If Q(0,-1,-3) is the image of the point

Question:

If $\mathrm{Q}(0,-1,-3)$ is the image of the point $\mathrm{P}$ in the plane $3 x-y+4 z=2$ and $R$ is the point $(3,-1,-2)$, then the area (in sq. units) of $\triangle \mathrm{PQR}$

  1. $\frac{\sqrt{65}}{2}$

  2. $\frac{\sqrt{91}}{4}$

  3. $2 \sqrt{13}$

  4. $\frac{\sqrt{91}}{2}$


Correct Option: , 4

Solution:

$\mathrm{R}$ lies on the plane.

$\mathrm{DQ}=\frac{|1-12-2|}{\sqrt{9+1+16}}=\frac{13}{\sqrt{26}}=\sqrt{\frac{13}{2}}$

$\Rightarrow P Q=\sqrt{26}$

Now, $\mathrm{RQ}=\sqrt{9+1}=\sqrt{10}$

$\Rightarrow \mathrm{RD}=\sqrt{10-\frac{13}{2}}=\sqrt{\frac{7}{2}}$

Hence, $\operatorname{ar}(\triangle \mathrm{PQR})=\frac{1}{2} \times \sqrt{26} \times \sqrt{\frac{7}{2}}=\frac{\sqrt{91}}{2}$.

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