Question
What is the angle between the following vectors?
→A=3^i−2^j+^k
→B=2^i+6^j−6^k
→A=3^i−2^j+^k
→B=2^i+6^j−6^k
Answer: Option B
:
B
→A.→B=(3×2)+(−2×6)+(1×−6)
= 6 - 12 - 6
= -12
|→A|=√32+(−2)2+12=√9+4+1=√14
|→B|=√22+62+(−6)2=√4+36+36=√76
→A.→B|→A||→B|=−12√14√76=−6√266
∴cosθ=−6√266
⇒θ=cos−1(−6√266)
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:
B
→A.→B=(3×2)+(−2×6)+(1×−6)
= 6 - 12 - 6
= -12
|→A|=√32+(−2)2+12=√9+4+1=√14
|→B|=√22+62+(−6)2=√4+36+36=√76
→A.→B|→A||→B|=−12√14√76=−6√266
∴cosθ=−6√266
⇒θ=cos−1(−6√266)
Was this answer helpful ?
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