Question

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

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

Answer 1

The correct option is A

2:3 externally

Let the ratio is $\lambda :1$ so the point on sphere is
$\frac{27\lambda +12}{\lambda +1},\frac{-9\lambda -4}{\lambda +1},{\frac{18\lambda +8}{\lambda +1}}^2$
Also, ${\left(\frac{27\lambda +12}{\lambda +1}\right)}^2+{\left(\frac{-9\lambda -4}{\lambda +1}\right)}^2+{\left(\frac{18\lambda +8}{\lambda +1}\right)}^2=504$
$\therefore \lambda =\frac{-2}{3}$, So that the ratio is 2:3 externally.