Cross Product (Algebraic)

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.

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 \((-54,30,-2)\).

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:

\(\vec{a}\times(\vec{b}+\vec{c}) = (\vec{a}\times\vec{b}) + (\vec{a}\times\vec{c})\)

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 \((-28,0,-112)\).

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.