Vector multiplication can be completed using the Dot Product or Cross Product. The result of the cross product is a vector. We can calculate the cross product using geometric or algebraic vectors.
The Algebraic Cross Product of two vectors \( \vec{a} = \left(a_{1},a_{2},a_{3}\right), \vec{b} = \left(b_{1},b_{2},b_{3}\right) \) is defined by:
\(\vec{a}\times\vec{b} = \left(a_{2}b_{3}-a_{3}b_{2},a_{3}b_{1}-a_{1}b_{3}, a_{1}b_{2}-a_{2}b_{1} \right)\)
First, we can calculate \(\vec{a}\times\vec{b}\):
\(\vec{a}\times\vec{b} = \left(a_{2}b_{3}-a_{3}b_{2},a_{3}b_{1}-a_{1}b_{3}, a_{1}b_{2}-a_{2}b_{1} \right)\)
\(\vec{a}\times\vec{b} = (4(-5)-6(2), 6(-1)-2(-5), 2(2)-4(-1))\)
\(\vec{a}\times\vec{b} = (-20-12, -6+10, 4+4)\)
\(\vec{a}\times\vec{b} = (-32, 4, 8) \)
Next, we can calculate \(\vec{b}\times\vec{a}\):
\(\vec{b}\times\vec{a} = \left(b_{2}a_{3}-b_{3}a_{2},b_{3}a_{1}-b_{1}a_{3}, b_{1}a_{2}-b_{2}a_{1} \right)\)
\(\vec{b}\times\vec{a} = 6(2)-4(-5), 2(-5)-6(-1), 4(-1)-2(2)\)
\( \vec{b}\times\vec{a} = (12+20, -10+6, -4-4)\)
\( \vec{b}\times\vec{a} = (32,-4,-8)\)
Since \( \vec{a}\times\vec{b} \neq \vec{b}\times\vec{a} \), we can determine that Cross Product is NOT commutative.
\(\vec{a}\times(\vec{b}+\vec{c})\)
First, we can add the vectors \(\vec{b}\) and \(\vec{c}\):
\(\vec{b}+\vec{c}= (-1+4, 2+3, -5+(-1))\)
\(= (3, 5, -6)\)
Next, we can perform the Cross Product between \(\vec{a}\) and \((\vec{b}+\vec{c})\):
\(\vec{a}\times(\vec{b}+\vec{c}) = (2,4,6) \times (3, 5, -6)\)
\(\vec{a}\times(\vec{b}+\vec{c}) = (4(-6)-6(5), 6(3)-2(-6), 2(5)-4(3))\)
\(\vec{a}\times(\vec{b}+\vec{c}) = (-24-30, 18+12, 10-12) \)
\(= (-54,30,-2)\)
Therefore, we can determine \(\vec{a}\times(\vec{b}+\vec{c})\) is \((-54,30,-2)\).
\( (\vec{a}\times\vec{b}) + (\vec{a}\times\vec{c}) \)
First calculate the cross product then add the vectors:
\( \vec{a}\times\vec{c} = (-4-18, 24+2, 6-16) \)
\(= (-22, -26, -10) \)
\(\therefore (\vec{a}\times\vec{b}) + (\vec{a}\times\vec{c}) \)
\(= (-32, 4, 8) + (-22, -26, -10)\)
\(= (-54,30,-2)\)
Since the result from i) and ii) is the same - it demonstrates that the cross product is distributive:
\(\vec{a}\times(\vec{b}+\vec{c}) = (\vec{a}\times\vec{b}) + (\vec{a}\times\vec{c})\)
\( (\vec{a}\times\vec{b})\times\vec{c}\)
We can calculate \((\vec{a}\times\vec{b})\) as such:
\((\vec{a}\times\vec{b})\times\vec{c} = (-32, 4, 8) \times (4,3,-1)\)
\((\vec{a}\times\vec{b})\times\vec{c} = (-4-24, -32-32, -96-16)\)
\(=(-28,0,-112)\)
Therefore, we can determine that the Cross Product of \((\vec{a}\times\vec{b})\) is \(\boldsymbol{(-28,0,-112)}\).
\(\vec{a}\times(\vec{b}\times\vec{c})\)
We can calculate \((\vec{b}\times\vec{c})\) as such:
\( \vec{a}\times(\vec{b}\times\vec{c}) = (2,4,6) \times (13,-21,-11) \)
\(\vec{a}\times(\vec{b}\times\vec{c}) = (-44+126, 78+22, -42-52)\)
\(= (82, 100, -94) \)
Since the results from the above questions are different, we can determine the cross product is not associative.