3D Geometry 3 Question 56
56. The distance of the point $(1,1,1)$ from the plane passing through the point $(-1,-2,-1)$ and whose normal is perpendicular to both the lines $L _1$ and $L _2$, is
(a) $2 / \sqrt{75}$ unit
(b) $7 / \sqrt{75}$ unit
(c) $13 / \sqrt{75}$ units
(d) $23 / \sqrt{75}$ units
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Answer:
Correct Answer: 56. (8)
Solution:
- The equation of the plane passing through the point $(-1,-2,-1)$ and whose normal is perpendicular to both the given lines $L _1$ and $L _2$ may be written as
$(x+1)+7(y+2)-5(z+1)=0 \Rightarrow x+7 y-5 z+10=0$
The distance of the point $(1,1,1)$ from the plane
$ =\left|\frac{1+7-5+10}{\sqrt{1+49+25}}\right|=\frac{13}{\sqrt{75}} \text { units } $
