Question

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are three mutually perpendicular vectors of equal magnitude, then the angle between the vectors $(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})and\overrightarrow{c}=$

• A. ${\cos }^{-1}(\frac{1}{3})$
• B. ${\cos }^{-1}(\frac{1}{\sqrt{3}})$
• C. ${\cos }^{-1}(\frac{3}{4})$
• D. ${\cos }^{-1}(\frac{\sqrt{3}}{4})$

Answer 1

The correct option is B ${\cos }^{-1}(\frac{1}{\sqrt{3}})$

Given $(\overline{a}+\overline{b}+\overline{c})\cdot \overline{c}$

$=|\overline{a}+\overline{b}+\overline{c}||\overline{c}|\cos \theta$

$a\cdot c+b\cdot c+c\cdot c=\left(\sqrt{a^2+b^2+c^2+2\sum \overline{a}\cdot \overline{b}}\right)|\overline{c}|\cos \theta$

$a\cdot c+b\cdot c+|c{|}^2=\left(\sqrt{a^2+b^2+c^2+2\sum \overline{a}\cdot \overline{b}}\right)|\overline{c}|\cos \theta$

$\overline{a}\cdot \overline{b}=\overline{b}\cdot \overline{c}=\overline{c}\cdot \overline{a}=0$ ($a,b,c$ are mutually $\perp$)

a,b,c are of equal magnitude $|a|=|b|=|c|$
$|\overline{c}{|}^2=\sqrt{3}|\overline{c}{|}^2\cos \theta$

$\cos \theta =\frac{1}{\sqrt{3}}$