Let ${{x - 2} \over 3} = {{y + 1} \over { - 2}} = {{z + 3} \over { - 1}}$ lie on the plane $px - qy + z = 5$, for some p, q $\in$ R. The shortest distance of the plane from the origin is :
Solution
$\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z+3}{-1}=\lambda$
<br/><br/>
$(3 \lambda+2,-2 \lambda-1,-\lambda-3)$ lies on plane $p x-q y+z=5$ <br/><br/>$p(3 \lambda+2)-q(-2 \lambda-1)+(-\lambda-3)=5$
<br/><br/>
$\lambda(3 p+2 q-1)+(2 p+q-8)=0$
<br/><br/>
$3 p+2 q-1=0\} p=15$
<br/><br/>
$2 p+q-8=0\} q=-22$
<br/><br/>
Equation of plane $15 x+22 y+z-5=0$
<br/><br/>
Shortest distance from origin $=\frac{|0+0+0-5|}{\sqrt{15^{2}+22^{2}+1}}$
<br/><br/>
$=\frac{5}{\sqrt{710}}$
<br/><br/>
$=\sqrt{\frac{5}{142}}$
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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