- A$48$
- B$42$
- ✓$44$
- D$24$
$\overrightarrow{ b }=3 \hat{ i }+5 \hat{ k }$
$\overrightarrow{ c }=\hat{ i }-\hat{ j }+2 \hat{ k }$
$\overrightarrow{ d }=\lambda(\overrightarrow{ a } \times \overrightarrow{ b })=\lambda\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & 7 & -1 \\ 3 & 0 & 5\end{array}\right|$
$\overrightarrow{ d }=\lambda(35 \hat{ i }-13 \hat{ j }-21 \hat{ k })$
$\lambda(35+13-42)=12$
$\lambda=2$
$\overrightarrow{ d }=2(35 \hat{ i }-13 \hat{ j }-21 \hat{ k })$
$(\hat{ i }+\hat{ j }-\hat{ k })(\overrightarrow{ c } \times \overrightarrow{ d })$
$=\left|\begin{array}{ccc}-1 & 1 & -1 \\ 1 & -1 & 2 \\ 70 & -26 & -42\end{array}\right|=44$
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$\overline{A B}=-2 \hat{i}+\hat{j}+3 \hat{k}$
$\overline{C B}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$
$\overline{C A}=4 \hat{i}+3 \hat{j}+\delta \hat{k}$
If $\delta > 0$ and the area of the triangle $ABC$ is $5 \sqrt{6}$, then $\overline{C B} \cdot \overline{C A}$ is equal to