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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the length and the foot of perpendicular from the point$\bigg(1,\large\frac{3}{2}$$,2\bigg)$ to the plane 2x-2y+4z+5=0

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  • The equation of a plane parallel to the plane
  • $\overrightarrow r.\overrightarrow n= d$ is $\overrightarrow r.\overrightarrow n= d_1$
  • The cartesian equation of a plane parallel to the plane $ax+by+cz+\lambda=0$
  • Length of the line segement $PQ$ is
  • $\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
Step 1:
Let $Q$ be the foot of the perpendicular from the point $P(1,\large\frac{3}{2}$$,2)$
Let $Q(x_1,y_1,z_1)$
Hence the direction ratios of $PQ$ is
Let $\overrightarrow n$ be the normal to the plane :$ 2x-2y+4z+5=0$
Now $\overrightarrow {PQ}$ and $\overrightarrow n$ are parallel to each other
The direction ratios of $\overrightarrow n$ are $(2,-2,4)$
Hence the cartesian equation of line $PQ$ is
$\large\frac{x_1-1}{2}=\frac{y_1-3/2}{-2}=\frac{z_1-2}{4}=\lambda$ (assume)
Hence $ x_1=2 \lambda+1,y_1=-2 \lambda+\large\frac{3}{2}$$, z_1=4 \lambda+2$
but $Q$ lies on the plane $2x-2y+4z+5=0$
Now substituting for $x_1,y_1$ and $z_1$ we get,
Step 2:
$2(2 \lambda+1)-2(-2 \lambda+\large\frac{3}{2}$$)+4(4\lambda+2)+5=0$
On simplifying we get,
$4 \lambda+2+4 \lambda-3+16\lambda+8+5=0$
On simplifying we get,
$24 \lambda+12=0$
=>$ \lambda=\large\frac{-12}{24}$
=>$\lambda =\large\frac{-1}{2}$
Now substituting for $\lambda$ we get the coordinate of Q as
Hence the coordinate of Q are $(0,\large\frac{5}{2},0)$
The length of this perpendicular $PQ$ is
$PQ=\sqrt {(1-0)^2+(\frac{3}{2}-\frac{5}{2})^2+(2-0)^2}$
$\qquad=\sqrt {1+1+4}$
$\qquad=\sqrt 6$
Hence $\sqrt 6$ is the required solution
answered Jun 11, 2013 by meena.p

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