Physics - Vectors Question with Solution | TestHub

The magnitude of $\vec{A}$ is $|\vec{A}|=\sqrt{6^2+(-8{)}^2+10^2}=\sqrt{36+64+100}=\sqrt{200}=10\sqrt{2}$. The unit vector is $\hat{A}=\frac{\vec{A}}{|\vec{A}|}=\frac{6\hat{i}-8\hat{j}+10\hat{k}}{10\sqrt{2}}=\frac{3}{5\sqrt{2}}\hat{i}-\frac{4}{5\sqrt{2}}\hat{j}+\frac{1}{\sqrt{2}}\hat{k}$. The component of $\vec{A}$ along the x-axis is $A_x=6$. The angle with the z-axis is given by $\cos \gamma =\frac{A_z}{|\vec{A}|}=\frac{10}{10\sqrt{2}}=\frac{1}{\sqrt{2}}$, so $\gamma =45^{\circ }$.