If | a|=4, | b|=2 and the angle between a and b is /6, then | a b|^2 is equal to: · questionr

questionrEarly access

Question 377 of 379EasyMCQ

If

|\vec a|=4

,

|\vec b|=2

and the angle between

\vec a

and

\vec b

is

\pi/6

, then

|\vec a\times\vec b|^2

is equal to:

Solutions · 1

Official solution

\vec a\cdot\vec b=|\vec a||\vec b|\cos\dfrac\pi6=4\cdot2\cdot\dfrac{\sqrt{3}}2=4\sqrt{3}

, so

(\vec a\cdot\vec b)^2=48

. Using

|\vec a\times\vec b|^2+(\vec a\cdot\vec b)^2=|\vec a|^2|\vec b|^2

: \[|\vec a\times\vec b|^2 = 16\times4-48=64-48=16\]

Answer the question to unlock