overrightarrow{A} and overrightarrow{B} are t | vedclass

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

Vedclass · Accepted Answer

$5 / \sqrt{2}$

Vedclass · Answer

(A) The component (projection) of vector $\overrightarrow{A}$ on vector $\overrightarrow{B}$ is given by the formula: $	ext{Component} = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{B}|}$.First,calculate the dot product $\overrightarrow{A} \cdot \overrightarrow{B} = (2)(1) + (3)(1) = 2 + 3 = 5$.Next,calculate the magnitude of vector $\overrightarrow{B}$,which is $|\overrightarrow{B}| = \sqrt{1^2 + 1^2} = \sqrt{2}$.Finally,divide the dot product by the magnitude: $	ext{Component} = \frac{5}{\sqrt{2}}$.

$\overrightarrow{A}$ 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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