Problems from Vector product

Development:

We apply the formula of the vector product:

$$$\vec{a}\times\vec{b}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 1 & -1 \\ 4 & 2 & -2 \end{vmatrix}= \begin{vmatrix} 1 & -1 \\ 2 & -2 \end{vmatrix} \vec{i} + (-1)\begin{vmatrix} 2 & -1 \\ 4 & -2 \end{vmatrix} \vec{j} + \begin{vmatrix} 1 & -1 \\ 2 & -2 \end{vmatrix} \vec{k}=(0,0,0)$$$

$$$\vec{b}\times\vec{c}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 4 & 2 & -2 \\ 0 & -1 & -1 \end{vmatrix}= \begin{vmatrix} 2 & -2 \\ -1 & -1 \end{vmatrix} \vec{i} + (-1)\begin{vmatrix} 4 & -2 \\ 0 & -1 \end{vmatrix} \vec{j} + \begin{vmatrix} 4 & -1 \\ 0 & -1 \end{vmatrix} \vec{k}=(-4,4,-4)$$$

$$$\vec{a}\times\vec{c}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 1 & -1 \\ 0 & -1 & -1 \end{vmatrix}= \begin{vmatrix} 1 & -1 \\ -1 & -1 \end{vmatrix} \vec{i} + (-1)\begin{vmatrix} 2 & -1 \\ 0 & -1 \end{vmatrix} \vec{j} + \begin{vmatrix} 1 & -1 \\ 0 & -1 \end{vmatrix} \vec{k}=(-2,2,-2)$$$

Solution:

$$(0,0,0)$$, $$(-4,4,-4)$$ and $$(-2,2,-2)$$

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Development:

We apply the formula of the vector product:

$$$\vec{a}\times\vec{b}=\vec{c}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{vmatrix}= \begin{vmatrix} 1 & 0 \\ 1 & 0 \end{vmatrix} \vec{i} + (-1)\begin{vmatrix} 0 & 0 \\ 1 & 0 \end{vmatrix} \vec{j} + \begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix} \vec{k}=(0,0,-1)$$$

$$$\vec{b}\times\vec{a}=\vec{c}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 1 & 0 \\ 0 & 1 & 0 \end{vmatrix}= \begin{vmatrix} 1 & 0 \\ 1 & 0 \end{vmatrix} \vec{i} + (-1)\begin{vmatrix} 1 & 0 \\ 0 & 0 \end{vmatrix} \vec{j} + \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} \vec{k}=(0,0,1)$$$

Solution:

$$(0,0,-1)$$ and $$(0,0,1)$$

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Development:

We apply the formula of the vector product:

$$$\vec{c}=(c_1,c_2,c_3)=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}= \begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix} \vec{i} + (-1)\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix} \vec{j} + \begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix} \vec{k}$$$

$$$\vec{a}\times\vec{b}=\vec{c}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 2 & 3 \\ -2 & 1 & 0 \end{vmatrix}= \begin{vmatrix} 2 & 3 \\ 1 & 0 \end{vmatrix} \vec{i} + (-1)\begin{vmatrix} 1 & 3 \\ -2 & 0 \end{vmatrix} \vec{j} + \begin{vmatrix} 1 & 2 \\-2 & 1 \end{vmatrix} \vec{k}=(-3,-6,5)$$$

$$$\vec{b}\times\vec{a}=\vec{c}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -2 & 1 & 0 \\ 1 & 2 & 2 \end{vmatrix}= \begin{vmatrix} 1 & 0 \\ 2 & 3 \end{vmatrix} \vec{i} + (-1)\begin{vmatrix} -2 & 0 \\ 1 & 3 \end{vmatrix} \vec{j} + \begin{vmatrix} -2 & 1 \\ 1 & 2 \end{vmatrix} \vec{k}=(3,6,-5)$$$

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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