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MathMedAdv 2010 • Paper 1

Scalar Triple Product Involving Cross and Linear Combinations

If a⃗\vec{a} and b⃗\vec{b} are vectors in space given by a⃗=i^−2j^5\vec{a} = \frac{\hat{i} - 2\hat{j}}{\sqrt{5}} and b⃗=2i^+j^+3k^14\vec{b} = \frac{2\hat{i} + \hat{j} + 3\hat{k}}{\sqrt{14}}, then the value of (2a⃗+b⃗)⋅[(a⃗×b⃗)×(a⃗−2b⃗)]\left(2\vec{a} + \vec{b}\right) \cdot \left[\left(\vec{a} \times \vec{b}\right) \times \left(\vec{a} - 2\vec{b}\right)\right] is

Official Numerical Answer5

Topics & Concepts

Vector AlgebraDot and Cross ProductScalar Triple Product

Step-by-Step Solution

To evaluate the expression
E=(2a⃗+b⃗)⋅[(a⃗×b⃗)×(a⃗−2b⃗)]E = \left(2\vec{a} + \vec{b}\right) \cdot \left[\left(\vec{a} \times \vec{b}\right) \times \left(\vec{a} - 2\vec{b}\right)\right]

we first analyze the properties of the given vectors a⃗\vec{a} and b⃗\vec{b}:
a⃗=i^−2j^5andb⃗=2i^+j^+3k^14\vec{a} = \frac{\hat{i} - 2\hat{j}}{\sqrt{5}} \quad \text{and} \quad \vec{b} = \frac{2\hat{i} + \hat{j} + 3\hat{k}}{\sqrt{14}}

Calculate their magnitudes:
∣a⃗∣2=12+(−2)25=55=1  ⟹  ∣a⃗∣=1|\vec{a}|^2 = \frac{1^2 + (-2)^2}{5} = \frac{5}{5} = 1 \implies |\vec{a}| = 1
∣b⃗∣2=22+12+3214=4+1+914=1414=1  ⟹  ∣b⃗∣=1|\vec{b}|^2 = \frac{2^2 + 1^2 + 3^2}{14} = \frac{4 + 1 + 9}{14} = \frac{14}{14} = 1 \implies |\vec{b}| = 1

Next, calculate the scalar product a⃗⋅b⃗\vec{a} \cdot \vec{b}:
a⃗⋅b⃗=(1)(2)+(−2)(1)+(0)(3)514=2−2+070=0\vec{a} \cdot \vec{b} = \frac{(1)(2) + (-2)(1) + (0)(3)}{\sqrt{5}\sqrt{14}} = \frac{2 - 2 + 0}{\sqrt{70}} = 0

Thus, a⃗\vec{a} and b⃗\vec{b} are mutually orthogonal unit vectors, meaning:
∣a⃗∣=1,∣b⃗∣=1,anda⃗⋅b⃗=0|\vec{a}| = 1, \quad |\vec{b}| = 1, \quad \text{and} \quad \vec{a} \cdot \vec{b} = 0

Now, consider the vector triple product:
(a⃗×b⃗)×(a⃗−2b⃗)\left(\vec{a} \times \vec{b}\right) \times \left(\vec{a} - 2\vec{b}\right)

Using the vector triple product identity (u⃗×v⃗)×w⃗=(u⃗⋅w⃗)v⃗−(v⃗⋅w⃗)u⃗(\vec{u} \times \vec{v}) \times \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u}, where u⃗=a⃗\vec{u} = \vec{a}, v⃗=b⃗\vec{v} = \vec{b}, and w⃗=a⃗−2b⃗\vec{w} = \vec{a} - 2\vec{b}, we obtain:
(a⃗×b⃗)×(a⃗−2b⃗)=[a⃗⋅(a⃗−2b⃗)]b⃗−[b⃗⋅(a⃗−2b⃗)]a⃗\left(\vec{a} \times \vec{b}\right) \times \left(\vec{a} - 2\vec{b}\right) = \left[\vec{a} \cdot \left(\vec{a} - 2\vec{b}\right)\right]\vec{b} - \left[\vec{b} \cdot \left(\vec{a} - 2\vec{b}\right)\right]\vec{a}

Evaluate the individual dot products:
a⃗⋅(a⃗−2b⃗)=∣a⃗∣2−2(a⃗⋅b⃗)=1−2(0)=1\vec{a} \cdot (\vec{a} - 2\vec{b}) = |\vec{a}|^2 - 2(\vec{a} \cdot \vec{b}) = 1 - 2(0) = 1
b⃗⋅(a⃗−2b⃗)=(b⃗⋅a⃗)−2∣b⃗∣2=0−2(1)=−2\vec{b} \cdot (\vec{a} - 2\vec{b}) = (\vec{b} \cdot \vec{a}) - 2|\vec{b}|^2 = 0 - 2(1) = -2

Substitute these back into the vector triple product:
(a⃗×b⃗)×(a⃗−2b⃗)=(1)b⃗−(−2)a⃗=2a⃗+b⃗\left(\vec{a} \times \vec{b}\right) \times \left(\vec{a} - 2\vec{b}\right) = (1)\vec{b} - (-2)\vec{a} = 2\vec{a} + \vec{b}

Finally, substituting this result into the original expression gives:
E=(2a⃗+b⃗)⋅(2a⃗+b⃗)=∣2a⃗+b⃗∣2E = \left(2\vec{a} + \vec{b}\right) \cdot \left(2\vec{a} + \vec{b}\right) = \left|2\vec{a} + \vec{b}\right|^2

Expanding the square:
∣2a⃗+b⃗∣2=4∣a⃗∣2+∣b⃗∣2+4(a⃗⋅b⃗)=4(1)+1+4(0)=5\left|2\vec{a} + \vec{b}\right|^2 = 4|\vec{a}|^2 + |\vec{b}|^2 + 4(\vec{a} \cdot \vec{b}) = 4(1) + 1 + 4(0) = 5

Therefore, the value of the given expression is 55.

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