MCQ
If the vectors $4\hat{\text{i}}+11\hat{\text{j}}+\text{m}\hat{\text{k}},7\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$ and $\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}$ are coplanar, then $m =$
  • A
    $0$
  • B
    $38$
  • C
    $-10$
  • $10$

Answer

Correct option: D.
$10$
Let
$\vec{\text{a}}=4\hat{\text{i}}+11\hat{\text{j}}+{\text{m}}\hat{\text{k}}$
$\vec{\text{b}}=7\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$
$\vec{\text{c}}=\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}$
We know that vectors $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ are coplanar if their scalar triple product is zero, i.e. $\big[\vec{\text{a}}\vec{\text{ b }}\vec{\text{c}}\big]=0$
$\Rightarrow\begin{vmatrix}4&11&\text{m}\\7&2&6\\1&5&4 \end{vmatrix}=0$
$\Rightarrow 4(8-30)-11(28-6)+\text{m}(35-2)=0$
$\Rightarrow-88-242+33\text{m}=0$
$\Rightarrow33\text{m}=330$
$\therefore\text{m}=10$

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