Three Dimensional Geometry Question 374
Question: The equation of the plane which passes through the line of intersection of planes $ \vec{r}.{{\vec{n}}_1}=q_1,\vec{r}.{{\vec{n}}_2}=q $ And is parallel to the line of intersection of planes $ \vec{r}.{{\vec{n}}_3}=q_3 $ and $ \vec{r}.{{\vec{n}}_4}=q_4 $ is
Options:
A) $ [{{\vec{n}}_2}{{\vec{n}}_3}{{\vec{n}}_4}]\vec{r}.{{\vec{n}}_1}-{{\vec{q}}_1}=[{{\vec{n}}_1}{{\vec{n}}_3}{{\vec{n}}_4}]\vec{r}.{{\vec{n}}_2}-{{\vec{q}}_2} $
B) $ [{{\vec{n}}_1}{{\vec{n}}_2}{{\vec{n}}_4}]\vec{r}.{{\vec{n}}_4}q_4=[{{\vec{n}}_4}{{\vec{n}}_3}{{\vec{n}}_1}]\vec{r}.{{\vec{n}}_2}-q_2 $
C) $ [{{\vec{n}}_4}{{\vec{n}}_3}{{\vec{n}}_1}]\vec{r}.{{\vec{n}}_4}-q_4=[{{\vec{n}}_1}{{\vec{n}}_2}\vec{n} 3]\vec{r}.{{\vec{n}}_2}=q_2 $
D) None of these
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Answer:
Correct Answer: A
Solution:
[a] $ (\vec{r}.{{\vec{n}}_1}+\lambda \vec{r}.{{\vec{n}}_2}=q_1+\lambda q_2) $ ??(i) Where $ \lambda $ is a parameter? So, $ {{\vec{n}}_1}+\lambda {{\vec{n}}_2} $ is normal to plane (i), now, any plane parallel to the line of intersection of the planes $ \vec{r}.{{\vec{n}}_3}=q_3 $ and $ \vec{r}.{{\vec{n}}_4}=q_4 $ is of the form $ \vec{r}.({{\vec{n}}_3}\times {{\vec{n}}_4})=d. $ Hence, we must have $ [{{\vec{n}}_1}+\lambda {{\vec{n}}_2}].[{{\vec{n}}_3}\times {{\vec{n}}_4}]=0 $ or $ [{{\vec{n}}_1}{{\vec{n}}_3}{{\vec{n}}_4}]+\lambda [{{\vec{n}}_2}{{\vec{n}}_3}{{\vec{n}}_4}]=0 $ or $ \lambda =\frac{-[{{{\vec{n}}}_1}{{{\vec{n}}}_3}{{{\vec{n}}}_4}]}{[{{{\vec{n}}}_2}{{{\vec{n}}}_3}{{{\vec{n}}}_4}]} $ on putting this value in Eq. (i), we have the equation of the required plane as $ \vec{r}.{{\vec{n}}_1}-q_1=\frac{[{{{\vec{n}}}_1}{{{\vec{n}}}_2}{{{\vec{n}}}_4}]}{[{{{\vec{n}}}_2}{{{\vec{n}}}_3}{{{\vec{n}}}_4}]}(r.{{\vec{n}}_2}-q_2) $ or $ [{{\vec{n}}_2}{{\vec{n}}_3}{{\vec{n}}_4}]\vec{r}.{{\vec{n}}_1}-q_1=[{{\vec{n}}_1}{{\vec{n}}_3}{{\vec{n}}_4}]\vec{r}.{{\vec{n}}_2}-q_2 $
BETA
