Question

The distance of the point (1, – 2, 4) from the plane passing through the point (1, 2, 2) and perpendicular to the planes x – y + 2x = 3 and 2x – 2y + z + 12 = 0 is:
(a) \[\dfrac{1}{\sqrt{2}}\]
(b) \[2\sqrt{2}\]
(c) 2
(d) \[\sqrt{2}\]

Answer 1

Hint: First of all find the equation of the plane as \[a\left( x-{{x}_{1}} \right)+b\left( y-{{y}_{1}} \right)+c\left( z-{{z}_{1}} \right)=0\text{ where }a\widehat{i}+b\widehat{j}+c\widehat{k}\] is a normal vector to it and \[\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\] is a point through which it passes. Now, since the remaining two planes are perpendicular to this plane, the dot product of the normal vector of this plane to the normal vector of the remaining plane would be zero individually. From this find the equation of the plane and then use \[\dfrac{\left| a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}\] to find the perpendicular distance of the plane ax + by + cz + d = 0 from the point \[\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\].

Complete step-by-step answer:
In this question, we are given a plane passing through the point (1, – 2, 4) from the plane passing through the point (1, 2, 2) and perpendicular to the planes x – y + 2x = 3 and 2x – 2y + z + 12 = 0. We have to find the distance of the point (1, – 2, 4) from this plane.
Let us find the equation of the plane. We know that if any plane P passes through point \[\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\] and vector normal to it is \[a\widehat{i}+b\widehat{j}+c\widehat{k}\] then its equation is given by:
\[P:a\left( x-{{x}_{1}} \right)+b\left( y-{{y}_{1}} \right)+c\left( z-{{z}_{1}} \right)=0\]
We are given that the plane passes through the point (1, 2, 2). So by substituting \[\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( 1,2,2 \right)\], we get,
\[P:a\left( x-1 \right)+b\left( y-2 \right)+c\left( z-2 \right)=0....\left( i \right)\]
We know that for any plane ax + by + cz + d = 0, \[a\widehat{i}+b\widehat{j}+c\widehat{k}\] is its normal vector. So, for the given plane x – y + 2z = 0, we get,
\[\text{Normal Vector }=\widehat{i}-\widehat{j}+2\widehat{k}.....\left( ii \right)\]
Also, for the given plane 2x – 2y + z + 12 = 0, we get,
\[\text{Normal Vector }=2\widehat{i}-2\widehat{j}+\widehat{k}.....\left( iii \right)\]
We know that when two planes are perpendicular to each other, their normal vectors are also perpendicular.
So, we are given that the plane P is perpendicular to the remaining two planes, so the normal vector of plane P would also be perpendicular to the normal vector of each of the planes.
We know that when two vectors are perpendicular to each other, their dot product is zero. Therefore, we get,
(Normal vector of plane P) . (Normal vector of plane x – y + 2z + 3) = 0
\[\left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right).\left( \widehat{i}-\widehat{j}+2\widehat{k} \right)=0\]
We know that,
\[\left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right).\left( p\widehat{i}+q\widehat{j}+r\widehat{k} \right)=ap+bq+cr\]
So, we get,
\[a-b+2c=0....\left( iv \right)\]
Also, (Normal vector of plane P) . (Normal vector of plane 2x – 2y + z + 12) = 0
\[\left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right).\left( 2\widehat{i}-2\widehat{j}+\widehat{k} \right)=0\]
\[2a-2b+c=0....\left( vi \right)\]
By multiplying equation (v) with 2, we get,
\[2a-2b+4c=0\]
By subtracting the above equation from equation (vi), we get,
\[\begin{align}
  & 2a-2b+c=0 \\
 & 2a-2b+4c=0 \\
 & \underline{-\text{ }+\text{ }-\text{ }} \\
 & c-4c=0 \\
 & -3c=0 \\
 & c=0 \\
\end{align}\]
By substituting c = 0 in equation (v), we get,
\[a-b+0=0\]
\[a=b\]
Now, by substituting a = b = k and c = 0 in equation (i), we get the equation of the plane P as,
\[P:k\left( x-1 \right)+k\left( y-2 \right)+0\left( z-2 \right)=0\]
\[k\left( x-1 \right)+k\left( y-2 \right)=0\]
\[k\left[ x-1+y-2 \right]=0\]
\[x+y-3=0\]
So, we get the equation of the plane P as x + y – 3 = 0.
Now, we know that from any point \[A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\], perpendicular distance d from the plane P: ax + by + cz + d = 0 is given by:
\[d=\dfrac{\left| a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}\]
So, by using this, we get the perpendicular distance of point P (1, – 2, 4) from the plane P: x + y – 3 = 0 is given by:
\[d=\dfrac{\left| \left( 1 \right)+\left( -2 \right)-\left( 3 \right) \right|}{\sqrt{1+1}}\]
\[d=\dfrac{\left| 1-5 \right|}{\sqrt{2}}\]
\[d=\dfrac{\left| -4 \right|}{\sqrt{2}}=\dfrac{4}{\sqrt{2}}=2\sqrt{2}\text{ units}\]
Hence, option (b) is the right answer.

Note: Students must note that to find the equation of the plane, absolute values of a, b and c are not required but we must know their ratios or we can express them in terms of common variables like in the above solution, we had a = b = k or a : b = 1. Also, students can cross-check the equation of the plane by substituting the point (1, 2, 2) in its equation and checking if LHS = RHS or not.