Question

The coplanar points $A,B,C,D$ are $(2 - x,2,2)$, $(2,2 - y,2)$, $(2,2,2 - z)$ and $(1,1,1)$ respectively, then
A) $\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 1$
B) $x + y + z = 1$
C) $\dfrac{1}{{1 - x}} + \dfrac{1}{{1 - y}} + \dfrac{1}{{1 - z}} = 1$
D) None of these

Answer 1

We have given that four points $A,B,C,D$ are coplanar and we know that if four points are coplanar then the vectors obtained by containing two of them points at a time have determinant = zero(0).
If four points are coplanar then the vectors obtained by containing two of them points at a time have determinant = zero(0).

Complete step-by-step answer:
We have given with the four points
$A$=$(2 - x,2,2)$
$B$=$(2,2 - y,2)$
$C$=(2,2,2 - z)$$$$A\vec C = \vec C - \vec A
$D$=$(1,1,1)$
four points are coplanar then the vectors obtained by containing two of them points represented by
$A\vec B$,$A\vec C$,$A\vec D$.
$A\vec B = \vec B - \vec A$
$A\vec D = \vec D - \vec A$
Vectors obtained by doing the mathematical calculations.
A\vec B = $$$$ - x\hat i + y\hat j
$A\vec C$=$- x\hat i + 0\hat j + z\hat k$
A\vec D = $$$$(1 - x)\hat i + \hat j + \hat k
Now using the formula , if four points are co-planar then the vectors obtained by containing two of them points at a time have determinant = zero(0).

{ - x}&y;&0 \\\ { - x}&0&z; \\\ {1 - x}&1&1 \end{array}} \right] = 0$$ Solving the determinant we get, $$ - x( - z) - y( - x - z + xz) + 0 = 0$$ $$xz + xy + yz = xyz$$ $$\dfrac{{xz + xy + yz}}{{xyz}} = 1$$ Simplifying the equation $$\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 1$$ **Hence option (A) is correct.** **Note:** A given number of vectors are said to be coplanar if their supports are parallel to the same plane. Always remember that two vectors are always coplanar. Also two vectors are said to be collinear if their supports are parallel disregards to their direction. Remember that collinear vectors are also called Parallel vectors . If they have the same direction they are named like vectors otherwise unlike vectors.