Physics - Vectors Question with Solution | TestHub

Given vectors are:
$\vec{A}=\hat{i}+\hat{j}+3\hat{k}$
$\vec{B}=-\hat{i}+\hat{j}+4\hat{k}$
$\vec{C}=2\hat{i}-2\hat{j}-8\hat{k}$

Calculating $\vec{P}=\vec{A}+\vec{B}+\vec{C}$:
$\vec{P}=(1-1+2)\hat{i}+(1+1-2)\hat{j}+(3+4-8)\hat{k}$
$\vec{P}=2\hat{i}-\hat{k}$

Calculating $\vec{Q}=\vec{A}\times \vec{B}$:
$\vec{Q}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 3\\ -1& 1& 4\end{array}\right|$
$\vec{Q}=\hat{i}(4-3)-\hat{j}(4-(-3))+\hat{k}(1-(-1))$
$\vec{Q}=\hat{i}-7\hat{j}+2\hat{k}$

To find the angle between $\vec{P}$ and $\vec{Q}$, we calculate their dot product:
$\vec{P}\cdot \vec{Q}=(2\hat{i}-\hat{k})\cdot (\hat{i}-7\hat{j}+2\hat{k})$
$\vec{P}\cdot \vec{Q}=(2)(1)+(0)(-7)+(-1)(2)$
$\vec{P}\cdot \vec{Q}=2-2=0$

Since the dot product is zero, the vectors $\vec{P}$ and $\vec{Q}$ are perpendicular to each other. Therefore, the angle between them is $90^{\circ }$.

Answer: $90^{\circ }$