Question

Find the angle between two vectors A =5i +j B= 2i+4j helpppp!!!

Answer 1

Answer:

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solution:

Explanation:

We're asked to find the angle between two vectors, given their unit vector notations.

To do this, we can use the equation

→

A

⋅

→

B

=

A

B

cos

θ

rearranging to solve for angle,

θ

:

cos

θ

=

→

A

⋅

→

B

A

B

θ

=

arccos

⎛

⎝

→

A

⋅

→

B

A

B

⎞

⎠

where

→

A

⋅

→

B

is the dot product of the two vectors, which is

→

A

⋅

→

B

=

A

x

B

x

+

A

y

B

y

+

A

z

B

z

....

= (5×2)+(4×1)=10+4=14

A =√5^2+1^2=√26

B =√2^2+4^2=√20

Therefore we have

cos

θ= 14

$14 \div \sqrt{26 \times 20}$

14/√520

therefore angle between two vectors is 70°36`

Answer 2

Answer:

A=2i^+3j^−4k^

And B=5i^+2j^+4k^

cosθ=ABA.B

cosθ=22+32+(−4)252+22+42(2i^+3j^−4k^).(5i^+2j^+4k^)

=22+32+(−4)252+22+42(10+6−16)=0

Therefore, cosθ=0

θ=90∘

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