Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
0 votes

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

Can you answer this question?

1 Answer

0 votes
  • 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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App