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  • Let Q be the foot of perpendicular from the origin to the plane  and R be a

Let Q be the foot of perpendicular from the origin to the plane 4x-3y+z+13=0 and R be a point (-1,1,-6) on the plane. Then length QR is :

Option: 1

\sqrt{14}


Option: 2

\sqrt\frac{19}{2}


Option: 3

3\sqrt\frac{7}{2}


Option: 4

\frac{3}{\sqrt{2}}


Answers (1)

best_answer

Let Coordinate of P will be the origin, i.e. (0, 0, 0) 

Point R is the image of point P with respect to the plane and point Q lie on the plane.

line PQ meets the plane 4x - 3y + z + 13 = 0 at point Q, direction ratio of normal to plane π are (4, --3, 1), since, PQ perpendicular to plane π.

So direction ratio of PQ are 4, -3, 1 

\\\frac{x-0}{4}=\frac{y-0}{-3}=\frac{z-0}{1}=r \\ x=4 r, y=-3 r, z=r

\\\text{Let the coordinate of }\mathrm{P}\text{ be }(4 r,-3 r, r)\text{ Since }Q \text{ be the mid point of OP} \\\therefore \quad \mathrm{Q}=\left(2 r,-\frac{3}{2} r, \frac{r}{2}\right)

\\\text{ Since Q lies in the given plane } \\4 x-3 y+z+13=0 \\ 8 r+\frac{9}{2} r+\frac{r}{2}+13=0

\\\Rightarrow \quad r=\frac{-13}{8+\frac{9}{2}+\frac{1}{2}}=\frac{-26}{26}=-1 \\ \therefore \quad \mathrm{Q}=\left(-2, \frac{3}{2},-\frac{1}{2}\right) \\ \quad \mathrm{Q} \mathrm{R}=\sqrt{(-1+2)^{2}+\left(1-\frac{3}{2}\right)^{2}+\left(-6+\frac{1}{2}\right)^{2}} \\ \quad=\sqrt{1+\frac{1}{4}+\frac{121}{4}}=3 \sqrt{\frac{7}{2}}

Posted by

Sanket Gandhi

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