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MathHardAdv 2014 • Paper 1

Evaluation of Vector Expression Involving Cross Products

Let a⃗,b⃗\vec{a}, \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 p,qp, q, and rr are scalars, determine the value of the expression p2+2q2+r2q2\frac{p^2 + 2q^2 + r^2}{q^2}.

Official Numerical Answer4

Topics & Concepts

Vector AlgebraDot and Cross Product

Step-by-Step Solution

To evaluate the expression p2+2q2+r2q2\frac{p^2 + 2q^2 + r^2}{q^2}, we use the properties of the dot product and scalar triple product.

Given that a⃗,b⃗\vec{a}, \vec{b}, and c⃗\vec{c} are non-coplanar unit vectors with an angle of π3\frac{\pi}{3} between any pair, we have: ∣a⃗∣=∣b⃗∣=∣c⃗∣=1|\vec{a}| = |\vec{b}| = |\vec{c}| = 1 a⃗⋅b⃗=b⃗⋅c⃗=c⃗⋅a⃗=1⋅1⋅cos⁡(π3)=12\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 1 \cdot 1 \cdot \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}

We are given the vector equation: 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}

Taking the dot product of both sides with a⃗\vec{a}: a⃗⋅(a⃗×b⃗+b⃗×c⃗)=a⃗⋅(pa⃗+qb⃗+rc⃗)\vec{a} \cdot (\vec{a} \times \vec{b} + \vec{b} \times \vec{c}) = \vec{a} \cdot (p\vec{a} + q\vec{b} + r\vec{c}) a⃗⋅(a⃗×b⃗)+a⃗⋅(b⃗×c⃗)=p(a⃗⋅a⃗)+q(a⃗⋅b⃗)+r(a⃗⋅c⃗)\vec{a} \cdot (\vec{a} \times \vec{b}) + \vec{a} \cdot (\vec{b} \times \vec{c}) = p(\vec{a} \cdot \vec{a}) + q(\vec{a} \cdot \vec{b}) + r(\vec{a} \cdot \vec{c})

Since a⃗⋅(a⃗×b⃗)=0\vec{a} \cdot (\vec{a} \times \vec{b}) = 0 and a⃗⋅(b⃗×c⃗)=[a⃗ b⃗ c⃗]\vec{a} \cdot (\vec{b} \times \vec{c}) = [\vec{a}\ \vec{b}\ \vec{c}], we get: [a⃗ b⃗ c⃗]=p(1)+q(12)+r(12)=p+12q+12r— (1)[\vec{a}\ \vec{b}\ \vec{c}] = p(1) + q\left(\frac{1}{2}\right) + r\left(\frac{1}{2}\right) = p + \frac{1}{2}q + \frac{1}{2}r \quad \text{--- (1)}

Taking the dot product of both sides with b⃗\vec{b}: b⃗⋅(a⃗×b⃗+b⃗×c⃗)=b⃗⋅(pa⃗+qb⃗+rc⃗)\vec{b} \cdot (\vec{a} \times \vec{b} + \vec{b} \times \vec{c}) = \vec{b} \cdot (p\vec{a} + q\vec{b} + r\vec{c}) b⃗⋅(a⃗×b⃗)+b⃗⋅(b⃗×c⃗)=p(b⃗⋅a⃗)+q(b⃗⋅b⃗)+r(b⃗⋅c⃗)\vec{b} \cdot (\vec{a} \times \vec{b}) + \vec{b} \cdot (\vec{b} \times \vec{c}) = p(\vec{b} \cdot \vec{a}) + q(\vec{b} \cdot \vec{b}) + r(\vec{b} \cdot \vec{c}) 0+0=p(12)+q(1)+r(12)0 + 0 = p\left(\frac{1}{2}\right) + q(1) + r\left(\frac{1}{2}\right) 12p+q+12r=0  ⟹  p+2q+r=0— (2)\frac{1}{2}p + q + \frac{1}{2}r = 0 \implies p + 2q + r = 0 \quad \text{--- (2)}

Taking the dot product of both sides with c⃗\vec{c}: c⃗⋅(a⃗×b⃗+b⃗×c⃗)=c⃗⋅(pa⃗+qb⃗+rc⃗)\vec{c} \cdot (\vec{a} \times \vec{b} + \vec{b} \times \vec{c}) = \vec{c} \cdot (p\vec{a} + q\vec{b} + r\vec{c}) c⃗⋅(a⃗×b⃗)+c⃗⋅(b⃗×c⃗)=p(c⃗⋅a⃗)+q(c⃗⋅b⃗)+r(c⃗⋅c⃗)\vec{c} \cdot (\vec{a} \times \vec{b}) + \vec{c} \cdot (\vec{b} \times \vec{c}) = p(\vec{c} \cdot \vec{a}) + q(\vec{c} \cdot \vec{b}) + r(\vec{c} \cdot \vec{c}) [a⃗ b⃗ c⃗]+0=p(12)+q(12)+r(1)[\vec{a}\ \vec{b}\ \vec{c}] + 0 = p\left(\frac{1}{2}\right) + q\left(\frac{1}{2}\right) + r(1) [a⃗ b⃗ c⃗]=12p+12q+r— (3)[\vec{a}\ \vec{b}\ \vec{c}] = \frac{1}{2}p + \frac{1}{2}q + r \quad \text{--- (3)}

Equating equations (1) and (3): p+12q+12r=12p+12q+rp + \frac{1}{2}q + \frac{1}{2}r = \frac{1}{2}p + \frac{1}{2}q + r 12p=12r  ⟹  p=r\frac{1}{2}p = \frac{1}{2}r \implies p = r

Substituting r=pr = p into equation (2): p+2q+p=0  ⟹  2p+2q=0  ⟹  q=−pp + 2q + p = 0 \implies 2p + 2q = 0 \implies q = -p

Thus, we have r=pr = p and q=−pq = -p.

Now, substituting these relations into the given expression: p2+2q2+r2q2=(−q)2+2q2+(−q)2q2=q2+2q2+q2q2=4q2q2=4\frac{p^2 + 2q^2 + r^2}{q^2} = \frac{(-q)^2 + 2q^2 + (-q)^2}{q^2} = \frac{q^2 + 2q^2 + q^2}{q^2} = \frac{4q^2}{q^2} = 4

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