Let three vectors a, b and c be such that a × b = c, b × c = a and |a| = 2. Then which one of the fo

Q: Wait, why do we assume the vectors are mutually perpendicular? Is that always true for cross products?

Great question! The cross product $\vec{u} \times \vec{v}$ is defined as a vector perpendicular to both $\vec{u}$ and $\vec{v}$. In this problem, we are given $\vec{a} \times \vec{b} = \vec{c}$. This means $\vec{c}$ is perpendicular to $\vec{a}$ and $\vec{b}$. Then, we are given $\vec{b} \times \vec{c} = \vec{a}$, which means $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{c}$. By transitivity, all three are mutually perpendicular. It is not a general rule for all vectors, but a specific consequence of these two equations combined!

Q: Why does the scalar triple product $[\vec{a} \vec{b} \vec{c}]$ equal $|\vec{a}|^2$?

The scalar triple product is defined as $\vec{a} \cdot (\vec{b} \times \vec{c})$. We are given $\vec{b} \times \vec{c} = \vec{a}$. Substituting this in, we get $\vec{a} \cdot \vec{a}$, which is by definition $|\vec{a}|^2$. Since $|\vec{a}| = 2$, the result is $2^2 = 4$. It is a beautiful simplification that happens when the vectors are related in this specific way.

Q: Why can't we just use the standard expansion for $|3\vec{a} + \vec{b} - 2\vec{c}|^2$?

You absolutely can! The standard expansion is $|\vec{u} + \vec{v} + \vec{w}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2 + 2(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u})$. However, because our vectors are mutually perpendicular, the dot products $\vec{a} \cdot \vec{b}, \vec{b} \cdot \vec{c},$ and $\vec{c} \cdot \vec{a}$ are all zero. This causes all those cross-terms to vanish, leaving you with only the sum of the squared magnitudes. It is not that you can't use the expansion; it is that the geometry makes the expansion collapse into something much simpler.

Question

Let three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ be such that $\vec{a} 	imes \vec{b} = \vec{c}, \vec{b} 	imes \vec{c} = \vec{a}$ and $|\vec{a}| = 2$. Then which one of the following is not true ?

Options:

(A) $\vec{a} 	imes ((\vec{b} + \vec{c}) 	imes (\vec{b} - \vec{c})) = \vec{0}$

(B) Projection of $\vec{a}$ on $(\vec{b} 	imes \vec{c})$ is 2

(C) $[\vec{a} \vec{b} \vec{c}] + [\vec{c} \vec{a} \vec{b}] = 8$

(D) $|3\vec{a} + \vec{b} - 2\vec{c}|^2 = 51$

Sigma Percentile · Accepted Answer

### The Geometry of the Cross Product: A Vector Dance

Welcome, fellow explorer of the mathematical universe. Today, we are not just solving a problem; we are uncovering a hidden symmetry.

When you look at the equations $\vec{a} 	imes \vec{b} = \vec{c}$ and $\vec{b} 	imes \vec{c} = \vec{a}$, do not see them as mere algebraic constraints. See them as a dance.

A cross product is a geometric machine that takes two vectors and spits out a third, one that is perfectly orthogonal to the first two. When you see these relations, you are looking at a system that is locked in a mutually perpendicular embrace, much like the axes of the Cartesian coordinate system itself.

### Phase 1: The Revelation of Orthogonality

Let us pause and reflect on the definition of the cross product. If $\vec{a} 	imes \vec{b} = \vec{c}$, then $\vec{c}$ must be perpendicular to both $\vec{a}$ and $\vec{b}$.

Similarly, if $\vec{b} 	imes \vec{c} = \vec{a}$, then $\vec{a}$ must be perpendicular to both $\vec{b}$ and $\vec{c}$. This is the "Aha!" moment.

We have three vectors, and each one is perpendicular to the other two. They are mutually perpendicular. This realization is the key that unlocks the entire problem.

### Phase 2: The Hunt for Magnitudes

Now that we know the geometry, let us quantify it. The magnitude of a cross product is given by:

$$|\vec{u} 	imes \vec{v}| = |\vec{u}||\vec{v}|\sin(	heta)$$

Since our vectors are mutually perpendicular, the angle $	heta$ between any two of them is $90^\circ$, and $\sin(90^\circ) = 1$. This simplifies our life immensely.

The magnitude equations become:

$$|\vec{a}||\vec{b}| = |\vec{c}| \quad 	ext{and} \quad |\vec{b}||\vec{c}| = |\vec{a}|$$

We are given $|\vec{a}| = 2$. Substituting this, we get $2|\vec{b}| = |\vec{c}|$ and $|\vec{b}||\vec{c}| = 2$.

Solving this system is trivial: $|\vec{b}|(2|\vec{b}|) = 2$, which leads to $|\vec{b}|^2 = 1$, so $|\vec{b}| = 1$. Consequently, $|\vec{c}| = 2$. We have our lengths: $2, 1,$ and $2$.

### Phase 3: Testing the Truth

Now, we evaluate the options. Option 1 asks us to look at $\vec{a} 	imes ((\vec{b} + \vec{c}) 	imes (\vec{b} - \vec{c}))$.

Expanding the inner cross product, we get:

$$\vec{b} 	imes \vec{b} - \vec{b} 	imes \vec{c} + \vec{c} 	imes \vec{b} - \vec{c} 	imes \vec{c}$$

Since $\vec{b} 	imes \vec{b} = \vec{0}$ and $\vec{c} 	imes \vec{c} = \vec{0}$, and $\vec{c} 	imes \vec{b} = -(\vec{b} 	imes \vec{c})$, this simplifies to $-2(\vec{b} 	imes \vec{c}) = -2\vec{a}$. Then, $\vec{a} 	imes (-2\vec{a}) = -2(\vec{a} 	imes \vec{a}) = \vec{0}$. Option 1 is true!

For Option 2, the projection of $\vec{a}$ on $\vec{b} 	imes \vec{c}$ is simply the projection of $\vec{a}$ on $\vec{a}$, which is $|\vec{a}| = 2$. True!

For Option 3, the scalar triple product is:

$$[\vec{a} \vec{b} \vec{c}] = \vec{a} \cdot (\vec{b} 	imes \vec{c}) = \vec{a} \cdot \vec{a} = |\vec{a}|^2 = 4$$

By the cyclic property, $[\vec{c} \vec{a} \vec{b}] = 4$ as well. Their sum is 8. True!

Finally, Option 4. We need $|3\vec{a} + \vec{b} - 2\vec{c}|^2$. Because the vectors are mutually perpendicular, the dot products vanish.

We are left with:

$$|3\vec{a}|^2 + |\vec{b}|^2 + |-2\vec{c}|^2 = 9|\vec{a}|^2 + |\vec{b}|^2 + 4|\vec{c}|^2$$

Substituting the values:

$$9(4) + 1 + 4(4) = 36 + 1 + 16 = 53$$

The option claims 51. This is the **false statement**!

### Conclusion

Mathematics is not about memorizing formulas; it is about seeing the structure beneath the surface. By recognizing the mutual perpendicularity, we turned a complex vector problem into a simple arithmetic exercise. Keep looking for that geometric soul in every problem you solve.

Animated Solution for Mathematics - Vector Algebra: Let three vectors $a, b$ and $c$ be such that $a \times b = c, b \times c = a$ and $∣ a ∣ = 2$ . Then which one of the following is not true ?

Select Answer:

Visualized Solution

Analyze Vector Relations

Establish Magnitude Relations

Substitute Known Magnitude

Solve for $∣ b ∣$ and $∣ c ∣$

Evaluate Option 1: Inner Cross Product

Evaluate Option 1: Simplification

Evaluate Option 2: Projection

Evaluate Option 3: Scalar Triple Product

Evaluate Option 4: Vector Magnitude Squared

Evaluate Option 4: Final Calculation

Final Conclusion

The Sigma Insight: Vector (Cross) Product

The Geometry of the Cross Product

Phase 1

Phase 2

Phase 3

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; } Let three vectors a,b and c be such that a×b=c,b×c=a and ∣a∣=2. Then which one of the following is not true ?