Question

The distance of origin from the image of (1, 2, 3) in plane $x-y+z=5$is
\[\begin{align}
  & \text{A})\,\,\,\sqrt{17} \\
 & \text{B})\,\,\,\sqrt{29} \\
 & \text{C})\,\,\,\sqrt{34} \\
 & \text{D})\,\,\,\sqrt{41} \\
\end{align}\]

Answer 1

Hint: First we need to find the coordinates of the image using the equation of normal because the line passing through both the point and its image is perpendicular to the plane. After that, we can find its distance from the origin (0,0,0) using distance formula.


Complete step by step Answer:

Let P and P’ be the points and its mirror image. And equation the plane
\[x-y+z=5\]
Since the point the P’ is the mirror image of P therefore PP’ should be perpendicular to the plane. Now if the line PP’ is perpendicular to the plane that means it is also parallel to the normal vector of the plane.
So we can write the equation of line PP’ as
\[PP\prime \,=\,\overset{\to }{\mathop{P}}\,\,+\,\overset{\to }{\mathop{\lambda n}}\,\]
Where $\vec{n}$ is the normal vector of the plane
\[PP\prime \,=\,(\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\,)\,+\,(\overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\,)\]
\[PP\prime \,=\,\,(\lambda +1)\,\,\overset{\wedge }{\mathop{i}}\,+(2-\lambda )\,\overset{\wedge }{\mathop{j}}\,+\,(3+\lambda )\overset{\wedge }{\mathop{k}}\,\]
Let X, Y, Z be a general point on the line PP’
     \[\begin{align}
  & X\,=\,\lambda +1 \\
 & Y\,=\,2-\lambda \\
 & Z\,=\,3+\lambda \\
\end{align}\]
Since the line PP’ intercepts the plane then the point (X, Y, Z) should satisfy it,
\[\therefore \,X-Y+Z=5\]

\[\begin{align}
  & (\lambda +1)+(2-\lambda )+(\lambda +3)=5 \\
 & \lambda +1-2+\lambda +\lambda +3=5 \\
 & 3\lambda +2=5 \\
 & 3\lambda =5-2 \\
 & 3\lambda =3 \\
 & \lambda =1 \\
\end{align}\]
For $\lambda =1$the point X, Y, Z lies on the plane.
Now we know that P and P’ are mirror images and they are at equal distance from the plane.
∴ Point M on the plane is the midpoint of PP’
Let $P\prime \,=\,(\alpha ,\beta ,\gamma )$
For $\lambda =1$ the values of X, Y and Z are
For $\lambda =1$
\[\begin{align}
  & \text{X}=1+1=2 \\
 & \text{Y}=2-1=1 \\
 & \text{Z}=3+1=4 \\
\end{align}\]
Point M is (2, 1, 4) using midpoint theorem on PP’ and M we get
\[\begin{align}
  & \dfrac{1+\alpha }{2}=2\alpha =3 \\
 & \dfrac{2+\beta }{2}=1\beta =0 \\
 & \dfrac{3+\gamma }{2}=4\gamma =5 \\
\end{align}\]
Point P’ is (3, 0, 5)
Now taking the distance of point P’ from origin (0, 0, 0) we get
distance$\sqrt{{{(3-0)}^{2}}+{{(0-0)}^{2}}+{{(5-0)}^{2}}}$
\[\begin{align}
  & =\sqrt{9+25} \\
 & =\sqrt{34} \\
\end{align}\]
The correct option is C.

Note: For any given point and its image along a plane, the line joining both the points should have a midpoint that lies on the plane. This also means that the distance of the point and its image from the plane is equal.