If -a-16-b-4- then- -a-b-2-a-b-2-

The correct option is B 64|a|=16, nbsp;|b|=4, then, √( |a×b|^2+|a.b|^2) =√(|a|^2|b|^2sin^2 θ+|a|^2|b|^2cos^2 θ) =√(|a|^2|b|^2)=|a|.|b|=16× 4 =64 nbsp; ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Definition of Vector

If |a⃗|=16,|b...

Question

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 64
$| \to a | = 16,$ $| \to b | = 4,$ then, $\sqrt{| \to a \times \to b |^{2} + | \to a . \to b |^{2}} = \sqrt{| \to a |^{2} | \to b |^{2} {sin}^{2} θ + | \to a |^{2} | \to b |^{2} {cos}^{2} θ}$ $= \sqrt{| \to a |^{2} | \to b |^{2}} = | \to a | . | \to b | = 16 \times 4 = 64$

Suggest Corrections

Similar questions

Nonzero vectors $\to a, \to b, \to c$ satisfy $\to a . \to b$ =0, $(\to b - \to a) . (\to b + \to c)$ = 0 and 2 $| \to b + \to c | = | \to b - \to a |$ .