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  • The plane intersects the line segment joining the points <img alt="A(a,-2,4)$ and $B(2, b,-3)" s

The plane 2 x-y+z=4 intersects the line segment joining the points A(a,-2,4)$ and $B(2, b,-3)at the point C in the ratio 2: 1 and the distance of the point C from the origin is \sqrt{5}. If a b<0 and \mathrm{P} is the point (a-b, b, 2 b-a) then C P^2 is equal to

Option: 1

\frac{97}{3}


Option: 2

\frac{17}{3}


Option: 3

\frac{16}{3}


Option: 4

\frac{73}{3}


Answers (1)

best_answer

        C divides AB in 2 : 1

                                            \begin{aligned} & C\left(\frac{4+a}{3}, \frac{2 b-2}{3}, \frac{-6+4}{3}\right) \\ & C\left(\frac{a+4}{3}, \frac{2 b-2}{3}, \frac{-2}{3}\right) \end{aligned}

                                               C lies in the plane

                                                         \begin{aligned} \quad \therefore & 2\left(\frac{a+4}{3}\right)-\left(\frac{2 b-2}{3}\right)+\left(\frac{-2}{3}\right)=4 \\ \Rightarrow & 2 a-2 b=4 \\ a-b & =2 \end{aligned}

\begin{array}{r} \because \mathrm{OC}=\sqrt{5} \\ \mathrm{OC}^2=5 \end{array}                                                                        \mathrm{C}\left(\frac{\mathrm{b}+6}{3}, \frac{2 b-2}{3}, \frac{-2}{3}\right)

\Rightarrow\left(\frac{\mathrm{b}+6}{3}\right)^2+\left(\frac{2 \mathrm{~b}-2}{3}\right)^2+\left(\frac{-2}{3}\right)^2=5 \quad \, \, \, \, \, \, \, \, \, \, \, \, \, \, \text { Now, }

\begin{array}{lr} \Rightarrow 5 b^2+4 b-1=0 & C\left(\frac{5}{3}, \frac{-4}{3}, \frac{-2}{3}\right) \\ \Rightarrow 5 b^2+5 b-b-1=0 & \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, P(a-b, b, 2 b-a) \end{array}         

\Rightarrow 5 \mathrm{~b}(\mathrm{~b}+1)-1(\mathrm{~b}+1)=0 \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \quad(2,-1,-3)

  \mathrm{b}=-1 \& \frac{1}{5} \quad \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \mathrm{CP}^2=\left(\frac{5}{3}-2\right)^2+\left(\frac{-4}{3}+1\right)^2+\left(\frac{-2}{3}+3\right)^2=\frac{17}{3}               

\begin{aligned} & \quad a=1 \\ & \because \mathrm{ab}<0 \\ & \therefore \mathrm{a}=1, \mathrm{~b}=-1 \end{aligned}                                                        

                                                                                             

 

 

            

                                                                               

Posted by

Pankaj

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