Question

Determine the points in yz-plane and are equidistant from the points A(1, –1, 0), B(2, 1, 2) and C(3, 2, –1).

Answer 1

We know that the x-coordinate of every point on the yz-plane is zero.
So, let P (0, y, z) be a point on the yz-plane such that PA = PB = PC
Now, PA = PB

\[\Rightarrow \left( 0 - 1 \right)^2 + \left( y + 1 \right)^2 + \left( z - 0 \right)^2 = \left( 0 - 2 \right)^2 + \left( y - 1 \right)^2 + \left( z - 2 \right)^2\]

\[\Rightarrow 1 + y^2 + 2y + 1 + z^2 = 4 + y^2 - 2y + 1 + z^2 - 4z + 4\]
\[ \Rightarrow 2y + 2 = - 2y - 4z + 9\]
\[ \Rightarrow 2y + 2y - 4z = 9 - 2\]
\[ \Rightarrow 4y - 4z = 7\]
\[ \Rightarrow y - z = \frac{7}{4} . . . \left( 1 \right)\]
\[PB = PC\]
\[ \Rightarrow P B^2 = P C^2 \]
\[ \Rightarrow \left( 0 - 2 \right)^2 + \left( y - 1 \right)^2 + \left( z - 2 \right)^2 = \left( 0 - 3 \right)^2 + \left( y - 2 \right)^2 + \left( z + 1 \right)^2 \]
\[ \Rightarrow 4 + y^2 - 2y + 1 + z^2 - 4z + 4 = 9 + y^2 - 4y + 4 + z^2 + 2z + 1\]
\[ \Rightarrow - 2y - 4z + 9 = - 4y + 2z + 14\]
\[ \Rightarrow - 2y + 4y - 4z - 2z = 14 - 9\]
\[ \Rightarrow 2y - 6z = 5\]
\[ \Rightarrow y - 3z = \frac{5}{2}\]
\[ \therefore y = \frac{5}{2} + 3z \left( 2 \right)\]
\[\text{ Putting the value of y in equation } \left( 1 \right): \]
\[ y - z = \frac{7}{4}\]
\[ \Rightarrow \frac{5}{2} + 3z - z = \frac{7}{4}\]
\[ \Rightarrow 2z = \frac{7}{4} - \frac{5}{2}\]
\[ \Rightarrow 2z = \frac{7 - 10}{4}\]
\[ \Rightarrow 2z = \frac{- 3}{4}\]
\[ \therefore z = \frac{- 3}{8}\]
\[\text{ Putting the value of z in equation } \left( 2 \right): \]
\[ y = \frac{5}{2} + 3z\]
\[ \Rightarrow y = \frac{5}{2} + 3\left( \frac{- 3}{8} \right)\]
\[ \Rightarrow y = \frac{5}{2} - \frac{9}{8}\]
\[ \Rightarrow y = \frac{20 - 9}{8}\]
\[ \therefore y = \frac{11}{8}\]

Hence, the required point is\[\left( 0, \frac{11}{8}, \frac{- 3}{8} \right)\]