If overrightarrow{A} = (2hat{i} + 3hat{j} - h | vedclass

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(A) Given vectors are $\overrightarrow{A} = (2\hat{i} + 3\hat{j} - \hat{k}) \; m$ and $\overrightarrow{B} = (\hat{i} + 2\hat{j} + 2\hat{k}) \; m$.The

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$2$

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(A) Given vectors are $\overrightarrow{A} = (2\hat{i} + 3\hat{j} - \hat{k}) \; m$ and $\overrightarrow{B} = (\hat{i} + 2\hat{j} + 2\hat{k}) \; m$.The component of vector $\overrightarrow{A}$ along vector $\overrightarrow{B}$ is given by the projection formula: $	ext{Component} = \overrightarrow{A} \cdot \hat{B} = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{B}|}$.First,calculate the dot product $\overrightarrow{A} \cdot \overrightarrow{B}$:$\overrightarrow{A} \cdot \overrightarrow{B} = (2)(1) + (3)(2) + (-1)(2) = 2 + 6 - 2 = 6$.Next,calculate the magnitude of vector $\overrightarrow{B}$:$|\overrightarrow{B}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.Finally,calculate the component:$	ext{Component} = \frac{6}{3} = 2 \; m$.

Column $-I$	Column $-II$
$(a) \vec A \cdot \vec B = 0$	$(i) \theta = 0^\circ$
$(b) \vec A \cdot \vec B = +8$	$(ii) \theta = 90^\circ$
$(c) \vec A \cdot \vec B = 4$	$(iii) \theta = 180^\circ$
$(d) \vec A \cdot \vec B = -8$	$(iv) \theta = 60^\circ$

If $\overrightarrow{A} = (2\hat{i} + 3\hat{j} - \hat{k}) \; m$ and $\overrightarrow{B} = (\hat{i} + 2\hat{j} + 2\hat{k}) \; m$,the magnitude of the component of vector $\overrightarrow{A}$ along vector $\overrightarrow{B}$ will be $...... \; m$.

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