Question
What is the angle between the following vectors?
⃗A=^i+^j+^k and ⃗B=−2^i−2^j−2^k.
⃗A=^i+^j+^k and ⃗B=−2^i−2^j−2^k.
Answer: Option D
:
D
⃗A.⃗B=|A||B|cosθor,cosθ=⃗A.⃗B|A||B| ----------------(i)
But, ⃗A.⃗B=(^i+^j+^k).(−2^i−2^j−2^k)
⃗A.⃗B=−2−2−2
⃗A.⃗B=−2−2−2=−6
Again A = |⃗A|=√(1)2+(1)2+(1)2=√3;
B = |⃗B|=√(−2)2+(−2)2+(−2)2=√12=2√3
Now,cosθ=−6√3×2√3=−1⇒θ=180∘
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:
D
⃗A.⃗B=|A||B|cosθor,cosθ=⃗A.⃗B|A||B| ----------------(i)
But, ⃗A.⃗B=(^i+^j+^k).(−2^i−2^j−2^k)
⃗A.⃗B=−2−2−2
⃗A.⃗B=−2−2−2=−6
Again A = |⃗A|=√(1)2+(1)2+(1)2=√3;
B = |⃗B|=√(−2)2+(−2)2+(−2)2=√12=2√3
Now,cosθ=−6√3×2√3=−1⇒θ=180∘
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