The area of the parallelogram for which the vectors $\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ are adjacent sides is equal to

Question

Aicharya · Accepted Answer

Let $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\mathbf{b}=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$

$\therefore$ Area of parallelogram $=|\mathbf{a} \times \mathbf{b}|$ $\mathbf{a} \times \mathbf{b}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 2 \\ 3 & -2 & 1\end{array}\right|$ $\begin{aligned} & =\hat{\mathbf{i}}[1-(-4)]-\hat{\mathbf{j}}(1-6)+\hat{\mathbf{k}}(-2-3) \\ & =5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}\end{aligned}$ $\therefore|\mathbf{a} \times \mathbf{b}|=\sqrt{5^2+5^2+(-5)^2}$ $=5 \sqrt{3}$

The area of the parallelogram for which the vectors $\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$…

Solution