Animated Solution for Mathematics - Vector Algebra: Let $\vec{u},\vec{v},\vec{w}$ be such that $|\vec{u}|=1,|\vec{v}|=2,|\vec{w}|=3$. If the projection $\vec{v}$ along $\vec{u}$ is equal to that of $\vec{w}$ along $\vec{u}$ and $\vec{v},\vec{w}$ are perpendicular to each other then $|\vec{u}-\vec{v}+\vec{w}|$ equals

A Journey into Geometric Elegance

Welcome, future engineer. Today, we are not just solving a vector problem; we are conducting a symphony. In the world of JEE Advanced, vectors are not just arrows on a page—they are the fundamental language of physics.

When you look at a problem involving magnitudes, projections, and perpendicularity, do not feel intimidated. Instead, look for the hidden symmetries. Let us break this down, step by step, and uncover the elegance hidden within the algebra.

Decoding the Projections

Imagine you are standing in a room with a single light source directly above a line defined by vector $\vec{u}$. If you hold vector $\vec{v}$ and vector $\vec{w}$ in the air, they will cast shadows on that line.

The problem tells us something profound: the shadow of $\vec{v}$ is identical to the shadow of $\vec{w}$. Mathematically, the projection of a vector $\vec{a}$ along $\vec{b}$ is defined as $\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}$.

Since the problem states that the projection of $\vec{v}$ along $\vec{u}$ equals the projection of $\vec{w}$ along $\vec{u}$, we write:

Because $|\vec{u}|=1$, the denominators vanish, leaving us with a beautiful, simple truth: $\vec{v}\cdot \vec{u}=\vec{w}\cdot \vec{u}$. This is our first key. Keep it safe; we will need it later.

The Gift of Perpendicularity

Next, the problem gives us a gift: $\vec{v}$ and $\vec{w}$ are perpendicular. In the language of dot products, perpendicularity is synonymous with zero.

When two vectors are at a $90^{\circ }$ angle, their dot product is zero because $\cos (90^{\circ })=0$. Thus, $\vec{v}\cdot \vec{w}=0$.

This is a massive simplification. Whenever you see 'perpendicular' in a vector problem, immediately think: 'This term is going to vanish.'

The Squaring Strategy

Now, we face the main challenge: finding the magnitude $|\vec{u}-\vec{v}+\vec{w}|$. Dealing with the magnitude of a sum of three vectors is daunting. We cannot simply add the lengths.

The standard JEE strategy here is to square the entire expression. Why? Because the square of a magnitude is the dot product of the vector with itself, which allows us to distribute the terms.

Let us expand $|\vec{u}-\vec{v}+\vec{w}{|}^2$ using the identity for the square of a trinomial:

Let us clean this up. We know that $|-\vec{v}{|}^2=|\vec{v}{|}^2$. Distributing the dot products, we get:

The Grand Cancellation

This is where the magic happens. Look at the terms we have generated. We know $\vec{v}\cdot \vec{w}=0$, so the term $-2(\vec{v}\cdot \vec{w})$ disappears entirely.

We are left with:

Recall our finding from Phase 1: $\vec{u}\cdot \vec{v}=\vec{w}\cdot \vec{u}$. If we substitute $\vec{w}\cdot \vec{u}$ for $\vec{u}\cdot \vec{v}$, the expression becomes:

Everything cancels out! The cross-terms are gone, leaving us with the sum of the squares of the magnitudes. This is the elegance of vector algebra. We are left with:

Substituting the given values $|\vec{u}|=1$, $|\vec{v}|=2$, and $|\vec{w}|=3$:

Finally, taking the square root, we arrive at our answer: $\sqrt{14}$. You have successfully navigated the complexity and found the simple, beautiful truth underneath. Keep this mindset—look for the cancellations, trust the algebra, and you will conquer any problem.