已知i j为互相垂直的单位向量,a=4i-j,b=i+2j,c=2i-3j 计算 向量A点乘向量A+3(A*B)-2(b*C)+1
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![已知i j为互相垂直的单位向量,a=4i-j,b=i+2j,c=2i-3j 计算 向量A点乘向量A+3(A*B)-2(b*C)+1](/uploads/image/z/6253242-42-2.jpg?t=%E5%B7%B2%E7%9F%A5i+j%E4%B8%BA%E4%BA%92%E7%9B%B8%E5%9E%82%E7%9B%B4%E7%9A%84%E5%8D%95%E4%BD%8D%E5%90%91%E9%87%8F%2Ca%3D4i-j%2Cb%3Di%2B2j%2Cc%3D2i-3j+%E8%AE%A1%E7%AE%97+%E5%90%91%E9%87%8FA%E7%82%B9%E4%B9%98%E5%90%91%E9%87%8FA%2B3%EF%BC%88A%2AB%EF%BC%89-2%EF%BC%88b%2AC%29%2B1)
已知i j为互相垂直的单位向量,a=4i-j,b=i+2j,c=2i-3j 计算 向量A点乘向量A+3(A*B)-2(b*C)+1
已知i j为互相垂直的单位向量,a=4i-j,b=i+2j,c=2i-3j 计算 向量A点乘向量A+3(A*B)-2(b*C)+1
已知i j为互相垂直的单位向量,a=4i-j,b=i+2j,c=2i-3j 计算 向量A点乘向量A+3(A*B)-2(b*C)+1
郭敦顒回答:
向量a•向量a=|a|•|a| cos0=17;
3(向量a•向量b)=3| a | • | b | cosθ1,θ1=∠AOB,
∵|AB|=√[(4-1)²+(-1-2)²]=3√2,
|a|=√17,|b|=√5,
∴cosθ1=(17+5-18)/(2√85)=2/√85=0.21693,
3(向量a•向量b)=3|a| • |b| cosθ
=3(√17 •√5)×2/√85=6;
2(向量b•向量c)=2|b| • |c| cosθ2,θ2=∠COB,|c|=√13,
∵|BC|=√[(1-2)²+(2+5)²]=5√2
∴cosθ2=(5+13-50)/(2√65)=-16/√65,
2(向量b•向量c)=2|b| • |c| cosθ2
=(2√65)(-16/√65)=-32,
∴向量A点乘向量A+3(A*B)-2(b*C)+1
=17+6+32+1=56.
B
b=i+2j
O
a=4i-j
A
c=2i-3j
C