Let vec{a} = hat{i} - hat{j},vec{b} = hat{j} | vedclass

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(A) Let $\vec{d} = x\hat{i} + y\hat{j} + z\hat{k}$. Since $\vec{d}$ is a unit vector,$x^2 + y^2 + z^2 = 1 \dots (i)$.Given $\vec{a} \cdot \vec{d} = 0$

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$\pm \frac{\hat{i} + \hat{j} - 2\hat{k}}{\sqrt{6}}$

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(A) Let $\vec{d} = x\hat{i} + y\hat{j} + z\hat{k}$. Since $\vec{d}$ is a unit vector,$x^2 + y^2 + z^2 = 1 \dots (i)$.Given $\vec{a} \cdot \vec{d} = 0$,we have $(\hat{i} - \hat{j}) \cdot (x\hat{i} + y\hat{j} + z\hat{k}) = 0$,which implies $x - y = 0$,so $x = y$.Given $[\vec{b} \, \vec{c} \, \vec{d}] = 0$,the scalar triple product is zero,meaning $\vec{d}$ is coplanar with $\vec{b}$ and $\vec{c}$. Thus,$\vec{d} = \lambda(\vec{b} 	imes \vec{c})$.Calculate $\vec{b} 	imes \vec{c} = (\hat{j} - \hat{k}) 	imes (\hat{k} - \hat{i}) = \hat{j} 	imes \hat{k} - \hat{j} 	imes \hat{i} - \hat{k} 	imes \hat{k} + \hat{k} 	imes \hat{i} = \hat{i} + \hat{j} + \hat{k}$.So,$\vec{d} = \lambda(\hat{i} + \hat{j} + \hat{k})$.Since $x=y$,this is consistent. Substituting into the unit vector condition: $\lambda^2(1^2 + 1^2 + 1^2) = 1 \implies 3\lambda^2 = 1 \implies \lambda = \pm \frac{1}{\sqrt{3}}$.However,checking the condition $\vec{a} \cdot \vec{d} = 0$ with $\vec{d} = \lambda(\hat{i} + \hat{j} + \hat{k})$ gives $(\hat{i} - \hat{j}) \cdot \lambda(\hat{i} + \hat{j} + \hat{k}) = \lambda(1 - 1 + 0) = 0$,which is satisfied for any $\lambda$.Wait,the vector $\vec{d}$ must be perpendicular to $\vec{a}$ and lie in the plane of $\vec{b}$ and $\vec{c}$. The vector $\vec{b} 	imes \vec{c} = \hat{i} + \hat{j} + \hat{k}$ is perpendicular to the plane of $\vec{b}$ and $\vec{c}$.Actually,$\vec{d}$ must be perpendicular to $\vec{a}$ and perpendicular to $(\vec{b} 	imes \vec{c})$.Thus,$\vec{d} = \pm \frac{\vec{a} 	imes (\vec{b} 	imes \vec{c})}{|\vec{a} 	imes (\vec{b} 	imes \vec{c})|}$.$\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\hat{i} - \hat{j}) 	imes (\hat{i} + \hat{j} + \hat{k}) = \hat{i} 	imes \hat{i} + \hat{i} 	imes \hat{j} + \hat{i} 	imes \hat{k} - \hat{j} 	imes \hat{i} - \hat{j} 	imes \hat{j} - \hat{j} 	imes \hat{k} = 0 + \hat{k} - \hat{j} + \hat{k} - 0 - \hat{i} = -\hat{i} - \hat{j} + 2\hat{k}$.The magnitude is $\sqrt{(-1)^2 + (-1)^2 + 2^2} = \sqrt{6}$.Therefore,$\vec{d} = \pm \frac{-\hat{i} - \hat{j} + 2\hat{k}}{\sqrt{6}} = \pm \frac{\hat{i} + \hat{j} - 2\hat{k}}{\sqrt{6}}$.

Let $\vec{a} = \hat{i} - \hat{j}$,$\vec{b} = \hat{j} - \hat{k}$,and $\vec{c} = \hat{k} - \hat{i}$. If $\vec{d}$ is a unit vector such that $\vec{a} \cdot \vec{d} = 0 = [\vec{b} \, \vec{c} \, \vec{d}]$,then find $\vec{d}$.

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