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lalola82
@lalola82
July 2022
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вычислить площадь параллелограмма,
построенного на векторах а= (2;4) и b=(5;-3)
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kirichekov
Verified answer
S=|a|*|b|*sinα
a{2;4}, |a|=√(2²+4²), |a|=√20, |a|=2√5
b{5;-3}, |b|=√(5²+(-3)²), |b|=√34
cosα=(a*b)/(|a|*|b|)
a*b=2*5+4*(-3)=-2
cosα=-2/(2√5*√34), cosα=-1/√170
sinα=√(1-(-1/√170)²), sinα=13/√170
S=2√5*√34*(13/√170),
S=26
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lalola82
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Answers & Comments
Verified answer
S=|a|*|b|*sinαa{2;4}, |a|=√(2²+4²), |a|=√20, |a|=2√5
b{5;-3}, |b|=√(5²+(-3)²), |b|=√34
cosα=(a*b)/(|a|*|b|)
a*b=2*5+4*(-3)=-2
cosα=-2/(2√5*√34), cosα=-1/√170
sinα=√(1-(-1/√170)²), sinα=13/√170
S=2√5*√34*(13/√170),
S=26