2)+c(z-5)=0$
<br/><br/>Dr's of plane : $3 \hat{i}-\hat{j}-2 \hat{k}$
<br/><br/>$$
\begin{aligned}
& <3,-1,-2> \\\\
& P: 3(x-3)-1(y+2)-2(z-5)=0 \\\\
& 3 x-9-y-2-2 z+10=0 \\\\
& 3 x-y-2 z=1 \\\\
& \therefore \alpha=3, \beta=-1, \gamma=-2 \\\\
& \Rightarrow \alpha \beta \gamma=6
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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