# Problem – Three unit vectors A, B, C are such that A x (B x C) = √3/2 C. Then angle which A makes with B and C, respectively?

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## Problem – Three unit vectors A, B, C are such that A x (B x C) = √3/2 C. Then angle which A makes with B and C, respectively?

Solution

Let’s use the tripple cross product identity here,

i.e.

$\vec{A} x (\vec{B} x \vec{C}) = (\vec{A}.\vec{C})\vec{B} – (\vec{A}.\vec{B})\vec{C}$ —- Eq. (1)

Now,

According to the problem problem,

$\vec{A} x (\vec{B} x \vec{C}) = \frac{\sqrt{3}}{2}\vec{C}$

so,

using Eq. (1)

($\vec{A}.\vec{C})\vec{B} – (\vec{A}.\vec{B})\vec{C} = \frac{\sqrt{3}}{2}\vec{C}$

On compairing both sides,

We get,

$\vec{A}.\vec{C}$ = 0

So,

$\vec{A}$ perpendicular to $\vec{C}$

i.e. angle between $\vec{A}$ and $\vec{C}$ is $90^{0}$

and comparing coefficient of $\vec{C}$

-$\vec{A}.\vec{B} = \frac{\sqrt{3}}{2}$

|A ||B| cos $\theta$ = -$\frac{\sqrt{3}}{2}$

And,

as $\vec{A},\vec{B},\vec{C}$ are unit vectors (given)

So,

or,

cos $\theta$ = -$\frac{\sqrt{3}}{2}$

$\theta = 150 ^{0}$

Hence,

angle between $\vec{A}$ and $\vec{B}$ is $150^{0}$

&

and, angle between $\vec{A}$ and $\vec{C}$ is $90^{0}$

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