Question

The ratio in which the surface $x^2+y^2+z^2=25$ divides the line joining $(0,1,2)$ and $(3,4,5)$ is $\frac{a\pm \sqrt{b}}{c}$ such that $a,c$ are co-primes, then

• A. $a=11$
• B. $a=9$
• C. $b=691$
• D. $c=35$

Answer 1

Answer: (A), (C)

The correct options are
A $a=11$
C $b=691$

Let the line joining the given points meets the given sphere $x^2+y^2+z^2=25$ at point $(x_1,y_1,z_1)$

Then

$x^2+y^2+z^2=25\left(1\right)$

Given points are $(0,1,2),(3,4,5)$

Suppose the point $(x_1,y_1,z_1)$ divides the join of the points $(0,1,2),(3,4,5)$ in ratio $\lambda :1$

Then

$x_1=\frac{3\lambda +0}{\lambda +1},y_1=\frac{4\lambda +1}{\lambda +1},z_1=\frac{5\lambda +2}{\lambda +1}$ (By section formula)

Putting these values in equation $\left(1\right)$

${\left(\frac{3\lambda +0}{\lambda +1}\right)}^2+{\left(\frac{4\lambda +1}{\lambda +1}\right)}^2+{\left(\frac{5\lambda +2}{\lambda +1}\right)}^2=259{\lambda }^2+16{\lambda }^2+1+18\lambda +25{\lambda }^2+1+20\lambda =25{(\lambda +1)}^{2}25{\lambda }^2+25{\lambda }^2+5+18\lambda +2\lambda =25{\lambda }^2+25+50\lambda 25{\lambda }^2-30\lambda -20=025{\lambda }^2+5+28\lambda +25{\lambda }^2=25{\lambda }^2+25+50\lambda 25{\lambda }^2-22\lambda -20=0$

On solving for $\lambda$ value by using $\lambda =\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ formula

$\lambda =\frac{-(-22)\pm \sqrt{{\left(22\right)}^2-4(-20)\left(25\right)}}{2\times 25}=\frac{22\pm \sqrt{484+2000}}{2\times 25}=\frac{2\times 11\pm \sqrt{2484}}{2\times 25}=\frac{11\pm \sqrt{691}}{25}$

By comparing $\frac{a\pm \sqrt{b}}{c}$ with $\frac{11\pm \sqrt{691}}{25}$

$\Rightarrow a=11,b=691,c=25$

Options: $A,C$