Question

If $|\vec{a}|=16,|\vec{b}|=4$ , then , $\sqrt{|\vec{a}\times \vec{b}{|}^2+|\vec{a}.\vec{b}{|}^2}$ =

• A. 16
• B. 4
• C. 64
• D. 8

Answer 1

Answer: (C)

$\sqrt{{\left(\vec{a}\times \vec{b}\right)}^2+{\left|\vec{a}.\vec{b}\right|}^2}=\sqrt{{\left|\vec{a}\right|}^2{\left|\vec{b}\right|}^2}=\left|\vec{a}\right|\left|\vec{b}\right|=\left(16\right)\left(4\right)=64$