Question
Let $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}$ and $\vec{\text{c}}=\text{c}_1\hat{\text{i}}+\text{c}_2\hat{\text{j}}+\text{c}_3\hat{\text{k}}.$ Then,
If $c_1= 1$ and $c_2= 2$, find $c_3$ which makes $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ coplanar.

Answer

If $c_1 = 1$ and $c_2= 2$, then $\vec{\text{a}}=\hat{\text{i}}+\vec{\text{j}}+\vec{\text{k}},\vec{\text{b}}=\hat{\text{i}}$ and $\vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}+\text{c}_3\hat{\text{k}}.$
We know that vectors $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar if $\big[\vec{\text{a}}\vec{\text{b}}{\text{c}}\big]=0.$
It is given that $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar.
$\therefore\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=0$
$\Rightarrow\begin{vmatrix}1&1&1\\1&0&0\\1&2&\text{c}_3 \end{vmatrix}=0$
$\Rightarrow1(0-0)-1(\text{c}_3-0)+1(2-0)=0$
$\Rightarrow-{\text{C}}_3+2=0$
$\Rightarrow\text{C}_3=2$

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