Question

1. Find the angle between the vectors ū= i – 2j – 2k and v = 6i + 3j + 2k.​

Answer 1

Answer:

u⃗=2i+2j−k⟶∥u⃗∥=4+4+1−−−−−−−√=3

v⃗=6i−3j+2k⟶∥v⃗∥=36+9+4−−−−−−−−√=7

u⃗⋅v⃗=(2)(6)+(2)(−3)+(−1)(2)=4

cosθ=u⃗⋅v⃗∥u⃗∥∥v⃗∥=43×7=421

θ=cos−1(421)≈79.02∘

Answer 2

Answer:

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Step-by-step explanation:

Let

a

=3i+j+2k and

b

=2i−2j+4k.

Now angle between

a

and

b

be

cos

−1

(

∣

a

∣∣

b

∣

a

.

b

)=cos

−1

(

14

.

24

6−2+8

)=cos

−1

(

28

3

).