Question
Find the angle between the vectors $\overrightarrow{ A }=\hat{ i }+2 \hat{ j }-\hat{ k }$ and $\overrightarrow{ B }=-\hat{ i }+\hat{ j }- 2 \hat{ k }$
✓
Answer
Let angle between the vectors be $\theta$
$\therefore \cos \theta=\frac{\vec{A} \cdot \vec{B}}{|\overrightarrow{ A }||\overrightarrow{ B }|}$
$|\overrightarrow{ A }|=\sqrt{1^2+2^2+(-1)^2}=\sqrt{6}$
$|\overrightarrow{ B }|=\sqrt{(-1)^2+1^2+(-2)^2}=\sqrt{6}$
$\overrightarrow{ A } \cdot \overrightarrow{ B }=1 \times(-1)+2 \times 1+(-1) \times(-2)$
$\quad=-1+2+2=3$
$\therefore \cos \theta=\frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|}=\frac{3}{\sqrt{6} \times \sqrt{6}}$
$ \cos \theta=\frac{3}{6}=\frac{1}{2}$
$\therefore \theta=\cos ^{-1}\left(\frac{1}{2}\right)=60^{\circ}$
$\therefore \theta=60^{\circ}$
The angle between the vectors is $60^\circ .$
$\therefore \cos \theta=\frac{\vec{A} \cdot \vec{B}}{|\overrightarrow{ A }||\overrightarrow{ B }|}$
$|\overrightarrow{ A }|=\sqrt{1^2+2^2+(-1)^2}=\sqrt{6}$
$|\overrightarrow{ B }|=\sqrt{(-1)^2+1^2+(-2)^2}=\sqrt{6}$
$\overrightarrow{ A } \cdot \overrightarrow{ B }=1 \times(-1)+2 \times 1+(-1) \times(-2)$
$\quad=-1+2+2=3$
$\therefore \cos \theta=\frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|}=\frac{3}{\sqrt{6} \times \sqrt{6}}$
$ \cos \theta=\frac{3}{6}=\frac{1}{2}$
$\therefore \theta=\cos ^{-1}\left(\frac{1}{2}\right)=60^{\circ}$
$\therefore \theta=60^{\circ}$
The angle between the vectors is $60^\circ .$
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