# In the following cases,find the distance of each of the given points from the corresponding given plane (a)Point(0,0,0),Plane $3x-4y+12z-3=0$

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$\begin{array}{1 1} \large\frac{3}{13} \\ 1 \\ \large\frac{2}{13} \\ \large\frac{13}{3} \ \end{array}$

Toolbox:
• Distance of a point $(x_1,y_1,z_1)$ from the plane $ax+by+cz+d=0$
• $D=\large\frac{\mid ax_1+by_1+cz_1+d\mid}{\sqrt{a^2+b^2+c^2}}$
Step 1:
Given point is $(0,0,0)$
Equation of the plane is $3x-4y+12z-3=0$
Direction cosines are $(3,-4,12)$
Step 2:
Distance of the point to the given plane is $D=\large\frac{\mid ax_1+by_1+cz_1+d\mid}{\sqrt{a^2+b^2+c^2}}$
$\Rightarrow \large\frac{\mid 0\times 3+0\times -4+0\times 12-3\mid}{\sqrt{3^2+(-4)^2+(12)^2}}$
$\Rightarrow \large\frac{3}{\sqrt{169}}=\large\frac{3}{13}$
$D=\large\frac{3}{13}$
edited Jun 3, 2013