Let vec{u} be a vector coplanar with the vect | vedclass

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(D) Since $\vec{u}$ is coplanar with $\vec{a}$ and $\vec{b}$,we can write $\vec{u} = x\vec{a} + y\vec{b}$.Given $\vec{u} \perp \vec{a}$,so $\vec{u} \c

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$336$

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(D) Since $\vec{u}$ is coplanar with $\vec{a}$ and $\vec{b}$,we can write $\vec{u} = x\vec{a} + y\vec{b}$.Given $\vec{u} \perp \vec{a}$,so $\vec{u} \cdot \vec{a} = 0$.$(x\vec{a} + y\vec{b}) \cdot \vec{a} = 0 \implies x|\vec{a}|^2 + y(\vec{a} \cdot \vec{b}) = 0$.Calculate $|\vec{a}|^2 = 2^2 + 3^2 + (-1)^2 = 4 + 9 + 1 = 14$.Calculate $\vec{a} \cdot \vec{b} = (2)(0) + (3)(1) + (-1)(1) = 3 - 1 = 2$.So,$14x + 2y = 0 \implies y = -7x$.Thus,$\vec{u} = x\vec{a} - 7x\vec{b} = x(\vec{a} - 7\vec{b})$.$\vec{a} - 7\vec{b} = (2\hat{i} + 3\hat{j} - \hat{k}) - 7(\hat{j} + \hat{k}) = 2\hat{i} - 4\hat{j} - 8\hat{k}$.Given $\vec{u} \cdot \vec{b} = 24$,so $x(\vec{a} - 7\vec{b}) \cdot \vec{b} = 24$.$(\vec{a} - 7\vec{b}) \cdot \vec{b} = \vec{a} \cdot \vec{b} - 7|\vec{b}|^2 = 2 - 7(1^2 + 1^2) = 2 - 14 = -12$.So,$x(-12) = 24 \implies x = -2$.Therefore,$\vec{u} = -2(2\hat{i} - 4\hat{j} - 8\hat{k}) = -4\hat{i} + 8\hat{j} + 16\hat{k}$.$|\vec{u}|^2 = (-4)^2 + 8^2 + 16^2 = 16 + 64 + 256 = 336$.

Let $\vec{u}$ be a vector coplanar with the vectors $\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{b} = \hat{j} + \hat{k}$. If $\vec{u}$ is perpendicular to $\vec{a}$ and $\vec{u} \cdot \vec{b} = 24$,then $|\vec{u}|^2 = \dots$

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