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  • Let the line <img alt="\ell: \mathrm{x}=\frac{1-\mathrm{y}}{-2}=\frac{\mathrm{z}-3}{\lambda}, \lambda \in \mathrm{R}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cell%3A%20%5Cmathrm%7Bx%

Let the line \ell: \mathrm{x}=\frac{1-\mathrm{y}}{-2}=\frac{\mathrm{z}-3}{\lambda}, \lambda \in \mathrm{R} meet the plane \mathrm{P}: \mathrm{x}+2 \mathrm{y}+3 \mathrm{z}=4 at the point (\alpha, \beta, \gamma).If the angle between the line \ell and the plane p is \cos ^{-1}\left(\sqrt{\frac{5}{14}}\right)$, then $\alpha+2 \beta+6 \gamma  is equal to_________.

Option: 1

11


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\ell: \mathrm{x}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-3}{\lambda}, \lambda \in \mathbb{R}
Dr's of line \ell(1,2, \lambda)

Dr's of normal vector of plane P: x+2 y+3 z=4
are (1,2,3)

Now, angle between line\ell and plane \mathrm{P}  is given by

\sin \theta=\left|\frac{1+4+3 \lambda}{\sqrt{5+\lambda^{2}} \cdot \sqrt{14}}\right|=\frac{3}{\sqrt{14}}\left(\right.$ given $\left.\cos \theta=\sqrt{\frac{5}{14}}\right)
\Rightarrow \lambda=\frac{2}{3}
Let variable point on line \ell is \left(\mathrm{t}, 2 \mathrm{t}+1, \frac{2}{3} \mathrm{t}+3\right)
line of plane \mathrm{P}.

\Rightarrow\left(-1,-1, \frac{7}{3}\right) \equiv(\alpha, \beta, \gamma)
\Rightarrow \alpha+2 \beta+6 \gamma=11
 

Posted by

SANGALDEEP SINGH

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