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  4. If | veca| = 16 , | vecb | = 4 , then , √ | veca × vecb|2 + | veca . vecb|2 =

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

KCETKCET 2019Vector Algebra

Solution:

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