Let A, B, C be vectors of length 3, 4, 5 respectively. Let A be perpendicular to B + C, B to C + A a

Q: Wait, why do we set the dot product to zero? Is it always zero?

Yes, absolutely! For any two non-zero vectors, if they are perpendicular, the angle $\theta$ between them is $90^\circ$. Since the dot product is defined as $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}| \cos(\theta)$, and $\cos(90^\circ) = 0$, the dot product must be zero. It is the fundamental signature of orthogonality.

Q: Can I just assume $\vec{A}, \vec{B}, \vec{C}$ are along the $x, y, z$ axes?

That is a very tempting trap! If you assume they are mutually perpendicular, you are imposing a constraint that the problem never gave you. The problem only states that each vector is perpendicular to the *sum* of the other two. Always stick to the algebraic conditions provided rather than assuming a specific coordinate geometry unless necessary.

Q: Why does the sum of dot products become zero?

It is a beautiful consequence of the symmetry in the problem. When you expand all three conditions, you get terms like $\vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C} = 0$. When you sum all three equations, every single dot product pair appears exactly twice. Since the sum of the equations is zero, the sum of the dot products must also be zero. It is not a coincidence; it is the structure of the problem!

Question

Let $\vec{A}, \vec{B}, \vec{C}$ be vectors of length 3, 4, 5 respectively. Let $\vec{A}$ be perpendicular to $\vec{B} + \vec{C}$, $\vec{B}$ to $\vec{C} + \vec{A}$ and $\vec{C}$ to $\vec{A} + \vec{B}$. Then the length of vector $\vec{A} + \vec{B} + \vec{C}$ is .........

Sigma Percentile · Accepted Answer

### The Geometry of Hidden Symmetry

Welcome, future engineer! Today, we are going to unravel a problem that looks like a tangled mess of vectors but is actually a masterclass in algebraic elegance.

We are given three vectors, $\vec{A}$, $\vec{B}$, and $\vec{C}$, with magnitudes $3$, $4$, and $5$ respectively. We are told they satisfy a peculiar set of perpendicularity conditions.

At first glance, you might be tempted to try and draw these in 3D space, perhaps placing them along the axes. But pause—stop! If you assume they are mutually perpendicular, you are adding constraints that aren't there. Let us instead rely on the raw, powerful language of vector algebra.

### The Power of the Dot Product

The problem gives us three conditions: $\vec{A} \perp (\vec{B} + \vec{C})$, $\vec{B} \perp (\vec{C} + \vec{A})$, and $\vec{C} \perp (\vec{A} + \vec{B})$. In the world of vectors, perpendicularity is synonymous with the dot product being zero.

So, we translate these geometric statements into the language of mathematics:

1. $\vec{A} \cdot (\vec{B} + \vec{C}) = 0$

2. $\vec{B} \cdot (\vec{C} + \vec{A}) = 0$

3. $\vec{C} \cdot (\vec{A} + \vec{B}) = 0$

Now, let us use the distributive property. This is where the magic happens. Expanding the first equation, we get $\vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C} = 0$. If we do this for all three, we get a system of equations that seems to be whispering a secret to us.

### The Algebraic Symphony

Let us write them all out:

- $\vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C} = 0$

- $\vec{B} \cdot \vec{C} + \vec{B} \cdot \vec{A} = 0$

- $\vec{C} \cdot \vec{A} + \vec{C} \cdot \vec{B} = 0$

Look closely at these terms. If we sum these three equations, notice that $\vec{A} \cdot \vec{B}$ appears twice, $\vec{B} \cdot \vec{C}$ appears twice, and $\vec{C} \cdot \vec{A}$ appears twice. We get:

$$2(\vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{C} + \vec{C} \cdot \vec{A}) = 0$$

This implies that the sum of the mutual dot products is exactly zero! This is a massive simplification. We have effectively eliminated the interaction terms between our vectors.

### The Grand Finale

Now, we need the magnitude of the sum, $|\vec{A} + \vec{B} + \vec{C}|$. We know the identity for the square of the sum of three vectors:

$$|\vec{A} + \vec{B} + \vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + |\vec{C}|^2 + 2(\vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{C} + \vec{C} \cdot \vec{A})$$

We already know the magnitudes: $|\vec{A}| = 3$, $|\vec{B}| = 4$, $|\vec{C}| = 5$. And thanks to our previous step, we know that the entire term $2(\vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{C} + \vec{C} \cdot \vec{A})$ is equal to zero.

Substituting these values in, we get:

$$|\vec{A} + \vec{B} + \vec{C}|^2 = 3^2 + 4^2 + 5^2 + 0$$

$$|\vec{A} + \vec{B} + \vec{C}|^2 = 9 + 16 + 25 = 50$$

Taking the square root, we find the magnitude is $\sqrt{50}$, which simplifies to **$5\sqrt{2}$**.

Isn't that beautiful? We didn't need to know the direction of a single vector. We only needed to trust the algebra. Keep this confidence in your toolkit—often, the most complex-looking problems are just waiting for you to apply the right identity.

Animated Solution for Mathematics - Vector Algebra: Let $A, B, C$ be vectors of length 3, 4, 5 respectively. Let $A$ be perpendicular to $B + C$ , $B$ to $C + A$ and $C$ to $A + B$ . Then the length of vector $A + B + C$ is .........

Visualized Solution

Given Magnitudes $∣ A ∣$ , $∣ B ∣$ , $∣ C ∣$

Condition of Perpendicularity

Perpendicularity Equations

Expanding $A \cdot (B + C) = 0$

Expanding Other Conditions

Summing the Equations

The Magnitude Identity

Substituting Known Values

Calculating the Sum of Squares

Final Result $52$

The Sigma Insight: Scalar (Dot) Product

The Geometry of Hidden Symmetry

The Power of the Dot Product

The Algebraic Symphony

The Grand Finale

Similar Questions

Let the vectors $a, b, c$ be such that $∣ a ∣ = 2, ∣ b ∣ = 4$ and $∣ c ∣ = 4$ . If the projection of $b$ on $a$ is equal to the projection of $c$ on $a$ and $b$ is perpendicular to $c$ , then the value of $∣ a + b - c ∣$ is

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