Physics - Vectors Question with Solution | TestHub

Since the vectors $\vec{P}$ and $\vec{Q}$ are perpendicular to each other, their dot product is zero.

$\vec{P}\cdot \vec{Q}=0$

$(a\hat{i}+a\hat{j}+3\hat{k})\cdot (a\hat{i}-2\hat{j}-\hat{k})=0$

$a^2-2a-3=0$

$(a-3)(a+1)=0$

$a=3$ or $a=-1$

The positive value of $a$ is $3$.

Answer: $3$