The angle between the two vectors vector a is equal to 3 icap + 4 j cap + 5 k cap and vector b is equal to 3 icap + 4 j cap + 5 k cap is
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let p⃗ =3i+4j+5kp→=3i+4j+5k, q⃗ =3i+4j−5kq→=3i+4j−5k, and the angle between p⃗ p→ and q⃗ q→ is θθ. We know that:
cos(θ)=p⃗ ⋅q⃗ |p⃗ |×|q⃗ |cos(θ)=p→⋅q→|p→|×|q→|
Taking dot product of the two vectors gives us:
p⃗ ⋅q⃗ =(3×3)+(4×4)+(5×−5)=9+16−25=0,p→⋅q→=(3×3)+(4×4)+(5×−5)=9+16−25=0,
hence:
cos(θ)=0|p⃗ |×|q⃗ |=0⇔θ=arccos(0)=π2
cos(θ)=p⃗ ⋅q⃗ |p⃗ |×|q⃗ |cos(θ)=p→⋅q→|p→|×|q→|
Taking dot product of the two vectors gives us:
p⃗ ⋅q⃗ =(3×3)+(4×4)+(5×−5)=9+16−25=0,p→⋅q→=(3×3)+(4×4)+(5×−5)=9+16−25=0,
hence:
cos(θ)=0|p⃗ |×|q⃗ |=0⇔θ=arccos(0)=π2
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