Q&A - Ask Doubts and Get Answers
Q

Solve! - Three Dimensional Geometry - JEE Main-6

Perpendicular distance of (1,1,1) from the line \frac{x-3}{1}=\frac{y-0}{1}=\frac{z-2}{-1}  is

  • Option 1)

    \sqrt{2}

  • Option 2)

    \sqrt{41}

  • Option 3)

    \sqrt{6}

  • Option 4)

    \sqrt{8}

 
Answers (1)
132 Views

As we have learned

Perpendicular distance of a point from a line, foot of perpendicular and image (Cartesian form ) - - wherein

L be the foot of perpendicular drawn from P(\alpha ,\beta ,\gamma ) on the line \frac{x-x_{1}}{a}= \frac{y-y_{1}}{b},\frac{z-z_{1}}{c}

Co-ordinates of L will be \left ( x_{1} +\lambda a,y_{1}+\lambda b,z_{1}+\lambda c\right )

DR's of PL will be \left ( x_{1} +\lambda a-\alpha ,y_{1}+\lambda b-\beta ,z_{1}+\lambda c-\gamma \right )

PL is perpendicular to line,\left ( x_{1} +\lambda a-\alpha \right )a+\left ( y_{1}+\lambda b-\beta \right )b+\left ( z_{1}+\lambda c-\gamma \right )c= 0

Find \gamma and put in coordinates of L Distance between P and L is perpendicular distance Image Q can be obtained by applying mid point formula in P and Q and equating it to L obtained.

 

 

 

Let foot of perpendicular be L 

So, L will be (\lambda+3,\lambda,2-\lambda)

\therefore DR of the PL will be \lambda +2, \lambda-1, 1-\lambda

\because PL is perpendicular to given line so 

a_{1}a_{2}+b_1b_2+c_1c_2=0 \Rightarrow 1(\lambda+2)+1(\lambda-1)-1(1-\lambda)=0

\Rightarrow \lambda+2+\lambda-1+\lambda-1=0

\Rightarrow \lambda=0

\therefore L(3,0,2)

\therefore PL = \sqrt{4+1+1}= \sqrt{6}

Option C

 

 

 

 


Option 1)

\sqrt{2}

Option 2)

\sqrt{41}

Option 3)

\sqrt{6}

Option 4)

\sqrt{8}

Exams
Articles
Questions