Let vec{a}=2hat{i}+hat{j}-2hat{k},vec{b}=hat{ | vedclass

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(C) First,calculate $\vec{c} = \vec{a} 	imes \vec{b}$:$\vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -2 \ 1 & 1 & 0 \end{vmatrix}

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(C) First,calculate $\vec{c} = \vec{a} 	imes \vec{b}$:$\vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -2 \ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1) = 2\hat{i} - 2\hat{j} + \hat{k}$.The magnitude is $|\vec{c}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4+4+1} = 3$.Given $|\vec{c} 	imes \vec{d}| = 3$,we have $|\vec{c}||\vec{d}| \sin(\frac{\pi}{4}) = 3$.Substituting $|\vec{c}| = 3$,we get $3|\vec{d}| \cdot \frac{1}{\sqrt{2}} = 3$,which implies $|\vec{d}| = \sqrt{2}$.Given $|\vec{d}-\vec{a}| = \sqrt{11}$,square both sides:$|\vec{d}|^2 + |\vec{a}|^2 - 2(\vec{a} \cdot \vec{d}) = 11$.We know $|\vec{a}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{9} = 3$,so $|\vec{a}|^2 = 9$.Substituting the values: $2 + 9 - 2(\vec{a} \cdot \vec{d}) = 11$.$11 - 2(\vec{a} \cdot \vec{d}) = 11$,which simplifies to $\vec{a} \cdot \vec{d} = 0$.

Let $\vec{a}=2\hat{i}+\hat{j}-2\hat{k}$,$\vec{b}=\hat{i}+\hat{j}$ and $\vec{c}=\vec{a}\times\vec{b}$. Let $\vec{d}$ be a vector such that $|\vec{d}-\vec{a}|=\sqrt{11}$,$|\vec{c}\times\vec{d}|=3$ and the angle between $\vec{c}$ and $\vec{d}$ is $\frac{\pi}{4}$. Then $\vec{a}\cdot\vec{d}$ is equal to

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