MCQ
Let $\vec{a}=2 \hat{i}+7 \hat{j}-\hat{k}, \vec{b}=3 \hat{i}+5 \hat{k}$ and $\vec{c}=\hat{i}-\hat{j}+2 \hat{k}$ Let $\vec{d}$ be a vector which is perpendicular to both $\overrightarrow{ a }$ and $\overrightarrow{ b }, \quad$ and $\quad \overrightarrow{ c } \cdot \overrightarrow{ d }=12$. Then $(-\hat{i}+\hat{j}-\hat{k}) \cdot(\vec{c} \times \vec{d})$ is equal to $........$.
  • A
    $48$
  • B
    $42$
  • $44$
  • D
    $24$

Answer

Correct option: C.
$44$
c
$\overrightarrow{ a }=2 \hat{ i }+7 \hat{ j }-\hat{ k }$

$\overrightarrow{ b }=3 \hat{ i }+5 \hat{ k }$

$\overrightarrow{ c }=\hat{ i }-\hat{ j }+2 \hat{ k }$

$\overrightarrow{ d }=\lambda(\overrightarrow{ a } \times \overrightarrow{ b })=\lambda\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & 7 & -1 \\ 3 & 0 & 5\end{array}\right|$

$\overrightarrow{ d }=\lambda(35 \hat{ i }-13 \hat{ j }-21 \hat{ k })$

$\lambda(35+13-42)=12$

$\lambda=2$

$\overrightarrow{ d }=2(35 \hat{ i }-13 \hat{ j }-21 \hat{ k })$

$(\hat{ i }+\hat{ j }-\hat{ k })(\overrightarrow{ c } \times \overrightarrow{ d })$

$=\left|\begin{array}{ccc}-1 & 1 & -1 \\ 1 & -1 & 2 \\ 70 & -26 & -42\end{array}\right|=44$

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