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