Find the magnitude of the vector $\vec{a}=3\hat{i}+4\hat{j}-12\hat{k}$.

Solution

1. The magnitude of a vector $\vec{a}=x\hat{i}+y\hat{j}+z\hat{k}$ is defined by the formula $|\vec{a}|=\sqrt{x^2+y^2+z^2}$.
2. Substitute the given components $x=3$, $y=4$, and $z=-12$ into the formula: $|\vec{a}|=\sqrt{3^2+4^2+(-12{)}^2}$.
3. Calculate the squares to get $$\begin{array}{rl}|\vec{a}|& =\sqrt{9+16+144}\\ & =\sqrt{169}\end{array}$$.
4. Simplify the square root to find the magnitude: $|\vec{a}|=13$.
5. Hence the answer is (A).