Search any question & find its solution
Question:
Answered & Verified by Expert
Let the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3 y-\alpha z+\beta=0$. Then $(\alpha, \beta)$ equals
Options:
Solution:
1158 Upvotes
Verified Answer
The correct answer is:
$(-6,7)$
$(-6,7)$
Dr's of line $=(3,-5,2)$
Dr's of normal to the plane $=(1,3,-\alpha)$
Line is perpendicular to normal $\Rightarrow 3(1)-5(3)+2(-\alpha)=0 \Rightarrow 3-15-2 \alpha=0 \Rightarrow 2 \alpha=-12 \Rightarrow \alpha=-6$
Also $(2,1,-2)$ lies on the plane
$$
\begin{aligned}
& 2+3+6(-2)+\beta=0 \Rightarrow \beta=7 \\
& \therefore(\alpha, \beta)=(-6,7)
\end{aligned}
$$
Dr's of normal to the plane $=(1,3,-\alpha)$
Line is perpendicular to normal $\Rightarrow 3(1)-5(3)+2(-\alpha)=0 \Rightarrow 3-15-2 \alpha=0 \Rightarrow 2 \alpha=-12 \Rightarrow \alpha=-6$
Also $(2,1,-2)$ lies on the plane
$$
\begin{aligned}
& 2+3+6(-2)+\beta=0 \Rightarrow \beta=7 \\
& \therefore(\alpha, \beta)=(-6,7)
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.