The angle made by the resultant vector of two | vedclass

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(B) Let the two vectors be $\overrightarrow{A} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ and $\overrightarrow{B} = 2 \hat{i} - 7 \hat{j} - 4 \hat{k}$.The r

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(B) Let the two vectors be $\overrightarrow{A} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ and $\overrightarrow{B} = 2 \hat{i} - 7 \hat{j} - 4 \hat{k}$.The resultant vector $\overrightarrow{R} = \overrightarrow{A} + \overrightarrow{B} = (2+2) \hat{i} + (3-7) \hat{j} + (4-4) \hat{k} = 4 \hat{i} - 4 \hat{j} + 0 \hat{k}$.The magnitude of the resultant vector is $R = \sqrt{(4)^2 + (-4)^2 + (0)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$.The angle $	heta$ made by the resultant vector with the $x$-axis is given by $\cos 	heta = \frac{R_x}{R}$.Here,$R_x = 4$.So,$\cos 	heta = \frac{4}{4\sqrt{2}} = \frac{1}{\sqrt{2}}$.Therefore,$	heta = 45^{\circ}$.

The angle made by the resultant vector of two vectors $2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $2 \hat{i}-7 \hat{j}-4 \hat{k}$ with the $x$-axis is (in $^{\circ}$)

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