Question

The equation of the plane which passes through the line of intersection of the planes $r\cdot n_1=q_{1}r\cdot n_2=q_2$ and is parallel to the line of intersection of the line of intersection of the planes $r\cdot n_3=q_2$ and $r\cdot n_4=q_4$ given that $\left[n_1{n}_3{n}_4\right)\left(n_2{n}_3{n}_4\right]$ is

• A. $r\cdot n_2-q_2=r\cdot n_3-q_3$
• B. $r\cdot n_1-q_1=\left[n_1{n}_2{n}_3\right](r\cdot n_2-q_2)$
• C. $r\cdot n_1-q_1=r\cdot n_2-q_2$
• D. $r\cdot n_3-q_3=r\cdot n_1-q_1$

Answer 1

Answer: (C)

$r\cdot n_1-q_1=r\cdot n_2-q_2$

Equation of any plane through the intersection of $r\cdot n_1=q_1,r_2\cdot n_2=q_2$ is of the form $r\cdot n_1+\lambda r\cdot n_2=q_1+\lambda q_2\left(1\right)$
where $\lambda$ is a parameter.

So $n_1+\lambda .n_2$ is normal to the plane $\left(1\right)$ Any plane containing planes $r\cdot n_3=q_{3}r\cdot n_4=q_4$ is of the form
$r\cdot (n_3+\mu n_4)=q_3+\mu q_4$ Hence we must have $n_1+\lambda n_2=k(n_3+\mu n_4)$ for some $k$
$\Rightarrow [n_1+\lambda .n_2]\cdot [n{,}_3\times n_4]=0$
$\Rightarrow [n_1,n_3,n_4]+\lambda [n_2,n_3,n_4]=0$
$\Rightarrow \lambda =-1$
Putting this value in $\left(1\right),$ we have equation of required plane as
$r\cdot n_1-q_1=(r\cdot n_2-q_2)$ or
$\left[n_2{n}_3{n}_4\right](r\cdot n_1-q_1)=\left[n_1{n}_3{n}_4\right](r\cdot n_2-q_2)$