Let vec{a} be a non-zero vector. If vec{x}=ha | vedclass

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(C) Using the vector triple product formula $\vec{A} 	imes (\vec{B} 	imes \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C}

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(C) Using the vector triple product formula $\vec{A} 	imes (\vec{B} 	imes \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C}$:$\vec{x} = \hat{i} 	imes (\vec{a} 	imes \hat{i}) = (\hat{i} \cdot \hat{i})\vec{a} - (\hat{i} \cdot \vec{a})\hat{i} = \vec{a} - (\hat{i} \cdot \vec{a})\hat{i}$$\vec{y} = \hat{j} 	imes (\vec{a} 	imes \hat{j}) - \vec{a} = ((\hat{j} \cdot \hat{j})\vec{a} - (\hat{j} \cdot \vec{a})\hat{j}) - \vec{a} = \vec{a} - (\hat{j} \cdot \vec{a})\hat{j} - \vec{a} = -(\hat{j} \cdot \vec{a})\hat{j}$$\vec{z} = \hat{k} 	imes (\vec{a} 	imes \hat{k}) - \vec{a} = ((\hat{k} \cdot \hat{k})\vec{a} - (\hat{k} \cdot \vec{a})\hat{k}) - \vec{a} = \vec{a} - (\hat{k} \cdot \vec{a})\hat{k} - \vec{a} = -(\hat{k} \cdot \vec{a})\hat{k}$Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$. Then:$\vec{x} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) - a_1\hat{i} = a_2\hat{j} + a_3\hat{k}$$\vec{y} = -a_2\hat{j}$$\vec{z} = -a_3\hat{k}$The scalar triple product is $\left[\begin{array}{lll}\vec{x} & \vec{y} & \vec{z}\end{array}ight] = \vec{x} \cdot (\vec{y} 	imes \vec{z})$.$\vec{y} 	imes \vec{z} = (-a_2\hat{j}) 	imes (-a_3\hat{k}) = a_2 a_3 (\hat{j} 	imes \hat{k}) = a_2 a_3 \hat{i}$.$\left[\begin{array}{lll}\vec{x} & \vec{y} & \vec{z}\end{array}ight] = (a_2\hat{j} + a_3\hat{k}) \cdot (a_2 a_3 \hat{i}) = 0$ (since $\hat{j} \cdot \hat{i} = 0$ and $\hat{k} \cdot \hat{i} = 0$).

Let $\vec{a}$ be a non-zero vector. If $\vec{x}=\hat{i} \times(\vec{a} \times \hat{i})$, $\vec{y}=\hat{j} \times(\vec{a} \times \hat{j})-\vec{a}$ and $\vec{z}=\hat{k} \times(\vec{a} \times \hat{k})-\vec{a}$, then $\left[\begin{array}{lll}\vec{x} & \vec{y} & \vec{z}\end{array}\right]=$

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