If the projection of 2 hat{i} + 4 hat{j} - 2 | vedclass

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(C) The projection of vector $\vec{A}$ on vector $\vec{B}$ is given by $\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$.Given that the projection is zero,we

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$5$

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(C) The projection of vector $\vec{A}$ on vector $\vec{B}$ is given by $\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$.Given that the projection is zero,we have $\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|} = 0$,which implies $\vec{A} \cdot \vec{B} = 0$.Let $\vec{A} = 2 \hat{i} + 4 \hat{j} - 2 \hat{k}$ and $\vec{B} = \hat{i} + 2 \hat{j} + \alpha \hat{k}$.Taking the dot product: $(2 \hat{i} + 4 \hat{j} - 2 \hat{k}) \cdot (\hat{i} + 2 \hat{j} + \alpha \hat{k}) = 0$.$(2)(1) + (4)(2) + (-2)(\alpha) = 0$.$2 + 8 - 2\alpha = 0$.$10 - 2\alpha = 0$.$2\alpha = 10$.$\alpha = 5$.

If the projection of $2 \hat{i} + 4 \hat{j} - 2 \hat{k}$ on $\hat{i} + 2 \hat{j} + \alpha \hat{k}$ is zero,then the value of $\alpha$ will be.

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