Physics - Vectors Question with Solution | TestHub

The area of a triangle with adjacent sides $\vec{P}$ and $\vec{Q}$ is given by $\frac{1}{2}|\vec{P}\times \vec{Q}|$.

Calculating the cross product of the given vectors:

$\vec{P}\times \vec{Q}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -4& 2\\ 1& 1& -2\end{array}\right|$

$\vec{P}\times \vec{Q}=\hat{i}(8-2)-\hat{j}(-6-2)+\hat{k}(3+4)=6\hat{i}+8\hat{j}+7\hat{k}$

Finding the magnitude of the cross product:

$|\vec{P}\times \vec{Q}|=\sqrt{6^2+8^2+7^2}=\sqrt{36+64+49}=\sqrt{149}$

The area of the triangle is:

$\text{Area}=\frac{1}{2}\sqrt{149}\approx \frac{12.2}{2}=6.1$

Approximating to the nearest integer, the numerical value of the area is $6$.

Answer: $6{\text{m}}^2$