Let a, b, c be three non-coplanar vectors such that a × b = 4c, b × c = 9a and c × a = αb, α 0. If

Q: Wait, how can we be so sure they are mutually perpendicular? Is it just a guess?

It is not a guess, it is a geometric necessity! If $\vec{a} \times \vec{b} = 4\vec{c}$, then $\vec{c}$ is orthogonal to both $\vec{a}$ and $\vec{b}$. Similarly, $\vec{b} \times \vec{c} = 9\vec{a}$ implies $\vec{a}$ is orthogonal to both $\vec{b}$ and $\vec{c}$. By transitivity, $\vec{a}, \vec{b}, \text{ and } \vec{c}$ must form an orthogonal triad. It is the beauty of the cross product!

Q: Why did we assume the angle is $90^\circ$ in the magnitude formula?

Because the cross product definition $\vec{a} \times \vec{b} = |\vec{a}||\vec{b}| \sin \theta \hat{n}$ implies that the resulting vector is perpendicular to the plane containing $\vec{a}$ and $\vec{b}$. Since the problem gives us a cyclic relationship where each vector is the cross product of the other two, they must be mutually orthogonal. Thus, $\theta = 90^\circ$ and $\sin 90^\circ = 1$.

Q: I am stuck on the algebra. Why did we multiply the equations?

Great question! When you have a system of equations where variables are multiplied (like $|\vec{a}||\vec{b}| = k_1|\vec{c}|$), multiplying the equations is a powerful technique to isolate the variables. It allows the terms to cancel out, leaving you with a single variable squared, which is much easier to solve.

Question

Let $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vectors such that $\vec{a} 	imes \vec{b} = 4\vec{c}, \vec{b} 	imes \vec{c} = 9\vec{a}$ and $\vec{c} 	imes \vec{a} = \alpha\vec{b}, \alpha > 0$. If $|\vec{a}| + |\vec{b}| + |\vec{c}| = \frac{1}{36}$, then $\alpha$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a 3D coordinate system. You have three vectors, $\vec{a}$, $\vec{b}$, and $\vec{c}$, dancing in space.

The equations $\vec{a} 	imes \vec{b} = 4\vec{c}$, $\vec{b} 	imes \vec{c} = 9\vec{a}$, and $\vec{c} 	imes \vec{a} = \alpha\vec{b}$ initially appear as a complex system of constraints. However, consider the first relation: $\vec{a} 	imes \vec{b} = 4\vec{c}$.

By the definition of the cross product, the resulting vector must be perpendicular to both $\vec{a}$ and $\vec{b}$. This implies that $\vec{c}$ is perpendicular to both $\vec{a}$ and $\vec{b}$.

Applying this logic to the second equation, $\vec{b} 	imes \vec{c} = 9\vec{a}$, we realize that $\vec{a}$ is perpendicular to both $\vec{b}$ and $\vec{c}$. These vectors form an orthogonal triad, similar to the unit vectors $\hat{i}, \hat{j}, \hat{k}$.

### The Algebraic Bridge

Since the angle between any two of these vectors is $90^{\circ}$, we can utilize the magnitude property $|\vec{u} 	imes \vec{v}| = |\vec{u}||\vec{v}| \sin 	heta$. Because $\sin 90^{\circ} = 1$, our vector equations transform into a system of scalar equations:

1. $|\vec{a}||\vec{b}| = 4|\vec{c}|$

2. $|\vec{b}||\vec{c}| = 9|\vec{a}|$

3. $|\vec{c}||\vec{a}| = \alpha|\vec{b}|$

To find $|\vec{b}|$, we multiply the first two equations:

$$ (|\vec{a}||\vec{b}|) \cdot (|\vec{b}||\vec{c}|) = (4|\vec{c}|) \cdot (9|\vec{a}|) $$

This simplifies to $|\vec{a}||\vec{b}|^2|\vec{c}| = 36|\vec{a}||\vec{c}|$. Assuming the vectors are non-zero, we divide by $|\vec{a}||\vec{c}|$ to obtain $|\vec{b}|^2 = 36$, which yields **$|\vec{b}| = 6$**.

### The Final Calculation

With $|\vec{b}| = 6$, we substitute this into the first equation: $|\vec{a}|(6) = 4|\vec{c}|$, which simplifies to:

$$ |\vec{a}| = \frac{4}{6}|\vec{c}| = \frac{2}{3}|\vec{c}| $$

Using the constraint $|\vec{a}| + |\vec{b}| + |\vec{c}| = 36$, we substitute our expressions:

$$ \frac{2}{3}|\vec{c}| + 6 + |\vec{c}| = 36 $$

This simplifies to $\frac{5}{3}|\vec{c}| = 30$, resulting in **$|\vec{c}| = 18$**. Consequently, $|\vec{a}| = \frac{2}{3} 	imes 18 = 12$.

Finally, we return to the third equation, $|\vec{c}||\vec{a}| = \alpha|\vec{b}|$. Plugging in our values:

$$ 18 	imes 12 = \alpha 	imes 6 $$

Solving for $\alpha$, we find:

$$ \alpha = \frac{18 	imes 12}{6} = 36 $$

The final value is **$\alpha = 36$**.

Animated Solution for Mathematics - Vector Algebra: Let $a, b, c$ be three non-coplanar vectors such that $a \times b = 4 c, b \times c = 9 a$ and $c \times a = α b, α > 0$ . If $∣ a ∣ + ∣ b ∣ + ∣ c ∣ = \frac{1}{36}$ , then $α$ is equal to

Enter Numerical Value:

Visualized Solution

Given Cross Product Relations

Deducing Mutual Orthogonality

Converting to Magnitude Equations

Finding the Magnitude of $b$

Relating $∣ a ∣$ and $∣ c ∣$

Using the Sum Condition

Calculating $∣ a ∣$

Solving for $α$

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Algebraic Bridge

The Final Calculation

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Enter Numerical Value:

Visualized Solution

Given Cross Product Relations

Deducing Mutual Orthogonality

Converting to Magnitude Equations

Finding the Magnitude of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Relating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣a∣ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣c∣

Using the Sum Condition

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

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α