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)

$\sqrt{8}$

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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)

$\sqrt{8}$

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