Problems from Vector product

If a→=(2,1,−1), b→=(4,2,−2), c→=(4,2,−2). Compute the vector product a→×b→, b→×c→ and a→×c→.

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Development:

We apply the formula of the vector product:

a→×b→=|i→j→k→21−142−2|=|1−12−2|i→+(−1)|2−14−2|j→+|1−12−2|k→=(0,0,0)

b→×c→=|i→j→k→42−20−1−1|=|2−2−1−1|i→+(−1)|4−20−1|j→+|4−10−1|k→=(−4,4,−4)

a→×c→=|i→j→k→21−10−1−1|=|1−1−1−1|i→+(−1)|2−10−1|j→+|1−10−1|k→=(−2,2,−2)

Solution:

(0,0,0), (−4,4,−4) and (−2,2,−2)

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If a→=(0,1,0), b→=(1,1,0). Compute the vector product a→×b→ and b→×a→.

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Development:

We apply the formula of the vector product:

a→×b→=c→=|i→j→k→010110|=|1010|i→+(−1)|0010|j→+|0111|k→=(0,0,−1)

b→×a→=c→=|i→j→k→110010|=|1010|i→+(−1)|1000|j→+|1101|k→=(0,0,1)

Solution:

(0,0,−1) and (0,0,1)

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If a→=(1,2,3), b→=(−2,1,0). Compute the vector product a→×b→ and b→×a→.

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Development:

We apply the formula of the vector product:

c→=(c1,c2,c3)=|i→j→k→a1a2a3b1b2b3|=|a2a3b2b3|i→+(−1)|a1a3b1b3|j→+|a1a2b1b2|k→

a→×b→=c→=|i→j→k→123−210|=|2310|i→+(−1)|13−20|j→+|12−21|k→=(−3,−6,5)

b→×a→=c→=|i→j→k→−210122|=|1023|i→+(−1)|−2013|j→+|−2112|k→=(3,6,−5)

We can see that if we switch the order in the vector product we obtain the same vector but it will be in the opposite sense (right-handed or left-handed).

Solution:

(−3,−6,5) and (3,6,−5)

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