arthham

JEE Advanced 2014 · Vector Algebra

Let a⃗\vec{a}, b⃗\vec{b} and c⃗\vec{c} be three non-coplanar unit vectors such that the angle between every pair of them is π3\frac{\pi}{3}. If a⃗×b⃗+b⃗×c⃗=pa⃗+qb⃗+rc⃗\vec{a} \times \vec{b} + \vec{b} \times \vec{c} = p\vec{a} + q\vec{b} + r\vec{c}, where pp, qq and rr are scalars, then the value of p2+2q2+r2q2\frac{p^{2} + 2q^{2} + r^{2}}{q^{2}} is

Answer: 44

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Written solution

Given: ∣a⃗∣=∣b⃗∣=∣c⃗∣=1\text{Given: } |\vec{a}| = |\vec{b}| = |\vec{c}| = 1

a⃗, b⃗, c⃗ are non-coplanar\vec{a},\ \vec{b},\ \vec{c} \text{ are non-coplanar}

∠(a⃗,b⃗)=∠(b⃗,c⃗)=∠(c⃗,a⃗)=π3\angle(\vec{a},\vec{b}) = \angle(\vec{b},\vec{c}) = \angle(\vec{c},\vec{a}) = \frac{\pi}{3}

a⃗⋅b⃗=b⃗⋅c⃗=c⃗⋅a⃗=cos⁡π3=12\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{c} = \vec{c}\cdot\vec{a} = \cos\frac{\pi}{3} = \frac{1}{2}

To find: p2+2q2+r2q2\text{To find: } \frac{p^{2} + 2q^{2} + r^{2}}{q^{2}}

Plan: dot both sides with a⃗, b⃗, c⃗ one at a time\text{Plan: dot both sides with } \vec{a},\ \vec{b},\ \vec{c} \text{ one at a time}

(a⃗×b⃗)⋅a⃗=0,(a⃗×b⃗)⋅b⃗=0(\vec{a}\times\vec{b})\cdot\vec{a} = 0, \quad (\vec{a}\times\vec{b})\cdot\vec{b} = 0

⇒ three scalar equations in p, q, r\Rightarrow \text{ three scalar equations in } p,\ q,\ r

Δ=a⃗⋅(b⃗×c⃗)=[a⃗,b⃗,c⃗]≠0\Delta = \vec{a}\cdot(\vec{b}\times\vec{c}) = [\vec{a},\vec{b},\vec{c}] \ne 0

(non-coplanar ⇒Δ≠0)\text{(non-coplanar } \Rightarrow \Delta \ne 0)

Dot with a⃗:0+Δ=p(a⃗⋅a⃗)+q(a⃗⋅b⃗)+r(a⃗⋅c⃗)\text{Dot with } \vec{a}: \quad 0 + \Delta = p(\vec{a}\cdot\vec{a}) + q(\vec{a}\cdot\vec{b}) + r(\vec{a}\cdot\vec{c})

Δ=p+q+r2(1)\Delta = p + \frac{q+r}{2} \qquad (1)

Dot with b⃗:p(a⃗⋅b⃗)+q(b⃗⋅b⃗)+r(c⃗⋅b⃗)=0\text{Dot with } \vec{b}: \quad p(\vec{a}\cdot\vec{b}) + q(\vec{b}\cdot\vec{b}) + r(\vec{c}\cdot\vec{b}) = 0

p2+q+r2=0(2)\frac{p}{2} + q + \frac{r}{2} = 0 \qquad (2)

Dot with c⃗:Δ+0=p(a⃗⋅c⃗)+q(b⃗⋅c⃗)+r(c⃗⋅c⃗)\text{Dot with } \vec{c}: \quad \Delta + 0 = p(\vec{a}\cdot\vec{c}) + q(\vec{b}\cdot\vec{c}) + r(\vec{c}\cdot\vec{c})

Δ=p+q2+r(3)\Delta = \frac{p+q}{2} + r \qquad (3)

Δ=p+q+r2(1)\Delta = p + \frac{q+r}{2} \qquad (1)

p2+q+r2=0(2)\frac{p}{2} + q + \frac{r}{2} = 0 \qquad (2)

Δ=p+q2+r(3)\Delta = \frac{p+q}{2} + r \qquad (3)

From (2):q=−p+r2\text{From } (2): \quad q = -\frac{p+r}{2}

(1)−(3):0=(p+q+r2)−(p+q2+r)(1) - (3): \quad 0 = \left(p + \frac{q+r}{2}\right) - \left(\frac{p+q}{2} + r\right)

0=p−r2⇒p=r0 = \frac{p-r}{2} \quad \Rightarrow \quad p = r

q=−p+r2=−p+p2=−pq = -\frac{p+r}{2} = -\frac{p+p}{2} = -p

(1):Δ=p+−p+p2=p(1): \quad \Delta = p + \frac{-p+p}{2} = p

⇒ p=r=Δ,q=−Δ\Rightarrow \ p = r = \Delta, \qquad q = -\Delta

p2+2q2+r2q2=Δ2+2(−Δ)2+Δ2(−Δ)2\frac{p^{2} + 2q^{2} + r^{2}}{q^{2}} = \frac{\Delta^{2} + 2(-\Delta)^{2} + \Delta^{2}}{(-\Delta)^{2}}

=Δ2+2Δ2+Δ2Δ2=4Δ2Δ2= \frac{\Delta^{2} + 2\Delta^{2} + \Delta^{2}}{\Delta^{2}} = \frac{4\Delta^{2}}{\Delta^{2}}

4Δ2Δ2=4\frac{4\Delta^{2}}{\Delta^{2}} = 4

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