The angle between two vectors 4hat i + 3hat j | vedclass

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Question

$\cos 	heta=\frac{\overrightarrow{\mathrm{A}} \overrightarrow{\mathrm{B}}}{\mathrm{AB}}=\frac{(4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}

Vedclass · Accepted Answer

$90$

Vedclass · Answer

$\cos 	heta=\frac{\overrightarrow{\mathrm{A}} \overrightarrow{\mathrm{B}}}{\mathrm{AB}}=\frac{(4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \cdot(-3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})}{\sqrt{26} \sqrt{49}}=0$

$	herefore 	heta=90^{\circ}$

The angle between two vectors $4\hat i + 3\hat j + \hat k$ and $-3\hat i + 2\hat j + 6\hat k$ is ....... $^o$

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