Question

If the plane $2ax - 3ay + 4az + 6 = 0$ passes through the midpoint of the line joining the centres of the spheres

$x^{2} + y^{2} + z^{2} + 6x - 8y - 2z = 13$ and

$x^{2} + y^{2} + z^{2} - 10x + 4y - 2z = 8$, then $a$ equals

• A. – 2
• B. 2
• C. – 1
• D. 1

Answer 1

Answer: (A)

– 2

Answer 2

$S _ { 1 } \equiv x ^ { 2 } + y ^ { 2 } + z ^ { 2 } + 6 x - 8 y - 2 z = 13$ $C _ { 1 } \equiv ( - 3,4,1 )$

$S _ { 2 } \equiv x ^ { 2 } + y ^ { 2 } + z ^ { 2 } - 10 x + 4 y - 2 z = 8$ $C _ { 2 } \equiv ( 5 , - 2,1 )$

So mid point of $C _ { 1 } C _ { 2 }$ (say P)

$\equiv P \left( \frac { 5 - 3 } { 2 } , \frac { 4 - 2 } { 2 } , \frac { 1 + 1 } { 2 } \right) = P ( 1,1,1 )$

Now the plane $2 a x - 3 a y + 4 a z + 6 = 0$ passes through the point P,

So, $2 a ( 1 ) - 3 a ( 1 ) + 4 a ( 1 ) + 6 = 0 = 2 a - 3 a + 4 a + 6 = 0$

⇒ $3 a + 6 = 0$⇒ $3 a = - 6 \Rightarrow a = - 2$ .