Three Dimensional Geometry Question 400
Question: The vector equation of the line of intersection of the planes $ \vec{r}=\vec{b}+{\lambda_1}(\vec{b}-\vec{a})+{\mu_1}(\vec{a}-\vec{c}) $ and $ \vec{r}=\vec{b}+{\lambda_2}(\vec{b}-\vec{c})+{\mu_2}(\vec{a}+\vec{c}) $, $\vec{a},\vec{b},\vec{c}$ being non-coplanar vectors, is
Options:
A) $ \vec{r}=\vec{b}+{\mu_1}(\vec{a}+\vec{c}) $
B) $ \vec{r}=\vec{b}+{\lambda_1}(\vec{a}-\vec{c}) $
C) $ \vec{r}=2\vec{b}+{\lambda_2}(\vec{a}-\vec{c}) $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
At the line of intersection of the two given planes, we have $ \vec{b}+{\lambda_1}(\vec{b}-\vec{a})+{\mu_1}(\vec{a}+\vec{c})=\vec{c}+{\lambda_2}(\vec{b}-\vec{c})+{\mu_2}(\vec{a}+\vec{b}) $
$ \Rightarrow (-\lambda +{\mu_1}-{\mu_2})\vec{a}+(1+{\lambda_1}-{\lambda_2}-{\mu_2})\vec{b}+({\mu_1}-1+{\lambda_2})\vec{c}=\vec{0} $
$ \Rightarrow -{\lambda_1}+{\mu_1}-{\mu_2}=0,\ 1+{\lambda_2}-{\lambda_2}{\mu_2}=0 $
and $ {\mu_1} - 1 + {\lambda_2} = 0 $
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$ \therefore $ $ \vec{a},\vec{b},\vec{c} $ are non-coplanar vectors.
From the last two equations, we obtain
$ {\lambda_1}+{\mu_1}-{\mu_2}=0 $
on solving this equation with $ -{\lambda_1}+{\mu_1}-{\mu_2}=0, $ we get
$ {\mu_1}={\mu_2} $ and $ {\lambda_1}=0 $
$ \therefore {\lambda_1}=0,{\mu_1}={\mu_2} $ and $ {\lambda_2}=1-{\mu_1} $ on substituting these values in either of the given equations, we obtain $ \vec{r}=\vec{b}+{\mu_1}(\vec{a}+\vec{c}) $ As the required line of intersection of the given plane.
BETA
