Let a, b and c be three non-coplanar unit vectors such that the angle between every pair of them is

Q: Why can't I just assume $\vec{a} = (1,0,0)$ and find the others? It feels faster.

That is a very tempting trap! While you could theoretically construct such vectors, it is algebraically exhausting and prone to error. The 'Dot Product Method' is the elegant, universal path. By using the dot product, you are working with the intrinsic properties of the vectors rather than their representation in a specific coordinate system. It is faster, cleaner, and much less likely to lead to a calculation error in the heat of the exam.

Q: Wait, why does the scalar triple product $[\vec{a} \vec{b} \vec{c}]$ appear in the equations for $\vec{a}$ and $\vec{c}$? Do I need to calculate it?

That is a brilliant observation! Yes, the scalar triple product does appear. However, notice that it appears in both equations for $\vec{a}$ and $\vec{c}$. When you equate the two expressions, the scalar triple product appears on both sides and cancels out perfectly. You do not need its value. This is the beauty of symmetry in vector problems!

Q: Why did we dot with $\vec{b}$ first? Is there a specific order?

Great question! We chose $\vec{b}$ first because it is the 'common' vector in the cross products on the left side ($\vec{a} \times \vec{b}$ and $\vec{b} \times \vec{c}$). By dotting with $\vec{b}$, we immediately eliminate both cross product terms, giving us a clean equation involving only $p, q,$ and $r$. It is the most efficient starting point to break the problem open.

Question

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-coplanar unit vectors such that the angle between every pair of them is $\frac{\pi}{3}$. If $\vec{a} 	imes \vec{b} + \vec{b} 	imes \vec{c} = p\vec{a} + q\vec{b} + r\vec{c}$, where $p, q$ and $r$ are scalars, then the value of $\frac{p^2 + 2q^2 + r^2}{q^2}$ is .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When dealing with non-coplanar unit vectors $\vec{a}, \vec{b},$ and $\vec{c}$ with a mutual angle of $\frac{\pi}{3}$, we avoid Cartesian coordinates. Instead, we leverage the properties of the vectors themselves.

Given $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$, the dot product of any pair is:
$$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = |\vec{a}||\vec{b}|\cos\left(\frac{\pi}{3}ight) = \frac{1}{2}$$

We are tasked with solving for scalars $p, q,$ and $r$ in the equation:
$$\vec{a} 	imes \vec{b} + \vec{b} 	imes \vec{c} = p\vec{a} + q\vec{b} + r\vec{c}$$

### The Vanishing Act

To isolate the scalars, we use the dot product as a surgical tool. Taking the dot product of the entire equation with $\vec{b}$:
$$(\vec{a} 	imes \vec{b}) \cdot \vec{b} + (\vec{b} 	imes \vec{c}) \cdot \vec{b} = p(\vec{a} \cdot \vec{b}) + q(\vec{b} \cdot \vec{b}) + r(\vec{c} \cdot \vec{b})$$

Since the cross product is perpendicular to its components, the left side vanishes to zero. Substituting our known dot products:
$$0 = p\left(\frac{1}{2}ight) + q(1) + r\left(\frac{1}{2}ight)$$

Multiplying by $2$, we obtain our first relation:
$$p + 2q + r = 0$$

### The Symmetry of the System

Next, we dot the original equation with $\vec{a}$. The term $(\vec{a} 	imes \vec{b}) \cdot \vec{a}$ vanishes, leaving the scalar triple product $[\vec{a} \vec{b} \vec{c}]$:
$$(\vec{b} 	imes \vec{c}) \cdot \vec{a} = p(1) + q\left(\frac{1}{2}ight) + r\left(\frac{1}{2}ight)$$
$$2[\vec{a} \vec{b} \vec{c}] = 2p + q + r$$

Repeating this process by dotting with $\vec{c}$, we find:
$$(\vec{a} 	imes \vec{b}) \cdot \vec{c} = p\left(\frac{1}{2}ight) + q\left(\frac{1}{2}ight) + r(1)$$
$$2[\vec{a} \vec{b} \vec{c}] = p + q + 2r$$

### Final Calculation

Equating the two expressions for $2[\vec{a} \vec{b} \vec{c}]$, we have $2p + q + r = p + q + 2r$, which simplifies to $p = r$. Substituting $r = p$ into our first relation $p + 2q + r = 0$:
$$2p + 2q = 0 \Rightarrow q = -p$$

We now evaluate the target expression $\frac{p^2 + 2q^2 + r^2}{q^2}$ by substituting $r = p$ and $q = -p$:
$$\frac{p^2 + 2(-p)^2 + p^2}{(-p)^2} = \frac{p^2 + 2p^2 + p^2}{p^2} = \frac{4p^2}{p^2}$$

The final result is **4**.

Enter Numerical Value:

Visualized Solution

Visualizing the Vectors $a, b, c$

The Strategy: Dot Product Method

Dot Product with $b$

Simplifying the First Equation

Dot Product with $a$

Dot Product with $c$

Finding the Relation between $p$ and $r$

Solving for $q$ in terms of $p$

Calculating the Final Value

Conclusion and Key Takeaways

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Vanishing Act

The Symmetry of the System

Final Calculation

Similar Questions

Let $a, b, c$ be three non-coplanar vectors and $p, q, r$ are vectors defined by the relations $p = \frac{b \times c}{[ a b c ]}, q = \frac{c \times a}{[ a b c ]}, r = \frac{a \times b}{[ a b c ]}$ then the value of the expression $(a + b) \cdot p + (b + c) \cdot q + (c + a) \cdot r$ is equal to

If $A, B, C$ are three non-coplanar vectors, then $\frac{A \cdot B \times C}{C \times A \cdot B} + \frac{B \cdot A \times C}{C \cdot A \times B} = .........$

If $u, v$ and $w$ are three non-coplanar vectors, then $(u + v - w) \cdot [(u - v) \times (v - w)]$ equals

If $a, b$ and $c$ are three non coplanar vectors, then $(a + b + c) \cdot [(a + b) \times (a + c)]$ equals

If $a = α \hat{i} + β \hat{j} + 3 \hat{k}$ , $b = β \hat{i} - α \hat{j} - \hat{k}$ and $c = \hat{i} - 2 \hat{j} - \hat{k}$ such that $a \cdot b = 1$ and $b \cdot c = - 3$ , then $\frac{1}{3} ((a \times b) \cdot c)$ is equal to

If $a = 2 \hat{i} + \hat{j} + 3 \hat{k}, b = 3 \hat{i} + 3 \hat{j} + \hat{k}$ and $c = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}$ are coplanar vectors and $a \cdot c = 5, b ⊥ c$ , then $122 (c_{1} + c_{2} + c_{3})$ is equal to

If $a, b$ and $c$ are unit coplanar vectors, then the scalar triple product $[2 a - b, 2 b - c, 2 c - a] =$

Enter Numerical Value:

Visualized Solution

Visualizing the Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a,b,c

The Strategy: Dot Product Method

Dot Product with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Simplifying the First Equation

Dot Product with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Dot Product with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Finding the Relation between .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p