Question

The ratio in which the sphere $x^{2} + y^{2} + z^{2} = 504$ divides the line segment AB joining the points $A\mspace{6mu}(12,\mspace{6mu} - 4,\mspace{6mu} 8)$ and $(27,\mspace{6mu} - 9,\mspace{6mu} 18)$ is given by

• A. $2:3$ externally
• B. $2:3$ internally
• C. $1:2$ externally
• D. None of these

Answer 1

Answer: (A)

$2:3$ externally

Answer 2

Let the ratio is $\lambda : 1$ so the point on sphere is $\frac { 27 \lambda + 12 } { \lambda + 1 } , \frac { - 9 \lambda - 4 } { \lambda + 1 } , \frac { 18 \lambda + 8 } { \lambda + 1 }$

Also, $\left( \frac { 27 \lambda + 12 } { \lambda + 1 } \right) ^ { 2 } + \left( \frac { 9 \lambda + 4 } { \lambda + 1 } \right) ^ { 2 } + \left( \frac { 18 \lambda + 8 } { \lambda + 1 } \right) ^ { 2 } = 504$

∴ $\lambda = \frac { - 2 } { 3 }$, So that the ratio is 2 : 3 externally.