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

Q: Wait, why do we reject the negative root here?

Actually, in this problem, we are dealing with magnitudes, which are always non-negative. The dot product definition $\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos(\theta)$ involves the absolute magnitudes. Since $|\vec{b}|$ and $|\vec{c}|$ are lengths, they must be positive. There is no negative root to reject because the equation $10 = 5 |\vec{c}| (1/2)$ yields a unique positive solution.

Q: Why is $\vec{a}$ in the plane of $\vec{b}$ and $\vec{c}$?

This is a beautiful geometric consequence! The cross product $\vec{b} \times \vec{c}$ is, by definition, perpendicular to the plane containing $\vec{b}$ and $\vec{c}$. If $\vec{a}$ is also perpendicular to this cross product vector, it must lie in the plane that is perpendicular to the cross product vector—which is the plane of $\vec{b}$ and $\vec{c}$ itself.

Q: Can we use the vector triple product expansion formula?

Yes, you absolutely can! The formula $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$ is a powerful tool. However, in this specific problem, we don't know the individual components of the vectors, so the geometric approach is much faster and more elegant. Always look for the geometric shortcut before diving into algebraic expansions!

Question

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

Sigma Percentile · Accepted Answer

### The Elegance of Vector Geometry

Welcome, fellow traveler of the JEE journey. Today, we are not just solving a problem; we are peeling back the layers of a beautiful geometric puzzle.

Vectors are the language of the physical universe, and understanding them requires more than just memorizing formulas—it requires a shift in perspective. Let us dive into this problem, where $\vec{a}$, $\vec{b}$, and $\vec{c}$ are not just letters, but entities living in a three-dimensional space.

### Phase 1: Unlocking the Hidden Magnitude

We begin with the basics. We are given $|\vec{a}| = \sqrt{3}$, $|\vec{b}| = 5$, and the dot product $\vec{b} \cdot \vec{c} = 10$. We also know the angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{3}$.

Our first mission is to find the magnitude of $\vec{c}$. Recall the fundamental definition of the dot product: $\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos(	heta)$. This is our golden key.

We know $\vec{b} \cdot \vec{c} = 10$, $|\vec{b}| = 5$, and $	heta = \frac{\pi}{3}$. Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$, we can write:

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

Simplifying this, we get $10 = \frac{5}{2} |\vec{c}|$. Multiplying both sides by $2$, we find $20 = 5 |\vec{c}|$, which leads us to $|\vec{c}| = 4$. Just like that, we have unlocked the length of our second vector.

### Phase 2: The Vertical Vector

Now, consider the cross product $\vec{b} 	imes \vec{c}$. Geometrically, this vector is the powerhouse of the operation. By the right-hand rule, $\vec{b} 	imes \vec{c}$ is a vector that stands perfectly perpendicular to the plane containing $\vec{b}$ and $\vec{c}$.

To work with it, we need its magnitude, $|\vec{b} 	imes \vec{c}|$. The formula is $|\vec{b} 	imes \vec{c}| = |\vec{b}| |\vec{c}| \sin(	heta)$. Substituting our known values: $|\vec{b}| = 5$, $|\vec{c}| = 4$, and $	heta = \frac{\pi}{3}$.

Since $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$, we have:

$$|\vec{b} 	imes \vec{c}| = 5 \cdot 4 \cdot \frac{\sqrt{3}}{2} = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3}$$

We now have the magnitude of this vertical vector. Let us call this vector $\vec{V} = \vec{b} 	imes \vec{c}$.

### Phase 3: The Geometric Insight

The problem gives us a crucial constraint: $\vec{a} \perp (\vec{b} 	imes \vec{c})$. This is the "Aha!" moment.

If $\vec{b} 	imes \vec{c}$ is pointing straight up from the plane, and $\vec{a}$ is perpendicular to it, then $\vec{a}$ must lie flat on the plane containing $\vec{b}$ and $\vec{c}$. This means the angle between $\vec{a}$ and the vector $\vec{V} = \vec{b} 	imes \vec{c}$ is exactly $\frac{\pi}{2}$.

### Phase 4: The Final Calculation

We are asked to find $|\vec{a} 	imes (\vec{b} 	imes \vec{c})|$, which is $|\vec{a} 	imes \vec{V}|$. Using the magnitude formula for the cross product again, we have:

$$|\vec{a} 	imes \vec{V}| = |\vec{a}| |\vec{V}| \sin(\phi)$$

where $\phi$ is the angle between $\vec{a}$ and $\vec{V}$. We know $|\vec{a}| = \sqrt{3}$, $|\vec{V}| = 10\sqrt{3}$, and $\phi = \frac{\pi}{2}$. Since $\sin(\frac{\pi}{2}) = 1$, the calculation becomes:

$$|\vec{a} 	imes \vec{V}| = \sqrt{3} \cdot 10\sqrt{3} \cdot 1 = 10 \cdot 3 = 30$$

And there it is! The **final answer is 30**. It is not just a number; it is the result of understanding the spatial relationship between these vectors.

Enter Numerical Value:

Visualized Solution

Visualizing the Vectors

The Dot Product Formula

Substituting Known Values

Solving for $∣ c ∣$

The Cross Product Vector

Magnitude of Cross Product

Substituting for Cross Product

Calculating $∣ b \times c ∣$

Positioning Vector $a$

The Final Cross Product

Substituting Final Values

Final Calculation

Conclusion & Key Takeaways

The Sigma Insight: Vector Triple Product

The Elegance of Vector Geometry

Phase 1

Phase 2

Phase 3

Phase 4

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