Three Dimensional Geometry Question 399
Question: Distance of the point $ P(\vec{p}) $ from the line $ \vec{r}=\vec{a}+\lambda \vec{b} $ is
Options:
A) $ | (\vec{a}-\vec{p})+\frac{((\vec{p}-\vec{a}).\vec{b})\vec{b}}{{{| {\vec{b}} |}^{2}}} | $
B) $ | (\vec{b}-\vec{p})+\frac{((\vec{p}-\vec{a}).\vec{b})\vec{b}}{{{| {\vec{b}} |}^{2}}} | $
C) $ | (\vec{a}-\vec{p})+\frac{((\vec{p}-\vec{b}).\vec{b})\vec{b}}{{{| {\vec{b}} |}^{2}}} | $
D) None of these
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Answer:
Correct Answer: B
Solution:
[c] Let Q $ (\vec{q}) $ be the foot of altitude drawn from $ P(\vec{p}) $ to the line $ \vec{r}=\vec{a}+\lambda \vec{b}, $
$ \Rightarrow (\vec{q}-\vec{p}).\vec{b}=0 $ and $ \vec{q}=\vec{a}+\lambda \vec{b} $
$ \Rightarrow (\vec{a}+\lambda \vec{b}-\vec{p}).\vec{b}=0 $ or $ | (\vec{a}-\vec{p}) |.\vec{b}+\lambda {{| {\vec{b}} |}^{2}}=0 $ or $ \lambda =\frac{(\vec{p}-\vec{a}).\vec{b})\vec{b}}{{{| {\vec{b}} |}^{2}}}-\vec{P} $
$ \Rightarrow | \vec{q}-\vec{p} |=| (\vec{a}-\vec{p})+\frac{(\vec{p}-\vec{a}).\vec{b})\vec{b}}{{{| {\vec{b}} |}^{2}}} | $
BETA
