Question

If the vectors $4\hat{i}+11\hat{j}+m\hat{k},7\hat{i}+2\hat{j}+6\hat{k}\mathrm{and}\hat{i}+5\hat{j}+4\hat{k}$ are coplanar, then m =
(a) 0
(b) 38
(c) −10
(d) 10

Answer 1

$$\begin{gathered} \left(\mathrm{d}\right)10 \\ \mathrm{Let}: \\ \overrightarrow{a}=4\hat{i}+11\hat{j}+m\hat{k} \\ \overrightarrow{b}=7\hat{i}+2\hat{j}+6\hat{k} \\ \overrightarrow{c}=\hat{i}+5\hat{j}+4\hat{k} \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{vectors}\mathit{}\overrightarrow{a}\mathit{,}\mathit{}\overrightarrow{b}\mathrm{and}\mathit{}\overrightarrow{c}\mathrm{are}\mathrm{coplanar}\mathrm{iff}\mathrm{their}\mathrm{scalar}\mathrm{triple}\mathrm{product}\mathrm{is}\mathrm{zero},\mathrm{i}.\mathrm{e}.\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=0 \\ \Rightarrow \left|\begin{array}{ccc}4& 11& m\\ 7& 2& 6\\ 1& 5& 4\end{array}\right|=0 \\ \Rightarrow 4\left(8-30\right)-11\left(28-6\right)+m\left(\mathit{35}\mathit{-}\mathit{2}\right)=0 \\ \Rightarrow -88-242+33m=0 \\ \Rightarrow 33m=330 \\ \therefore m=10 \end{gathered}$$