Let vec{c} be a vector coplanar with the unit | vedclass

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(C) Given that $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$, the scalar triple product $[\vec{a} \vec{b} \vec{c}] = 0$.Since $\vec{d}$ is perpen

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) Given that $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$, the scalar triple product $[\vec{a} \vec{b} \vec{c}] = 0$.Since $\vec{d}$ is perpendicular to $\vec{a}$, $\vec{b}$, and $\vec{c}$, it is the unit vector along the direction of $\vec{a} 	imes \vec{b}$.Thus, $[\vec{a} \vec{b} \vec{d}] = \vec{a} \cdot (\vec{b} 	imes \vec{d}) = (\vec{a} 	imes \vec{b}) \cdot \vec{d} = |\vec{a} 	imes \vec{b}| |\vec{d}| \cos 0^{\circ} = |\vec{a}| |\vec{b}| \sin 30^{\circ} (1) = (1)(1)(\frac{1}{2}) = \frac{1}{2}$.The given equation becomes $\frac{1}{2} \vec{c} - 0 \cdot \vec{d} = \hat{i} + 2\hat{j} + 2\hat{k}$.So, $\vec{c} = 2\hat{i} + 4\hat{j} + 4\hat{k}$.Therefore, $|\vec{c}| = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.

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