Question

The intercepts made on the axes by the plane which bisects the line joining the points (1, 2, 3) and (–3, 4, 5,) at right angles are

• A. $(-\frac{9}{2},9,9)$
• B. $(\frac{9}{2},9,9)$
• C. $(9,-\frac{9}{2},9)$
• D. $(9,\frac{9}{2},9)$

Answer 1

The correct option is A $(-\frac{9}{2},9,9)$

Let plane is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1....\left(i\right)$
mid point of P(1, 2, 3) and Q(-3, 4, 5)
ie, -1, 3, 4 lie on Eq (i)
$\therefore -\frac{1}{a}+\frac{3}{b}+\frac{4}{c}....\left(ii\right)$
Also, PQ is parallel to normal of the plane (i)
$\frac{\frac{1}{a}}{-4}=\frac{\frac{1}{b}}{2}=\frac{\frac{1}{c}}{2}\Rightarrow \frac{1}{-2a}=\frac{1}{b}=\frac{1}{c}=\lambda \left(say\right)\therefore \frac{1}{a}=-2\lambda ,\frac{1}{b}=\lambda ,\frac{1}{c}=\lambda \therefore FromEq.\left(ii\right),2\lambda +3\lambda +4\lambda =1\therefore \lambda =\frac{1}{9}a=-\frac{1}{2{\lambda }^{\prime }}b=\frac{1}{{\lambda }^{\prime }}c=\frac{1}{\lambda }\therefore a=-\frac{9}{2^{\prime }}b=9,c=9Interceptsare(-\frac{9}{2},9,9)$