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If U = 10±0.05 and V=20±0.01, find the relative |
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Answer» Answer: A⃗ =−2i^+3j^ and B⃗ =3i^+2j^ Dot product of TWO vector is A⃗ ⋅B⃗ =(−2i^+3j^)⋅(B⃗ =3i^+2j^) A⃗ ⋅B⃗ =(−2)(3)+(3)(2)=0 The dot product of two vector is ZERO which implies that they are perpendicular to each other. This can be obtained as fallows A⃗ ⋅B⃗ =0 ABcosθ=0 Where A and B are the magnitude of two vectors A⃗ and B⃗ and θ is the angle between them. ⇒cosθ=0 ⇒θ=cos−1(0) ⇒θ=900 vector product is given by A×B=∣∣∣∣∣i^−23j^32k^00∣∣∣∣∣ A×B=i^(0–0)−j^(0–0)+k^(−4–9) A×B=−13k^ or you can do using the properties i^×j^=j^×j^=k^×k^=0 i^×j^=k^,j^×k^=i^andk^×i^=j^(clockwise) i^×k^=−j^,j^×i^=−k^andk^×j^=−i^(anticlockwise) A⃗ ×B⃗ =(−2i^+3j^)×(3i^+2j^) ⇒A×B=−4(i^×j^)+9(j^×i^) ∴A×B=−4k^+9(−k^)=−13k^ |
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