If the magnitudes of two vectors A and B are 3 and 4 respectively and their scalar product is 6 find the angle between them and also find their vector product
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Answer:
60° , 6√3
Explanation:
for angle
A.B= ABCOSΦ
6÷12=COSΦ
Φ= COS INVERSE (1÷2)
Φ=60°
VECTR PRODUCT
A×B=ABSIN60
A×B = 4×3√3/2
A×B=6√3
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