Physics - Vectors Question with Solution | TestHub

$\vec{a}+3\vec{b}=(\hat{i}+\hat{j}+2\hat{k})+3(3\hat{i}+2\hat{j}-\hat{k})=10\hat{i}+7\hat{j}-\hat{k}$

$2\vec{a}-\vec{b}=(2\hat{i}+2\hat{j}+4\hat{k})-(3\hat{i}+2\hat{j}-\hat{k})=-\hat{i}+0\hat{j}+5\hat{k}$

$(\vec{a}+3\vec{b})\cdot (2\vec{a}-\vec{b})=(10)(-1)+(7)(0)+(-1)(5)=-10+0-5=-15$

The magnitude of a scalar is its absolute value.

$\left|(\vec{a}+3\vec{b})\cdot (2\vec{a}-\vec{b})\right|=|-15|=15$