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Question: Are the four points \[A\left( {1, - 1,1} \right),B\left( { - 1,1,1} \right),C\left( {1,1,1} \right)\...

Are the four points A(1,−1,1),B(−1,1,1),C(1,1,1)  and  D(2,−3,4)A\left( {1, - 1,1} \right),B\left( { - 1,1,1} \right),C\left( {1,1,1} \right)\,\,{\text{and}}\,\,D\left( {2, - 3,4} \right) are coplanar? Justify your answer.

Explanation

Solution

In this problem, first we need to find the position vectors of the given points. Next, find the vectors joining the points AB,AC  and  ADAB,AC\,\,{\text{and}}\,\,AD. Then, apply the condition of the points to be coplanar. Four points will be coplanar if the volume generated by the points is zero.

Complete step by step answer:
The given points are A(1,−1,1),B(−1,1,1),C(1,1,1)  and  D(2,−3,4)A\left( {1, - 1,1} \right),B\left( { - 1,1,1} \right),C\left( {1,1,1} \right)\,\,{\text{and}}\,\,D\left( {2, - 3,4} \right).
The position vectors of the points A,B,C,DA,B,C,D are shown below.

a⃗=i^−j^+k^ b⃗=−i^+j^+k^ c⃗=i^+j^+k^ d⃗=2i^−3j^+4k^  \vec a = \hat i - \hat j + \hat k \\\ \vec b = - \hat i + \hat j + \hat k \\\ \vec c = \hat i + \hat j + \hat k \\\ \vec d = 2\hat i - 3\hat j + 4\hat k \\\

The vector joining the points AA and BB is calculated as follows:

     AB→=b⃗−a⃗ ⇒AB→=−i^+j^+k^−i^+j^−k^ ⇒AB→=−2i^+2j^  \,\,\,\,\,\overrightarrow {AB} = \vec b - \vec a \\\ \Rightarrow \overrightarrow {AB} = - \hat i + \hat j + \hat k - \hat i + \hat j - \hat k \\\ \Rightarrow \overrightarrow {AB} = - 2\hat i + 2\hat j \\\

The vector joining the points AA and CC is calculated as follows:

     AC→=c⃗−a⃗ ⇒AC→=i^+j^+k^−i^+j^−k^ ⇒AC→=2j^  \,\,\,\,\,\overrightarrow {AC} = \vec c - \vec a \\\ \Rightarrow \overrightarrow {AC} = \hat i + \hat j + \hat k - \hat i + \hat j - \hat k \\\ \Rightarrow \overrightarrow {AC} = 2\hat j \\\

The vector joining the points AA and DD is calculated as follows:

     AD→=d⃗−a⃗ ⇒AD→=2i^−3j^+4k^−i^+j^−k^ ⇒AD→=i^−2j^+3k^  \,\,\,\,\,\overrightarrow {AD} = \vec d - \vec a \\\ \Rightarrow \overrightarrow {AD} = 2\hat i - 3\hat j + 4\hat k - \hat i + \hat j - \hat k \\\ \Rightarrow \overrightarrow {AD} = \hat i - 2\hat j + 3\hat k \\\

For, points A,B,C,DA,B,C,D to be coplanar, there exist scalars x,yx,y such that,

      AB→=x⋅AC→+y⋅AD→ ⇒−2i^+2j^=x(2j^)+y(i^−2j^+3k^) ⇒−2i^+2j^=yi^+(2x−2y)j^+3yk^  \,\,\,\,\,\,\overrightarrow {AB} = x \cdot \overrightarrow {AC} + y \cdot \overrightarrow {AD} \\\ \Rightarrow - 2\hat i + 2\hat j = x\left( {2\hat j} \right) + y\left( {\hat i - 2\hat j + 3\hat k} \right) \\\ \Rightarrow - 2\hat i + 2\hat j = y\hat i + \left( {2x - 2y} \right)\hat j + 3y\hat k \\\

By equation the vectors,

y=−2                                            ......(1) 2x−2y=2                                  ......(2) 3y=0                                             ......(3)  y = - 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 1 \right) \\\ 2x - 2y = 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 2 \right) \\\ 3y = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 3 \right) \\\

From equation (1), y=−2y = - 2, and from equation (3), y=0y = 0 which is not possible.
Thus, the points A,B,C,DA,B,C,D are not coplanar.

Note: In this problem, we need to form three vectors using the given points. Further, we need to find if there exist any linear relationship or not.