Let a, b and c be three vectors such that |a| = √3, |b| = 5, b.c = 10 and the angle between b and c

Q: Why do we need to calculate $|\vec{c}|$ first? Can't we just use the dot product directly in the cross product formula?

That is a great question! The dot product gives you a scalar value, but the cross product magnitude depends on the individual lengths of the vectors. You cannot bypass finding $|\vec{c}|$ because the cross product formula $|\vec{b} \times \vec{c}| = |\vec{b}| |\vec{c}| \sin(\theta)$ requires the specific magnitude of each vector involved. Think of it as gathering all your ingredients before you start cooking.

Q: The problem says $\vec{a} \perp (\vec{b} \times \vec{c})$. Does this mean $\vec{a}$ is parallel to $\vec{b}$ or $\vec{c}$?

Not necessarily! It just means $\vec{a}$ lies in the plane defined by $\vec{b}$ and $\vec{c}$. Because $\vec{b} \times \vec{c}$ is the normal vector to that plane, any vector in that plane is perpendicular to the normal. It is a beautiful geometric constraint that simplifies the final cross product calculation significantly.

Q: Why is the final answer just a scalar? I thought the cross product resulted in a vector.

You are absolutely right that a cross product results in a vector, but the question asks for the *magnitude* of that resulting vector, denoted by the vertical bars $|...|$. The magnitude is always a scalar value, representing the length of that vector in space.

Question

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three vectors such that $|\vec{a}| = \sqrt{3}, |\vec{b}| = 5, \vec{b}.\vec{c} = 10$ and the angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{3}$. If $\vec{a}$ is perpendicular to vector $\vec{b} 	imes \vec{c}$, then $|\vec{a} 	imes (\vec{b} 	imes \vec{c})|$ is equal to ________

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers and physicists. Today, we are not just solving a problem; we are constructing a mental model of space. When you look at vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$, do not see them as mere symbols on a page. See them as arrows piercing through the fabric of three-dimensional space.

Our journey begins with a classic setup: we have two vectors, $\vec{b}$ and $\vec{c}$, and we know their relationship through a dot product. The dot product, $\vec{b} \cdot \vec{c} = 10$, is our key. It tells us how much these two vectors 'agree' with each other.

We know $|\vec{b}| = 5$ and the angle between them is $\frac{\pi}{3}$. To unlock the magnitude of $\vec{c}$, we reach for the fundamental definition:

$$\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos\left(\frac{\pi}{3}ight)$$

Substituting our knowns, we get:

$$10 = 5 	imes |\vec{c}| 	imes \frac{1}{2}$$

This simplifies beautifully to $10 = \frac{5}{2} |\vec{c}|$, which reveals that $|\vec{c}| = 4$. We have our first piece of the puzzle.

### The Vertical Ascent: The Cross Product

Now, we shift our perspective. We need to understand the vector $\vec{b} 	imes \vec{c}$. Imagine $\vec{b}$ and $\vec{c}$ lying flat on your desk. Their cross product is a vector that shoots straight up, perpendicular to the desk surface.

Its magnitude is given by:

$$|\vec{b} 	imes \vec{c}| = |\vec{b}| |\vec{c}| \sin\left(\frac{\pi}{3}ight)$$

We have all the ingredients: $|\vec{b}| = 5$, $|\vec{c}| = 4$, and $\sin\left(\frac{\pi}{3}ight) = \frac{\sqrt{3}}{2}$. Plugging these in, we find:

$$|\vec{b} 	imes \vec{c}| = 5 	imes 4 	imes \frac{\sqrt{3}}{2} = 10\sqrt{3}$$

This is the 'height' of our new vector in space. It is a powerful, tangible value.

### The Final Convergence: The Triple Product

We are now at the climax of our problem. We are asked for the magnitude of $\vec{a} 	imes (\vec{b} 	imes \vec{c})$. Let us treat $(\vec{b} 	imes \vec{c})$ as a single vector, let's call it $\vec{v}$. We need to find $|\vec{a} 	imes \vec{v}|$.

The definition of the magnitude of a cross product is $|\vec{a}| |\vec{v}| \sin(\phi)$, where $\phi$ is the angle between $\vec{a}$ and $\vec{v}$. The problem gives us a crucial hint: $\vec{a}$ is perpendicular to $(\vec{b} 	imes \vec{c})$. This means $\phi = \frac{\pi}{2}$, and $\sin\left(\frac{\pi}{2}ight) = 1$.

The expression simplifies to:

$$|\vec{a}| 	imes |\vec{b} 	imes \vec{c}| 	imes 1$$

We know $|\vec{a}| = \sqrt{3}$ and we just calculated $|\vec{b} 	imes \vec{c}| = 10\sqrt{3}$. The final calculation is elegant:

$$\sqrt{3} 	imes 10\sqrt{3} = 10 	imes 3 = 30$$

There it is. The complexity collapses into a clean, integer result. The final answer is **30**. This is the beauty of vector algebra—when you respect the geometry, the math rewards you with simplicity.

Enter Numerical Value:

Visualized Solution

Visualizing the Vectors $b$ and $c$

The Dot Product Formula

Substituting Known Values

Evaluating the Cosine Term

Solving for $∣ c ∣$

Magnitude of Cross Product $b \times c$

Substituting Values for Cross Product

Calculating $∣ b \times c ∣$

Analyzing the Perpendicularity

Vector Triple Product Magnitude Formula

Final Substitution

The Final Result

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Vertical Ascent

The Final Convergence

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