Question

3$\overrightarrow{i}$ +3$\overrightarrow{j}$ +$\sqrt{3}$ $\overrightarrow{k}$, $\overrightarrow{i}$+$\overrightarrow{k}$, $\sqrt{3}$ $\overrightarrow{i}$ + $\sqrt{3}$ $\overrightarrow{j}$ + p $\overrightarrow{k}$ are coplanar vectors then value of p will be:

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

$3\overrightarrow{i}+3\overrightarrow{j}+\sqrt{3}\overrightarrow{k},\overrightarrow{i}+\overrightarrow{k},\sqrt{3}\overrightarrow{i}+\sqrt{3}\overrightarrow{j}+p\overrightarrow{k}$ are coplanar means there determinant will be zero.

$\left[\begin{array}{ccc}3& 3& \sqrt{3}\\ 1& 0& 1\\ \sqrt{3}& \sqrt{3}& p\end{array}\right]=0$

$⟹3(-\sqrt{3})-3(p-\sqrt{3})+\sqrt{3}\left(\sqrt{3}\right)=0$

$⟹3p-3=0$

$⟹3p=3$

$⟹p=1$