Get Answers to all your Questions

header-bg qa

If the image of the point P(1, −2, 3) in the plane, 2x+3y−4z+22=0 measured parallel to the line, \frac{x}{1}= \frac{y}{4}= \frac{z}{5}    is Q, then PQ is equal to :
Option: 1 2\sqrt{42}

Option: 5 \sqrt{42}

Option: 9 6\sqrt{5}

Option: 13 3\sqrt{5}

Answers (1)



Intersection of line and plane -

Let the line

\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c} plane

a_{1}x+b_{1}y+c_{1}z+d=0 intersect at P

to find P assume general point on line as \left ( x_{1}+\lambda a_{1}y_{1} +\lambda b_{1}z_{1}+\lambda c_{1}\right )

now put it in plane to find \lambda,

a_{1}\left ( x_{1}+\lambda a \right )+b_{1}\left ( y_{1}+\lambda b \right )+c_{1}\left ( z_{1}+\lambda c \right )+d=0




Distance formula -

The distance between two points A(x_{1},y_{1},z_{1})\, and \, B(x_{2},y_{2},z_{2}) is =\sqrt{\left ( x_{2}-x_{1} \right )^{2}+{\left ( y_{2}-y_{1} \right )^{2}}+{\left ( z_{2}-z_{1} \right )^{2}}}


- wherein

\vec{OA}= x_{1}\hat{i}+y_{1}\hat{j}+z_{1}\hat{k}

\vec{OB}= x_{2}\hat{i}+y_{2}\hat{j}+z_{2}\hat{k}

\underset{AB}{\rightarrow}    = \underset{OB}{\rightarrow} - \underset{AB}{\rightarrow}

\underset{AB}{\rightarrow}  = \left ( x_{2}-x_{1} \right )\hat{i}+\left ( y_{2}-y_{1} \right )\hat{j}+\left ( z_{2}-z_{1} \right )\hat{k}

AB= \left | \underset{AB}{\rightarrow} \right |

AB= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}+\left ( z_{2}-z_{1} \right )^{2}}






Now, for finding M


\Rightarrow x=k,y=4k,z=5k

Putting in equation of plane,


So, M=\left (\frac{11}{3} ,\frac{44}{3},\frac{55}{3} \right )

\therefore PM=\sqrt{\frac{64}{9}+\frac{2500}{9}+\frac{2116}{9}}= 2\sqrt{130}

\Rightarrow PQ=4\sqrt{130}






Posted by

vishal kumar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE