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MathHardAdv 2014 • Paper 1

Vector Relations with Cross Products and Orthogonality Conditions

Assume x⃗\vec{x}, y⃗\vec{y}, and z⃗\vec{z} are three vectors, each having magnitude 2\sqrt{2}, with an angle of π3\frac{\pi}{3} between any pair. If a⃗\vec{a} is a non-zero vector orthogonal to x⃗\vec{x} and y⃗×z⃗\vec{y} \times \vec{z}, and b⃗\vec{b} is a non-zero vector orthogonal to y⃗\vec{y} and z⃗×x⃗\vec{z} \times \vec{x}, then which of the following holds true?

Options

A

b⃗=(b⃗⋅z⃗)(z⃗−x⃗)\vec{b} = (\vec{b} \cdot \vec{z})(\vec{z} - \vec{x})

Correct
B

a⃗=(a⃗⋅y⃗)(y⃗−z⃗)\vec{a} = (\vec{a} \cdot \vec{y})(\vec{y} - \vec{z})

Correct
C

a⃗⋅b⃗=−(a⃗⋅y⃗)(b⃗⋅z⃗)\vec{a} \cdot \vec{b} = -(\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z})

Correct
D

a⃗=(a⃗⋅y⃗)(z⃗−y⃗)\vec{a} = (\vec{a} \cdot \vec{y})(\vec{z} - \vec{y})

Topics & Concepts

Vector AlgebraDot and Cross ProductScalar Triple Product

Step-by-Step Solution

To determine which of the given relations hold true, we first calculate the scalar dot products between the vectors x⃗\vec{x}, y⃗\vec{y}, and z⃗\vec{z}.

Given: ∣x⃗∣=∣y⃗∣=∣z⃗∣=2|\vec{x}| = |\vec{y}| = |\vec{z}| = \sqrt{2} The angle between any pair of these vectors is π3\frac{\pi}{3}. Thus, the pairwise dot products are: x⃗⋅y⃗=y⃗⋅z⃗=z⃗⋅x⃗=∣x⃗∣∣y⃗∣cos⁡(π3)=2⋅2⋅12=1\vec{x} \cdot \vec{y} = \vec{y} \cdot \vec{z} = \vec{z} \cdot \vec{x} = |\vec{x}||\vec{y}| \cos\left(\frac{\pi}{3}\right) = \sqrt{2} \cdot \sqrt{2} \cdot \frac{1}{2} = 1

Step 1: Finding the vector a⃗\vec{a}

The vector a⃗\vec{a} is given to be orthogonal to y⃗×z⃗\vec{y} \times \vec{z}. This implies that a⃗\vec{a} lies in the plane containing y⃗\vec{y} and z⃗\vec{z}. Thus, a⃗\vec{a} can be expressed as a linear combination of y⃗\vec{y} and z⃗\vec{z}: a⃗=c1y⃗+c2z⃗\vec{a} = c_1 \vec{y} + c_2 \vec{z}

We are also given that a⃗\vec{a} is orthogonal to x⃗\vec{x}, which means a⃗⋅x⃗=0\vec{a} \cdot \vec{x} = 0: a⃗⋅x⃗=(c1y⃗+c2z⃗)⋅x⃗=c1(y⃗⋅x⃗)+c2(z⃗⋅x⃗)=0\vec{a} \cdot \vec{x} = (c_1 \vec{y} + c_2 \vec{z}) \cdot \vec{x} = c_1 (\vec{y} \cdot \vec{x}) + c_2 (\vec{z} \cdot \vec{x}) = 0 Substituting y⃗⋅x⃗=1\vec{y} \cdot \vec{x} = 1 and z⃗⋅x⃗=1\vec{z} \cdot \vec{x} = 1: c1(1)+c2(1)=0  ⟹  c2=−c1c_1 (1) + c_2 (1) = 0 \implies c_2 = -c_1

So, vector a⃗\vec{a} becomes: a⃗=c1(y⃗−z⃗)\vec{a} = c_1 (\vec{y} - \vec{z})

Taking the dot product of a⃗\vec{a} with y⃗\vec{y}: a⃗⋅y⃗=c1(y⃗−z⃗)⋅y⃗=c1(∣y⃗∣2−z⃗⋅y⃗)=c1(2−1)=c1\vec{a} \cdot \vec{y} = c_1 (\vec{y} - \vec{z}) \cdot \vec{y} = c_1 (|\vec{y}|^2 - \vec{z} \cdot \vec{y}) = c_1 (2 - 1) = c_1

Substituting c1=a⃗⋅y⃗c_1 = \vec{a} \cdot \vec{y} back into the expression for a⃗\vec{a}: a⃗=(a⃗⋅y⃗)(y⃗−z⃗)\vec{a} = (\vec{a} \cdot \vec{y})(\vec{y} - \vec{z})

This proves that Option B is correct, while Option D is incorrect.

Step 2: Finding the vector b⃗\vec{b}

Similarly, b⃗\vec{b} is orthogonal to z⃗×x⃗\vec{z} \times \vec{x}, meaning b⃗\vec{b} lies in the plane spanned by z⃗\vec{z} and x⃗\vec{x}: b⃗=k1z⃗+k2x⃗\vec{b} = k_1 \vec{z} + k_2 \vec{x}

Given that b⃗\vec{b} is orthogonal to y⃗\vec{y}, we have b⃗⋅y⃗=0\vec{b} \cdot \vec{y} = 0: b⃗⋅y⃗=k1(z⃗⋅y⃗)+k2(x⃗⋅y⃗)=k1(1)+k2(1)=0  ⟹  k2=−k1\vec{b} \cdot \vec{y} = k_1 (\vec{z} \cdot \vec{y}) + k_2 (\vec{x} \cdot \vec{y}) = k_1(1) + k_2(1) = 0 \implies k_2 = -k_1

So, vector b⃗\vec{b} becomes: b⃗=k1(z⃗−x⃗)\vec{b} = k_1 (\vec{z} - \vec{x})

Taking the dot product of b⃗\vec{b} with z⃗\vec{z}: b⃗⋅z⃗=k1(z⃗−x⃗)⋅z⃗=k1(∣z⃗∣2−x⃗⋅z⃗)=k1(2−1)=k1\vec{b} \cdot \vec{z} = k_1 (\vec{z} - \vec{x}) \cdot \vec{z} = k_1 (|\vec{z}|^2 - \vec{x} \cdot \vec{z}) = k_1 (2 - 1) = k_1

Substituting k1=b⃗⋅z⃗k_1 = \vec{b} \cdot \vec{z} back into the expression for b⃗\vec{b}: b⃗=(b⃗⋅z⃗)(z⃗−x⃗)\vec{b} = (\vec{b} \cdot \vec{z})(\vec{z} - \vec{x})

This proves that Option A is correct.

Step 3: Evaluating a⃗⋅b⃗\vec{a} \cdot \vec{b}

Now, we compute the dot product a⃗⋅b⃗\vec{a} \cdot \vec{b}: a⃗⋅b⃗=[(a⃗⋅y⃗)(y⃗−z⃗)]⋅[(b⃗⋅z⃗)(z⃗−x⃗)]=(a⃗⋅y⃗)(b⃗⋅z⃗)[(y⃗−z⃗)⋅(z⃗−x⃗)]\vec{a} \cdot \vec{b} = \left[ (\vec{a} \cdot \vec{y})(\vec{y} - \vec{z}) \right] \cdot \left[ (\vec{b} \cdot \vec{z})(\vec{z} - \vec{x}) \right] = (\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z}) \left[ (\vec{y} - \vec{z}) \cdot (\vec{z} - \vec{x}) \right]

Evaluating the factor (y⃗−z⃗)⋅(z⃗−x⃗)(\vec{y} - \vec{z}) \cdot (\vec{z} - \vec{x}): (y⃗−z⃗)⋅(z⃗−x⃗)=y⃗⋅z⃗−y⃗⋅x⃗−∣z⃗∣2+z⃗⋅x⃗(\vec{y} - \vec{z}) \cdot (\vec{z} - \vec{x}) = \vec{y} \cdot \vec{z} - \vec{y} \cdot \vec{x} - |\vec{z}|^2 + \vec{z} \cdot \vec{x} =1−1−2+1=−1= 1 - 1 - 2 + 1 = -1

Therefore: a⃗⋅b⃗=−(a⃗⋅y⃗)(b⃗⋅z⃗)\vec{a} \cdot \vec{b} = -(\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z})

This proves that Option C is correct.

Conclusion

The correct options are A, B, and C.

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