Vector Algebra Question 144
Question: If the equation of a line through a point a and parallel to vector b is $ \mathbf{r}=\mathbf{a}+t,\mathbf{b}, $ where t is a parameter, then its perpendicular distance from the point c is
[MP PET 1998]
Options:
A) $ |(\mathbf{c}-\mathbf{b})\times \mathbf{a}|\div |\mathbf{a}| $
B) $ |(\mathbf{c}-\mathbf{a})\times \mathbf{b}|\div |\mathbf{b}| $
C) $ |(\mathbf{a}-\mathbf{b})\times \mathbf{c}|\div |\mathbf{c}| $
D) $ |(\mathbf{a}-\mathbf{b})\times \mathbf{c}|\div |\mathbf{a}+\mathbf{c}| $
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Answer:
Correct Answer: B
Solution:
- For point $ P $ on the line $ r=\mathbf{a}+t\mathbf{b} $
$ \therefore ,\overrightarrow{PC}=(\mathbf{c}-\mathbf{a})-t\mathbf{b} $ , $ \because ,\overrightarrow{PC},\bot ,\mathbf{b} $
$ \therefore ,|(\mathbf{c}-\mathbf{a})-t\mathbf{b}|,.,\mathbf{b}=0 $ or $ t=\frac{(\mathbf{c}-\mathbf{a}),.,\mathbf{b}}{{{\mathbf{b}}^{2}}} $ ?..(i) Distance of $ \mathbf{c} $ from line $ |\overrightarrow{PC}|\ = $ $ d=|\mathbf{c}-\mathbf{a}-t\mathbf{b}| $ $ d=| \mathbf{c}-\mathbf{a}-\frac{(\mathbf{c}-\mathbf{a}),.,\mathbf{bb}}{{{\mathbf{b}}^{2}}} |=| \frac{(\mathbf{c}-\mathbf{a}),\mathbf{b},.,\mathbf{b}-(\mathbf{c}-\mathbf{a}),.,\mathbf{bb}}{{{\mathbf{b}}^{2}}} | $ $ d=| \frac{\mathbf{b}\times (\mathbf{c}-\mathbf{a})\times \mathbf{b}}{{{\mathbf{b}}^{2}}} |=\frac{|\mathbf{b}||(\mathbf{c}-\mathbf{a})\times \mathbf{b}|\sin 90{}^\circ }{|\mathbf{b}{{|}^{2}}} $ , $ (\because ,\mathbf{b},\bot ,(\mathbf{c}-\mathbf{a})\times \mathbf{b}) $ $ d=\frac{|(\mathbf{c}-\mathbf{a})\times \mathbf{b}|}{|\mathbf{b}|} $ .
BETA
