Three Dimensional Geometry Question 11
Question: Equation of the plane through the mid-point of the line segment joining the points P(4, 5, -10) and Q(-1, 2, 1) and perpendicular to PQ is
Options:
A) $ \vec{r}.( \frac{3}{2}\hat{i}+\frac{7}{2}\hat{j}-\frac{9}{2}\hat{k} )=45 $
B) $ \vec{r}.( -\hat{i}+2\hat{j}-\hat{k} )=\frac{135}{2} $
C) $ \vec{r}.(5\hat{i}+3\hat{j}-11\hat{k})+\frac{135}{2}=0 $
D) $ \vec{r}.(5\hat{i}+3\hat{j}-11\hat{k})=\frac{135}{2} $
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Answer:
Correct Answer: D
Solution:
[d] Mid-point of PQ is $ =( \frac{3}{2},\frac{7}{2},\frac{-9}{2} ) $ DR of the normal is $ =(4-(-1),5-2,-10-1) $ $ =5,3,-11 $
$ \therefore $ Eqn. of plane is $ 5( x-\frac{3}{2} )+3( y-\frac{7}{2} )-11( z+\frac{9}{2} )=0 $
$ \Rightarrow 5x+3y-11z=\frac{135}{2} $
$ \Rightarrow r.,(5\hat{j}+3\hat{j}-11\hat{k})=\frac{135}{2} $
BETA
