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  • Let a line  pass through the point <img alt="P(2,3,1)" src="https://entrancecorner.oncodecogs.com/gif.latex?P%282%2C3%2C1

Let a line L pass through the point P(2,3,1) and be parallel to the line x+3 y-2 z-2=0=x-y+2z. If the distance of L from the point (5,3,8) is \alpha, then 3 \alpha^{2} is equal to_______. 

Option: 1

158


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

The Direction ratio of line
\left|\begin{array}{ccc} \mathrm{i} & \mathrm{j} & \mathrm{k} \\ 1 & 3 & -2 \\ 1 & -1 & 2 \end{array}\right|=\mathrm{i}(6-2)-\mathrm{j}(2+2)+\mathrm{k}(-1-3)\\
                                \mathrm{= 4i-4j-4k}
Equation of line L

\frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-3}{-1}=\frac{\mathrm{z}-1}{-1}=\lambda(\alpha \mathrm{d})

Let \mathrm{M}(\lambda+2,-\lambda+3,-\lambda+1)

DR's of MQ is <\lambda+2-5,-\lambda+3-3,-\lambda+1-8>

< \lambda-3,-\lambda \cdot-\lambda>7>

\because \mathrm{L} \perp \mathrm{MQ}
\Rightarrow(\lambda-3)(1)+(-\lambda)(-1)+(-\lambda-7)(-1)=0
\Rightarrow \lambda-3+\lambda+\lambda+7=0
\Rightarrow 3 \lambda=-4 \Rightarrow \lambda=-\frac{4}{3}
\therefore \mathrm{M}\left(-\frac{4}{3}+2, \frac{+4}{3}+3, \frac{4}{3}+1\right)=\left(\frac{2}{3}, \frac{13}{3}, \frac{7}{3}\right)
\mathrm{MQ}=\alpha
\therefore 3 \alpha^{2}=3 \times\left(\left(5-\frac{2}{3}\right)^{2}+\left(3-\frac{13}{3}\right)^{2}+\left(8-\frac{7}{5}\right)^{2}\right)
=3\left(\frac{169}{9}+\frac{16}{9}+\frac{289}{9}\right) \Rightarrow \frac{474}{9}=158
 

Posted by

himanshu.meshram

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