Q.Let $\vec{a}$, $\vec{b}$, $\vec{c}$ be three vectors such that $|\vec{a}|=3$, $|\vec{b}|=4$, $|\vec{c}|=5$ and each one of them being perpendicular to the sum of the other two, find $|\vec{a}+\vec{b}+\vec{c}|$.

Concept: Perpendicular Vectors Condition — if $\vec{x} \perp \vec{y}$, then $\vec{x} \cdot \vec{y} = 0$.

Given: each vector is perpendicular to the sum of the other two.

So:

From the first: $\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 0$.

From the second: $\vec{b} \cdot \vec{c} + \vec{b} \cdot \vec{a} = 0$.

From the third: $\vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} = 0$.

Adding all three gives:

Now, …

The condition “each vector is perpendicular to the sum of the other two” is a compact way of saying three things at once:

1. $\vec{a} \perp (\vec{b}+\vec{c})$
2. $\vec{b} \perp (\vec{c}+\vec{a})$
3. $\vec{c} \perp (\vec{a}+\vec{b})$

When two vectors are perpendicular, their dot product is zero. So this condition translates into three dot-product equations. Let’s see what they reveal.

Step 1: Write the perpendicularity conditions as dot products

These are three equations in the three unknown dot products. Let’s label them for clarity:

Let $x = \vec{a}\cdot\vec{b}$, $y = \vec{b}\cdot\vec{c}$, $z = \vec{c}\cdot\vec{a}$.

Then the equations become:

Step 2: Solve for the dot products

From (1): $z = -x$

From (2): $y = -x$

Substitute into (3): $(-x) + (-x) = 0 \Rightarrow -2x = 0 \Rightarrow x = 0$

Then $y = 0$ and $z = 0$.

So all three dot products are zero:

Step 3: Use the pairwise perpendicularity to find $|\vec{a}+\vec{b}+\vec{c}|$

When vectors are mutually perpendicular, the square of the magnitude of their sum is simply the sum of the squares of their magnitudes. This is the vector version of the Pythagorean theorem.

Expand: …

Method: Magnitude of a vector sum from perpendicularity conditions

Convert every "perpendicular" statement into a dot-product-equals-zero equation, deduce the pairwise dot products, then expand $|\vec{a}+\vec{b}+\vec{c}|^2$.

Steps

Step 1: Turn each perpendicularity into a dot equation

"$\vec{a}$ perpendicular to $\vec{b}+\vec{c}$" means $\vec{a}\cdot(\vec{b}+\vec{c}) = 0$, i.e. $\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} = 0$. Write one such equation for each condition.

Step 2: Solve for the pairwise dot products

Adding or comparing the equations typically forces each pairwise dot product to zero — meaning the three vectors are mutually perpendicular.

Step 3: Expand the squared magnitude of the sum …

Common Mistakes

Mistake 1: Adding the magnitudes directly, $|\vec{a}+\vec{b}+\vec{c}| = |\vec{a}|+|\vec{b}|+|\vec{c}|$.

Why it's wrong: magnitude is not additive; the length of a sum depends on directions, captured by the cross terms. Correct approach: expand $|\vec{a}+\vec{b}+\vec{c}|^2$ and take the square root at the end.

Mistake 2: Dropping the $2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a})$ cross terms.

Why it's wrong: the square of a three-term sum contains every pairwise product; ignoring them gives the wrong value whenever the vectors are not perpendicular. Correct approach: always write the full expansion, then use the perpendicularity results to zero out the cross terms. …