Question-29 If the points A(2 − x, 2, 2), B(2, 2 − y, 2), C(2, 2, 2 − z) and D(1, 1, 1) are coplanar, then the locus of point 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
Answer: (A)
Explanation:
\overrightarrow{AB}=x\hat{i}-y\hat{j} \overrightarrow{AC}=x\hat{i}-z\hat{k} \overrightarrow{AD}=(x-1)\hat{i}-\hat{j}-\hat{k}A, B, C, D are coplanar.
⇒ \begin{vmatrix}AB&AC&AD\end{vmatrix}=0
\Rightarrow\begin{vmatrix}x&-y&0\\x&0&-z\\x-1&-1&-1\end{vmatrix}=0⇒ x(-z)+y(-x+z(x-1))=0
⇒ -xz-xy+xyz-zy=0
⇒ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1