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  • Let the line   <img alt="L: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}" src="https://entrancecorner.oncodecogs.com/gif.latex?L%3A%20%5Cfrac%7Bx-1%7D%7B2%7D%3D%5Cfrac%7By&plus;1%7D%7B-

Let the line   L: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}     intersect the plane  2x+y+3z=16  at the point  P . Let the point Q be the foot of perpendicular from the point  R(1,-1,-3)  on the line L. If α is the area of triangle PQR, then \alpha ^{2}  is equal to

 

Option: 1

180


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Point on line L is  (2\lambda+1,-\lambda-1,\lambda+3)  

If above point is intersection point of line L and plane then

2(2\lambda+1)+,(-\lambda-1),3(\lambda+3)=16

\lambda=1

Point P = (3, –2, 4)

Dr of QR =< 2\lambda,-\lambda,\lambda+6>
Dr of L = < 2 –1, 1 >

4\lambda+\lambda+\lambda+6=0\\ \lambda=-1

Q = (–1, 0, 2)

\begin{aligned} & \overrightarrow{\mathrm{QR}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-5 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{QP}}=4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{QR}} \times \overrightarrow{\mathrm{QP}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & -1 & -5 \\ 4 & -2 & 2 \end{array}\right|=-12 \hat{\mathrm{i}}-24 \hat{\mathrm{j}} \\ & \alpha=\frac{1}{2} \times \sqrt{144+576} \end{aligned}

\begin{aligned} & \alpha^2=\frac{720}{4}=180 \\ & \alpha^2=180 \end{aligned}

 

 

Posted by

Ritika Harsh

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