Question

Let \(\vec a, \vec b\) and \(\vec c\) be three vector having magnitudes 1, 1 and 2 respectively, If \(\vec a \times (\vec a \times \vec c) - \vec b = 0\) then the acute angle between \(\vec a\) and \(\vec c\) is
1. π / 4
2. π / 6
3. π / 3
4. None

Answer 1

Correct Answer - Option 2 : π / 6

Concept:

Vector Triple Product: Vector Triple Product is a vector quantity.

Vector triple product of three vectors a, b, c is defined as the cross product of vector a with the cross product of vectors b and c, i.e. a × (b × c)

a × (b × c) = (a . c) b – (a . b) c

a.b = |a||b|cos θ

 

Calculation: 

Here, |a| = 1, |b| = 1, |c| = 2

\(\rm \vec a × (\vec a × \vec c) - \vec b = 0\)

\(\rm (\vec a.\vec c)\vec a-(\vec a.\vec a)\vec c-\vec b=0\)

\( \rm (\vec a.\vec c)\vec a=(\vec a.\vec a)\vec c+\vec b\)

\(\rm (|a||c|cosθ)\vec a=(|a||a|cos 0)\vec c+\vec b\)

\(\rm 2\cos θ\; \vec{a} = \vec{c}+\vec{b}\)

\(\rm 2\cos θ\; \vec{a} - \vec{c}=\vec{b}\)

Taking magnitude both sides, we get

4cos² θ |a|² + |c|² - 2 × 2cos θ \(\rm \vec a \cdot \vec c\) = |b|²

4cos2 θ + 4 - 2 × 2cos θ × |a||c|cos θ = 1

4cos2 θ + 4 - 8cos2 θ  = 1

4cos2 θ = 3

cos2 θ = \(\frac 3 4\)

cos θ = \(\frac {\sqrt{3}}{2}\)

∴ θ = π / 6

Hence, option (2) is correct.