Find the angle between two vectors vec A | vedclass

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Question

$\cos \,	heta  = \frac{{\vec A.\vec B}}{{AB}} = \frac{{2 + 1}}{{\sqrt 6  + \sqrt 2 }} = \frac{3}{{\sqrt 6 \sqrt 2 }} = \frac{{\sqrt 3 }}{2}

Vedclass · Accepted Answer

$30$

Vedclass · Answer

$\cos \,	heta  = \frac{{\vec A.\vec B}}{{AB}} = \frac{{2 + 1}}{{\sqrt 6  + \sqrt 2 }} = \frac{3}{{\sqrt 6 \sqrt 2 }} = \frac{{\sqrt 3 }}{2}$

$	herefore$  $	heta = 30^o$

colum $I$	colum $II$
$(A)$ $\|A+B\|$	$(p)$ $\frac{\sqrt{3}}{2} x$
$(B)$ $\|A-B\|$	$(q)$ $x$
$(C)$ $A \cdot B$	$(r)$ $\sqrt{3} x$
$(D)$ $\|A \times B\|$	$(s)$ None

Find the angle between two vectors $\vec A = 2\hat i + \hat j - \hat k$ and $\vec B = \hat i - \hat k$ ....... $^o$

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