Let hat{u} = u_1 hat{i} + u_2 hat{j} + u_3 ha | vedclass

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(D) Given $\hat{u} 	imes \vec{w} = \hat{v}$. Since $\hat{u}$ and $\vec{w}$ are unit vectors,$|\hat{u} 	imes \vec{w}| = |\hat{u}| |\vec{w}| \sin 	he

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$B, C$

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(D) Given $\hat{u} 	imes \vec{w} = \hat{v}$. Since $\hat{u}$ and $\vec{w}$ are unit vectors,$|\hat{u} 	imes \vec{w}| = |\hat{u}| |\vec{w}| \sin 	heta = \sin 	heta$. Also,$|\hat{v}| = \frac{1}{\sqrt{6}} \sqrt{1^2 + 1^2 + 2^2} = 1$. Thus,$\sin 	heta = 1$,which implies $	heta = 90^{\circ}$.Since $\hat{u} 	imes \vec{w} = \hat{v}$,$\hat{v}$ is perpendicular to both $\hat{u}$ and $\vec{w}$.For a fixed $\hat{u}$,there are infinitely many vectors $\vec{w}$ perpendicular to $\hat{u}$ such that their cross product with $\hat{u}$ is $\hat{v}$. Thus,option $B$ is correct.Since $\hat{v} \cdot \hat{u} = 0$,we have $\frac{1}{\sqrt{6}}(u_1 + u_2 + 2u_3) = 0$,so $u_1 + u_2 + 2u_3 = 0$.If $\hat{u}$ lies in the $xy$-plane,$u_3 = 0$,then $u_1 + u_2 = 0$,so $|u_1| = |u_2|$. Thus,option $C$ is correct.If $\hat{u}$ lies in the $xz$-plane,$u_2 = 0$,then $u_1 + 2u_3 = 0$,so $|u_1| = 2|u_3|$. Thus,option $D$ is incorrect as it states $2|u_1| = |u_3|$.

Let $\hat{u} = u_1 \hat{i} + u_2 \hat{j} + u_3 \hat{k}$ be a unit vector in $\mathbb{R}^3$ and $\hat{v} = \frac{1}{\sqrt{6}}(\hat{i} + \hat{j} + 2 \hat{k})$. Given that there exists a unit vector $\vec{w}$ such that $\hat{u} \times \vec{w} = \hat{v}$,which of the following is(are) correct?

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