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

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(B) The projection of vector $\overrightarrow{A}$ on vector $\overrightarrow{B}$ is given by the formula: $	ext{Projection} = \frac{\overrightarrow{A

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$\frac{3}{\sqrt{26}}$

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(B) The projection of vector $\overrightarrow{A}$ on vector $\overrightarrow{B}$ is given by the formula: $	ext{Projection} = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{B}|}$.First,calculate the dot product $\overrightarrow{A} \cdot \overrightarrow{B}$:$\overrightarrow{A} \cdot \overrightarrow{B} = (2)(-1) + (3)(3) + (-1)(4) = -2 + 9 - 4 = 3$.Next,calculate the magnitude of vector $\overrightarrow{B}$:$|\overrightarrow{B}| = \sqrt{(-1)^2 + 3^2 + 4^2} = \sqrt{1 + 9 + 16} = \sqrt{26}$.Finally,calculate the projection:$	ext{Projection} = \frac{3}{\sqrt{26}}$.

If $\overrightarrow{A} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\overrightarrow{B} = -\hat{i} + 3\hat{j} + 4\hat{k}$,then the projection of $\overrightarrow{A}$ on $\overrightarrow{B}$ will be:

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$\overrightarrow{A}$ and $\overrightarrow{B}$ are two vectors given by $\overrightarrow{A} = 2\widehat{i} + 3\widehat{j}$ and $\overrightarrow{B} = \widehat{i} + \widehat{j}$. The magnitude of the component (projection) of $\overrightarrow{A}$ on $\overrightarrow{B}$ is

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