NIMCET 2017 — Mathematics PYQ

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)

$a\times (b\times c)=(a\cdot c)b-(a\cdot b)c$

$a.b=|a||b|\cos \theta$

Calculation:

Here, |$\vec{a}$| = 1, |$\vec{b}$| = 1, |$\vec{c}$| = 2

$\vec{a}\times (\vec{a}\times \vec{c})-\vec{b}=0$

$(\vec{a}\cdot \vec{c})\vec{a}-(\vec{a}\cdot \vec{a})\vec{c}-\vec{b}=0$

$(\vec{a}\cdot \vec{c})\vec{a}=(\vec{a}\cdot \vec{a})\vec{c}+\vec{b}$

$(|a||c|\cos \theta )\vec{a}=(|a||a|\cos 0)\vec{c}+\vec{b}$

$2\cos \theta \vec{a}=\vec{c}+\vec{b}$

$2\cos \theta \vec{a}-\vec{c}=\vec{b}$

Taking magnitude both sides, we get

$4{\cos }^2\theta |a{|}^2+|c{|}^2-2\times 2\cos \theta \vec{a}\cdot \vec{c}=|\vec{b}{|}^2$

$4{\cos }^2\theta +4-2\times 2\cos \theta |a||c|\cos \theta =1$

$4{\cos }^2\theta +4-8{\cos }^2\theta =1$

$4{\cos }^2\theta =3$

${\cos }^2\theta =\frac{3}{4}$

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

$\therefore \theta =\pi /6$