Question

Equation of the plane containing the straight line $\frac{x}{2}=\frac{y}{3}= \frac{z}{4}$ and perpendicular to the plane containing the staight lines $\frac{x}{2}=\frac{y}{4}= \frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}= \frac{z}{3}$ is

• A. $x + 2y - 2z = 0$
• B. $3x + 2y - 2z = 0$
• C. $x - 2y + 2 = 0$
• D. $5x + 2y - 42 = 0$

Answer 1

Answer: (C)

$x - 2y + 2 = 0$

Answer 2

The DR's of normal to the plane containing $\frac{x}{3}=\frac{y}{4}= \frac{z}{2}$
and $\frac{x}{4}=\frac{y}{z}= \frac{z}{3}.$
\hspace20mm n_1= \begin{vmatrix} \widehat {i} &\widehat {j} &\widehat {k} \\\ 3 & 4 & 2 \\\ 4 & 2 & 3 \\\ \end{vmatrix}= (8 \widehat {i}+ \widehat {j} +10\widehat {k})
Also, equation of plane
containing $\frac{x}{2}=\frac{y}{3}= \frac{z}{4}$ and
DR's of normal to be
$n_1 = a \widehat {i}+ b\widehat {j} +c\widehat {k}$
$\Rightarrow \ \ ax+by+cz = 0$.
\hspace30mm ...(i)
Where, $n_1 : n_2 = 0$
\Rightarrow \hspace 30mm 8a-b-10c = 0 \hspace 20 mm ...(ii)
and $n_2 \perp (2\widehat{i}+3\widehat{j}+4\widehat{k})$
\Rightarrow \hspace30mm 2a+3b+4c = 0 \hspace20mm ...(iii)
From Eqs (ii) and (iii),
\hspace20mm \frac{a}{-4+30} = \frac{b}{-20-32} = \frac{c}{24+2}
\Rightarrow\hspace 30mm \frac{a}{26} = \frac{b}{-52} = \frac{c}{26}
\Rightarrow\hspace30mm \frac{a}{1} = \frac{b}{-2} = \frac{c}{1}\hspace30mm ...(iv)
Form Eqs. (i) and (iv), required equation of plane is $x - 2y + z = 0$