Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ and…
Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$ be two vectors. If $A_1$ is the area of the quadrilateral having $\vec{a}, \vec{b}$ as its diagonals and $A_2$ is the area of the parallelogram having $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ as its two adjacent sides, then $\mathrm{A}_1 \cdot \mathrm{A}_2=$
26
$\frac{27}{2}$
52
27
Solution
$\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}$ and $\vec{b}=\hat{i}-2 \hat{j}-3 \hat{k}$
$A_1=$ Area of the quadrilateral having $\vec{a}, \vec{b}$ as diagonals
$A_1=\frac{1}{2}|\vec{a} \times \vec{b}|$
Now,
$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 1 & -2 & -3\end{array}\right|$
$\begin{aligned} & =\hat{i}(-6+6)-\hat{j}(-3-3)+\hat{k}(-2-2) \\ & =0 \hat{i}+6 \hat{j}-4 \hat{k} \\ & \Rightarrow \quad \vec{a} \times \vec{b}=6 \hat{j}-4 \hat{k} \\ & \Rightarrow|\vec{a} \times \vec{b}|=\sqrt{36+16}=\sqrt{52}\end{aligned}$
Then $A_1=\frac{1}{2} \times \sqrt{52}$
Now, $A_2=$ Area of parallelogram having $\vec{a}, \vec{b}$ as its adjacent sides
$\Rightarrow \quad A_2=|\vec{a} \times \vec{b}|=\sqrt{52}$
So, $A_1 \cdot A_2=\frac{1}{2} \times \sqrt{52} \times \sqrt{52}=\frac{52}{2}=26$.