The angle between the two vectors $\vec{A}=3 \hat{i}+4 \hat{j}+5 \hat{k}$ and $\vec{B}=3 \hat{i}+4 \

Question

The angle between the two vectors $\vec{A}=3 \hat{i}+4 \hat{j}+5 \hat{k}$ and $\vec{B}=3 \hat{i}+4 \hat{j}-5 \hat{k}$ will be\

Professor PC Thomas Classes · Answer

(a) : $\vec{A}=3 \hat{i}+4 \hat{j}+5 \hat{k}$ and $\vec{B}=3 \hat{i}+4 \hat{j}-5 \hat{k}$\ $\cos 	heta=\frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|}=\frac{(3 \hat{i}+4 \hat{j}+5 \hat{k}) \cdot(3 \hat{i}+4 \hat{j}-5 \hat{k})}{\left[\sqrt{(3)^{2}+(4)^{2}+(5)^{2}}ight] 	imes\left[\sqrt{(3)^{2}+(4)^{2}+(5)^{2}}ight]}$ $$ =\frac{9+16-25}{50}=0 	ext { or } 	heta=90^{\circ} . $$