Question

The coplanar points A, B, C, D are $\left( {2 - x,2,2} \right)$, $\left( {2,2 - y,2} \right)$,$\left( {2,2,2 - z} \right)$ and $\left( {1,1,1} \right)$ 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

Hint: We have 4 points and we can make 3 vectors by taking a common point. If the 4 points are coplanar, the 3 vectors also will be coplanar. We know that volume enclosed by 3 vectors is given by $V = \left| {\vec a.\left( {\vec b \times \vec c} \right)} \right|$. If the vectors are coplanar, their volume enclosed will be zero. Using the relation, we can find the required relation.

Complete step by step solution: We have points A $\left( {2 - x,2,2} \right)$, B $\left( {2,2 - y,2} \right)$,C $\left( {2,2,2 - z} \right)$ and D $\left( {1,1,1} \right)$.
Now we can find the vectors connecting these points. Vector joining 2 points $\overrightarrow P = {a_1}\widehat i + {b_1}\widehat j + {c_1}\widehat k$ and \[\overrightarrow Q = {a_2}\widehat i + {b_2}\widehat j + {c_2}\widehat k\] is given by $\overrightarrow {PQ} = \overrightarrow Q - \overrightarrow P $.
Thus, we get,
\[
  \overrightarrow {AB} = \overrightarrow B - \overrightarrow A \\
   = 2\widehat i + \left( {2 - y} \right)\widehat j + 2\widehat k - [\left( {2 - x} \right)\widehat i + 2\widehat j + 2\widehat k] \\
 \]
On taking like terms together and solving we get,
\[
   = \left( {2 - 2 + x} \right)\widehat i + \left( {2 - y - 2} \right)\widehat j + \left( {2 - 2} \right)\widehat k \\
   = x\widehat i - y\widehat j \\
 \]
Similarly,for vector AC, we get,
\[
  \overrightarrow {AC} = \overrightarrow C - \overrightarrow A \\
   = 2\widehat i + 2\widehat j + \left( {2 - z} \right)\widehat k - [\left( {2 - x} \right)\widehat i + 2\widehat j + 2\widehat k] \\
 \]
On taking like terms together and solving we get,
\[
   = \left( {2 - 2 + x} \right)\widehat i + \left( {2 - 2} \right)\widehat j + \left( {2 - z - 2} \right)\widehat k \\
   = x\widehat i - z\widehat k \\
 \]
And for vector AD,
\[
  \overrightarrow {AD} = \overrightarrow D - \overrightarrow A \\
   = \widehat i + \widehat j + \widehat k - [\left( {2 - x} \right)\widehat i + 2\widehat j + 2\widehat {k]} \\
 \]
On taking like terms together and solving we get,
\[
   = \left( { - 1 + x} \right)\widehat i + \left( {1 - 2} \right)\widehat j + \left( {1 - 2} \right)\widehat k \\
   = \left( {x - 1} \right)\widehat i - \widehat j - \widehat k \\
 \]
Now we have vectors, \[\overrightarrow {AB} = x\widehat i - y\widehat j\]\[\overrightarrow {AC} = x\widehat i - z\widehat k\]and \[\overrightarrow {AD} = \left( {x - 1} \right)\widehat i - \widehat j - \widehat k\]. To get the volume enclosed by the vectors, we can find the box product.
\[
   \Rightarrow V = \left| {\vec a.\left( {\vec b \times \vec c} \right)} \right| \\
   = \left| {\overrightarrow {AB} .\left( {\overrightarrow {AC} \times \overrightarrow {AD} } \right)} \right| \\
   = \left| {\begin{array}{*{20}{c}}
  x&{ - y}&0 \\
  x&0&{ - z} \\
  {x - 1}&{ - 1}&{ - 1}
\end{array}} \right| \\
   = x\left( {0 - z} \right) + y\left( { - x + zx - z} \right) + 0 \\
 \]
$ \Rightarrow V = - xz - yz + xyz - zy$
If the points are coplanar, the vectors also will be coplanar. Then the volume will be zero.
$ \Rightarrow V = - xz - yz + xyz - zy = 0$
On rearranging, we get
$yz + xz + xy = xyz$
Dividing throughout with $xyz$ , we get,
$\dfrac{{yz}}{{xyz}} + \dfrac{{xz}}{{xyz}} + \dfrac{{xy}}{{xyz}} = 1$
$ \Rightarrow \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 1$
Therefore, the required relation is $\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 1$

So, the correct answer is option A.

Note: Coplanar points are points that lie on the same plane. The vectors joining coplanar points will be coplanar vectors. Box product or scalar triple product of three vectors $\overrightarrow A = {a_1}\widehat i + {b_1}\widehat j + {c_1}\widehat k$,\[\overrightarrow B = {a_2}\widehat i + {b_2}\widehat j + {c_2}\widehat k\]and \[\overrightarrow C = {a_3}\widehat i + {b_3}\widehat j + {c_3}\widehat k\]is defined as the dot product of one vector with the cross product of the other two vectors. It is given by,
$\overrightarrow A .\left( {\overrightarrow B \times \overrightarrow C } \right) = \det \left[ {\begin{array}{*{20}{c}}
  {{a_1}}&{{b_1}}&{{c_1}} \\
  {{a_2}}&{{b_2}}&{{c_2}} \\
  {{a_3}}&{{b_3}}&{{c_3}}
\end{array}} \right]$
The modulus of scalar triple product gives the volume of the parallelepiped formed by the three vectors. If the 3 vectors lie on the same plane, the volume of the parallelepiped will be zero. So, for 3 coplanar vectors their box product is zero. We don’t have any condition for directly proving 4 points are coplanar, we need to find three vectors from the four points and then prove its box product is zero. While forming the vectors, we must make sure that all the given points must be included. For that we usually keep the initial point of the vector the same and the vector from this point to the other 3 points are taken.