Question

If 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 $(a,0,0),(0,b,0)$ and $(0,0,c)$ then $(a,b,c)$ is

• A. $(-\frac{9}{2},9,9)$
• B. $(\frac{1}{2},1,1)$
• C. $(1,-\frac{1}{2},1)$
• D. $(1,\frac{1}{2},1)$

Answer 1

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

Given points are (1,2,3) and (-3,4,5)
Mid point of this segment is, $(-1,3,4)=M$(say)
and direction ratio are, $(4,-2,-2)$
Therefore, normal vector perpendicular to required plane is $\overrightarrow{n}=4\hat{i}-2\hat{j}-2\hat{k}$
Since required plane is bisecting given points perpendicularly, so point $M$ will lie in the plane.
Therefore equation of plane is given by,
$\left(\right(x+1)\hat{i}+(y-3)\hat{j}+(z-4\left)\hat{k}\right)\cdot \overrightarrow{n}=0$
$\Rightarrow \left(\right(x+1)\hat{i}+(y-3)\hat{j}+(z-4\left)\hat{k}\right)\cdot (4\hat{i}-2\hat{j}-2\hat{k})=0$
$\Rightarrow 2x-y-z+9=0$
Hence, intercepts made on the axes are $(-\frac{9}{2},9,9)$