A vector has a magnitude the same as that of | vedclass

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(A) First,calculate the magnitude of vector $\overrightarrow{A} = 3 \hat{i} + 4 \hat{j}$.$|\overrightarrow{A}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \s

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$4$

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(A) First,calculate the magnitude of vector $\overrightarrow{A} = 3 \hat{i} + 4 \hat{j}$.$|\overrightarrow{A}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.Next,find the unit vector in the direction of $\overrightarrow{B} = 4 \hat{i} + 3 \hat{j}$.$|\overrightarrow{B}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5$.$\hat{B} = \frac{\overrightarrow{B}}{|\overrightarrow{B}|} = \frac{4 \hat{i} + 3 \hat{j}}{5}$.The required vector $\overrightarrow{V}$ has magnitude $5$ and is parallel to $\overrightarrow{B}$,so $\overrightarrow{V} = |\overrightarrow{A}| \hat{B} = 5 	imes \frac{4 \hat{i} + 3 \hat{j}}{5} = 4 \hat{i} + 3 \hat{j}$.Comparing this with the given components $X \hat{i} + 3 \hat{j}$,we find $X = 4$.

$A$ vector has a magnitude the same as that of $\overrightarrow{A} = 3 \hat{i} + 4 \hat{j}$ and is parallel to $\overrightarrow{B} = 4 \hat{i} + 3 \hat{j}$. The $x$ and $y$ components of this vector in the first quadrant are $X$ and $3$ respectively,where $X = $ . . . . . . .

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