Question

Find the distance of the plane 3x-4y+12z=3 from the origin.

Answer 1

Hint: The shortest distance from a point to a plane is along a line perpendicular to the plane.
Let the plane be Ax+By+Cz=D and the point from which the distance is to be measured be (a, b, c) .
So the formula for measuring the distance between the plane and the point is given as follows
\[d=\dfrac{\left| Aa+Bb+Cc-D \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}+{{C}^{2}}}}\]

Complete step-by-step answer:
As mentioned in the question, we are asked to find the distance between the plane 3x-4y+12z=3 from the point (0, 0, 0) or the origin and this distance is nothing but the shortest distance between the plane and the point or origin.
Now, using the formula given in the question, we can find the distance between the plane which is 3x-4y+12z=3 and the origin as follows
We can the formula that is
\[d=\dfrac{\left| Aa+Bb+Cc-D \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}+{{C}^{2}}}}\]
By just replacing the constants in this formula with their respective values that are
A=3, B=-4, C=12, D=-3 and a=0, b=0 and c=0.
Hence, the formula will become
\[\begin{align}
  & d=\dfrac{\left| 3\cdot 0+-4\cdot 0+12\cdot 0-\left( -3 \right) \right|}{\sqrt{{{3}^{2}}+{{\left( -4 \right)}^{2}}+{{12}^{2}}}} \\
 & d=\dfrac{3}{\sqrt{169}} \\
 & d=\dfrac{3}{13} \\
\end{align}\]
Hence, the distance between the plane 3x-4y+12z=3 and the origin is \[\dfrac{3}{13}\] .

Note: -The students can make an error if they don’t know the formula for finding the distance between a plane and a point which is as follows.Let the plane be Ax+By+Cz=D and the point from which the distance is to be measured be (a, b, c) .So the formula for measuring the distance between the plane and the point is given as follows
\[d=\dfrac{\left| Aa+Bb+Cc-D \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}+{{C}^{2}}}}\] .