Question

If a b c r the unit vectors such that a.b=a.c=0.n the angle between b.c is pi by 6 prove that a=+_2(b×c)

Answer 1

Answer:

Here's ur answer

Answer 2

Answer:

We are given a,b,c three unit vectors with the property such that:

$a\cdot b=a\cdot c=0$ .

Now, the angle between $b\cdot c$ is $\dfrac{\pi}{6}$

i.e. we can write the expression as:

$b\cdot c=|b|\cdot |c|\cdot \cos (P)\\\\b\cdot c=\cos P=\cos \dfrac{\pi}{6}=\dfrac{\sqrt{3}}{2}$

( |b|=|c|=1

since b and c are unit vectors)

Now, we calculate the vector b×c as:

$b\times c=|b||c|\sin (\dfrac{\pi}{6})\cdot n\\\\b\times c=\dfrac{1}{2}n$

where n is the unit vector in the direction of vector a.

Hence, we have:

$n=\pm a$

Hence,

$b\times c=\dfrac{1}{2}\pm a\\\\2(b\times c)=\pm a\\\\a=\pm 2(b\times c)$