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The distance of the line $\overrightarrow r=2\hat i-2\hat j+3\hat k+\lambda(\hat i-\hat j+4\hat k)$ from the plane $x+5y+z=5$ is ?

$\begin{array}{1 1} (a)\:\:\large\frac{10}{3\sqrt 3}\:\:\:\qquad\:\:(b)\:\:\frac{10}{9}\:\:\:\qquad\:\:(c)\:\:\frac{10}{3}\:\:\:\qquad\:\:(d)\:\:\frac{3}{10} \end{array} $

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1 Answer

  • Distance between any line (parallel to the plane) and the plane is $\perp$ distance of any point on the line from the plane.
  • $\perp$ distance of any point $(x_1,y_1,z_1)$ from the plane $ax+by+cz+d=0$ is $\bigg|\large\frac{ax_1+by_1+cz_1+d}{\sqrt {a^2+b^2+c^2}}\bigg|$.
A point on the given line $\overrightarrow r=(2\hat i-2\hat j+3\hat k)+\lambda(\hat i-\hat j+4\hat k)$ is $A(2,-2,3)$
$\perp$ distance of $P$ from the plane $x+5y+z-5=0$ is $\bigg|\large\frac{2-10+3-5}{\sqrt {1+25+1}}\bigg|$
$=\large\frac{10}{3\sqrt 3}$
answered Dec 28, 2013 by rvidyagovindarajan_1

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