If a→,b→,c→are three unit vectors, such that a→.b→=0=a→.c→ and the angle between b→ and c→ is π/3 then the value of |a→×b→-a→×c→|is

# If $\stackrel{\to }{a},\stackrel{\to }{b},\stackrel{\to }{c}$are three unit vectors, such that $\stackrel{\to }{a}.\stackrel{\to }{b}=0=\stackrel{\to }{a}.\stackrel{\to }{c}$ and the angle between $\stackrel{\to }{b}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\stackrel{\to }{c}\text{\hspace{0.17em}is\hspace{0.17em}}\pi /3$ then the value of $|\stackrel{\to }{a}×\stackrel{\to }{b}-\stackrel{\to }{a}×\stackrel{\to }{c}|$is

1. A
2. B
3. C
4. D

Register to Get Free Mock Test and Study Material

+91

Live ClassesRecorded ClassesTest SeriesSelf Learning

Verify OTP Code (required)

### Solution:

$\begin{array}{l}{|\stackrel{\to }{a}×\stackrel{\to }{b}-\stackrel{\to }{a}×\stackrel{\to }{c}|}^{2}\\ ={|\stackrel{\to }{a}×\left(\stackrel{\to }{b}-\stackrel{\to }{c}\right)|}^{2}\\ ={|\stackrel{\to }{a}|}^{2}{|\stackrel{\to }{b}-\stackrel{\to }{c}|}^{2}-{\left(\stackrel{\to }{a}.\left(\stackrel{\to }{b}-\stackrel{\to }{c}\right)\right)}^{2}={|\stackrel{\to }{b}-\stackrel{\to }{c}|}^{2}\\ ={|\stackrel{\to }{b}|}^{2}+{|\stackrel{\to }{c}|}^{2}-2|\stackrel{\to }{b}||\stackrel{\to }{c}|cos\frac{\pi }{3}=1\end{array}$