For non-zero vectors a, b, c, |(a x b) . c| = |a| |b| |c| holds if and only if (a) a . b = 0, b . c

Q: Wait, why can we just cancel $|\vec{a}| |\vec{b}| |\vec{c}|$ from both sides?

The problem explicitly states that $\vec{a}, \vec{b}, \vec{c}$ are non-zero vectors. Since their magnitudes are non-zero, their product is non-zero, allowing us to safely divide both sides of the equation without worrying about division by zero.

Q: Why does $|\sin \theta| |\cos \alpha| = 1$ imply both must be $1$?

Remember that for any real angle, $|\sin \theta| \le 1$ and $|\cos \alpha| \le 1$. If you multiply two numbers that are at most $1$, the only way to get a product of exactly $1$ is if both numbers are exactly $1$. If either were less than $1$, the product would inevitably be less than $1$.

Q: How does $\vec{c}$ being parallel to $\vec{a} \times \vec{b}$ make it perpendicular to $\vec{a}$ and $\vec{b}$?

By definition, the cross product $\vec{a} \times \vec{b}$ creates a vector that is perpendicular to the plane containing both $\vec{a}$ and $\vec{b}$. If $\vec{c}$ is parallel to this normal vector, it must also be perpendicular to that same plane, and therefore perpendicular to both $\vec{a}$ and $\vec{b}$.

Question

For non-zero vectors $\vec{a}, \vec{b}, \vec{c}$, $|(\vec{a} 	imes \vec{b}) \cdot \vec{c}| = |\vec{a}| |\vec{b}| |\vec{c}|$ holds if and only if

Options:

(A) $\vec{a} \cdot \vec{b} = 0, \vec{b} \cdot \vec{c} = 0$

(B) $\vec{b} \cdot \vec{c} = 0, \vec{c} \cdot \vec{a} = 0$

(C) $\vec{c} \cdot \vec{a} = 0, \vec{a} \cdot \vec{b} = 0$

(D) $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The equation $|(\vec{a} 	imes \vec{b}) \cdot \vec{c}| = |\vec{a}| |\vec{b}| |\vec{c}|$ represents a profound geometric condition. The left-hand side is the **Scalar Triple Product**, which geometrically defines the volume of a parallelepiped formed by vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$.

The right-hand side is the product of the lengths of these vectors. When the volume of a parallelepiped equals the product of its edge lengths, it implies that the edges must be oriented in a specific, highly symmetric configuration.

### The Trigonometric Expansion

Consider the cross product $\vec{a} 	imes \vec{b}$. Its magnitude is given by $|\vec{a}| |\vec{b}| \sin 	heta$, where $	heta$ is the angle between $\vec{a}$ and $\vec{b}$.

Now, we take the dot product of this result with $\vec{c}$. If $\alpha$ is the angle between the normal vector $(\vec{a} 	imes \vec{b})$ and $\vec{c}$, the dot product becomes:

$$|(\vec{a} 	imes \vec{b}) \cdot \vec{c}| = |\vec{a}| |\vec{b}| |\vec{c}| |\sin 	heta| |\cos \alpha|$$

Equating this to the original expression $|\vec{a}| |\vec{b}| |\vec{c}|$, we arrive at the condition:

$$|\sin 	heta| |\cos \alpha| = 1$$

### The "Aha!" Moment

Because the maximum value of both the sine and cosine functions is $1$, the only way their product can equal $1$ is if both absolute values are exactly $1$. This forces the conditions $|\sin 	heta| = 1$ and $|\cos \alpha| = 1$.

From $|\sin 	heta| = 1$, we deduce that $	heta = 90^{\circ}$, which implies that $\vec{a}$ is perpendicular to $\vec{b}$.

From $|\cos \alpha| = 1$, we deduce that $\alpha = 0^{\circ}$ or $180^{\circ}$. This means $\vec{c}$ is parallel to the normal vector of the plane containing $\vec{a}$ and $\vec{b}$.

### Final Conclusion

Since $\vec{c}$ is parallel to the normal of the $\vec{a}-\vec{b}$ plane, it follows that $\vec{c}$ is perpendicular to both $\vec{a}$ and $\vec{b}$.

Thus, we conclude that the vectors are **mutually perpendicular**, satisfying the condition:

$$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0$$

It is a stunning result: the only way to maximize the volume of a parallelepiped relative to its edge lengths is to construct a **perfect rectangular box**.

Animated Solution for Mathematics - Vector Algebra: For non-zero vectors $a, b, c$ , $∣ (a \times b) \cdot c ∣ = ∣ a ∣∣ b ∣∣ c ∣$ holds if and only if

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Visualized Solution

The Given Condition

Volume of a Parallelepiped

Magnitude of $a \times b$

The Dot Product with $c$

Combining the Expansions

Setting Up the Equality

The Maximum Value Condition

$a$ is Perpendicular to $b$

$c$ is Parallel to $a \times b$

Mutually Perpendicular Vectors

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Trigonometric Expansion

The "Aha!" Moment

Final Conclusion

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Animated Solution for Mathematics - Vector Algebra: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For non-zero vectors a,b,c, ∣(a×b)⋅c∣=∣a∣∣b∣∣c∣ holds if and only if

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Visualized Solution

The Given Condition

Volume of a Parallelepiped

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

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

Combining the Expansions

Setting Up the Equality

The Maximum Value Condition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a is Perpendicular to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c is Parallel to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a×b