Question

Find the direction cosines and the length of the perpendicular drawn to the origin to the plane $2x-3y+6z+14=0$.

Answer 1

$O(0,0,0)=(x,y,z)$
And from the given equation of plane we have $a_1=2,b=-3,c=6$ and $d=-14$
p$=$ Perpendicular distance of plane from origin
$\therefore p=\frac{|-d|}{\sqrt{a^2+b^2+c^2}}$
$=\frac{|-14|}{\sqrt{4+9+36}}$
$=\frac{14}{\sqrt{49}}$
$=\frac{14}{7}=2$ unit
Equation of the plane with perpendicular distance p from origin is x $\cos \alpha +y\cos \beta +z\cos \gamma =p$
Here $2x=3y=6z=-14$
$\therefore \left(\frac{2}{7}\right)x+\left(\frac{3}{7}\right)y+\frac{6}{7}z=-2$
$\therefore \left(\frac{-2}{7}\right)x+\left(\frac{3}{7}\right)y+(-\frac{6}{7})z=2$
$\therefore \cos \alpha =-\frac{2}{7},\cos \beta =\frac{3}{7}$ and $\cos \gamma =-\frac{6}{7}$
are perpendicular direction cosines
Note:
$2x-3y+6z=-14$
$\therefore -2x+3y-6z=14$
$\therefore -\frac{a}{|n|}x+\frac{d}{|n|}y-\frac{cz}{|n|}=14$
$\therefore$ where $\overline{n}=(-2,3,-6)$
$\therefore |\overline{n}|=\sqrt{49}=7$
$\therefore -\frac{2}{7}x+\frac{3}{7}y-\frac{6}{7}z=2$
$\therefore$ Direction cosine are $=(-\frac{2}{7},\frac{3}{7},-\frac{6}{7})$.