# Check whether the 3 vectors are co-planar or not A= i+j+k , B= i+3j+k , C= 2i+2j+2k

## Question:

Check whether the 3 vectors are co-planar or not

A= i+j+k ,

B= i+3j+k ,

C= 2i+2j+2k

## Coplanar Vectors:

Points in space are said to be coplanar if there exists a geometric plane that contains them all. Similarly, Coplanar vectors are vectors that lie on the same plane or parallel to the same plane. Moreover, vectors are coplanar when their scalar triple product is zero, that is, {eq}\vec{a}\cdot(\vec{b}\times\vec{c})=0. {/eq}

The given vectors in the problem are the following:

{eq}\begin{align*} A&= \hat{i}+\hat{j}+\hat{k} \\ \ \\ B&= \hat{i}+3\hat{j}+\hat{k} \\ \ \\ C&= 2\hat{i}+2\hat{j}+\hat{2k} \end{align*}{/eq}

To be able to know if these three vectors are coplanar, we need to solve for the scalar triple product. That is:

{eq}\begin{align*} \vec{a}\cdot(\vec{b}\times\vec{c}) &=(\hat{i}+\hat{j}+\hat{k})\cdot[(\hat{i}+3\hat{j}+\hat{k})\times(2\hat{i}+2\hat{j}+2\hat{k})] \\ \ \\ &=(\hat{i}+\hat{j}+\hat{k})\cdot[(\hat{i}\times2\hat{i})+(\hat{i}\times2\hat{j})+(\hat{i}\times2\hat{k})] \\ &+[(3\hat{j}\times2\hat{i})+(3\hat{j}\times2\hat{j})+(3\hat{j}\times2\hat{k})] \\ &+[(\hat{k}\times2\hat{i})+(\hat{k}\times2\hat{j})+(\hat{k}\times2\hat{k})] \\ \ \\ &=(\hat{i}+\hat{j}+\hat{k})\cdot[(0+2\hat{k}-2\hat{j})+(-6\hat{k}+0+6\hat{i})+(2\hat{j}-2\hat{i}+0)] \\ \ \\ &=(\hat{i}+\hat{j}+\hat{k})\cdot [4\hat{i}-4\hat{k} ] \\ \ \\ &=[(\hat{i}\cdot4\hat{i})+(\hat{i}\cdot-4\hat{k})]+[(\hat{j}\cdot4\hat{i})+(\hat{j}\cdot-4\hat{k})]+[(\hat{k}\cdot4\hat{i})+(\hat{k}\cdot-4\hat{k})] \\ \ \\ &=[(4+0)+(0+0)+(0-4)] \\ \ \\ &=4-4 \\ \ \\ &=0 \end{align*}{/eq}

Therefore the three vectors are coplanar. In the calculation above, we used the following vector identities:

{eq}\begin{align*} \hat{i}\times\hat{i}&= \hat{j}\times\hat{j} = \hat{k}\times\hat{k}=0 \\ \hat{i}\times\hat{j}&=\hat{k} \\ \hat{i}\times\hat{k}&=-\hat{j} \\ \hat{j}\times\hat{i}&=-\hat{k} \\ \hat{j}\times\hat{k}&=\hat{i} \\ \hat{k}\times\hat{i}&=\hat{j} \\ \hat{k}\times\hat{j}&=-\hat{i} \\ \hat{i}\cdot\hat{i}&= \hat{j}\cdot\hat{j} = \hat{k}\cdot\hat{k}=1 \\ \hat{i}\cdot\hat{j}&= \hat{i}\cdot\hat{k} = \hat{j}\cdot\hat{i} = \hat{j}\cdot\hat{k} = \hat{k}\cdot\hat{i} = \hat{k}\cdot\hat{j}=0 \end{align*}{/eq}