Question

If the points $\text{A}(2 - x, 2, 2), \text{B}(2, 2 - y, 2), \text{C}(2, 2, 2 - z)$ and $\text{D}(1, 1, 1)$ are coplanar, then the locus of point $\text{P}(x, y, z)$ is

• A. $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$
• B. $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$
• C. $\frac{1}{1+x} + \frac{1}{1+y} + \frac{1}{1+z} = 1$
• D. $\frac{1}{x} + \frac{1}{2y} + \frac{1}{3z} = 0$

Hint:

Coplanarity $\Rightarrow$ scalar triple product = 0.

Answer 1

The Correct Option is A

Concept:
Four points are coplanar if the scalar triple product is zero: \[ [\vec{AB}, \vec{AC}, \vec{AD}] = 0 \] Step 1: Find vectors. \[ \vec{AB} = B - A = (2-(2-x),\, (2-y)-2,\, 2-2) = (x,\,-y,\,0) \] \[ \vec{AC} = C - A = (2-(2-x),\, 2-2,\, (2-z)-2) = (x,\,0,\,-z) \] \[ \vec{AD} = D - A = (1-(2-x),\, 1-2,\, 1-2) = (x-1,\,-1,\,-1) \]
Step 2: Form determinant. \[ \begin{vmatrix} x & -y & 0 \\ x & 0 & -z \\ x-1 & -1 & -1 \end{vmatrix} = 0 \]
Step 3: Expand determinant. \[ = x \begin{vmatrix} 0 & -z \\ -1 & -1 \end{vmatrix} + y \begin{vmatrix} x & -z \\ x-1 & -1 \end{vmatrix} + 0 \] \[ = x(0 + z) + y(-x + z(x-1)) \] \[ = xz + y[-x + zx - z] \] \[ = xz + yx(z-1) - yz \]
Step 4: Simplify. Rearranging gives: \[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1 \]
Step 5: Conclusion. \[ {\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1} \]