Animated Solution for Mathematics - Vector Algebra: Let a unit vector $\hat{u}=x\hat{i}+y\hat{j}+z\hat{k}$ make angles $\frac{\pi }{2},\frac{\pi }{3}$ and $\frac{2\pi }{3}$ with the vectors $\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k}$, $\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{\sqrt{2}}\hat{k}$ and $\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}$ respectively. If $\vec{v}=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{\sqrt{2}}\hat{k}$, then $|\hat{u}-\vec{v}{|}^2$ is equal to

A Journey into 3D Space

Welcome, future engineers! Today, we are going to peel back the layers of a beautiful problem in three-dimensional geometry. Often, when students see a problem involving a unit vector $\hat{u}=x\hat{i}+y\hat{j}+z\hat{k}$ making specific angles with other vectors, they panic.

They look for shortcuts or complex geometric theorems. But the secret to mastering JEE Advanced physics and mathematics is not finding a shortcut; it is finding the most elegant path. In vector algebra, that path is almost always the dot product.

The Dot Product Bridge

Imagine you are standing in a 3D coordinate system. You have an unknown vector $\hat{u}$ pointing somewhere in space. We know it is a unit vector, which means its magnitude is exactly $1$.

We are given three other vectors, let's call them ${\vec{p}}_1,{\vec{p}}_2,$ and ${\vec{p}}_3$. The problem provides the angles between $\hat{u}$ and these vectors. How do we connect an angle $\theta$ to the components of a vector?

The answer lies in the definition of the dot product:

Since $\hat{u}$ is a unit vector, $|\hat{u}|=1$. If the vectors ${\vec{p}}_n$ were also unit vectors, the dot product would simply be $\cos (\theta )$. Let's check: ${\vec{p}}_1=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k}$.

The magnitude is:

They are all unit vectors! This simplifies our life immensely. Our bridge is simply $\hat{u}\cdot {\vec{p}}_n=\cos ({\theta }_n)$.

The Three Pillars of the System

Now, let's build our system of equations. For the first vector, ${\vec{p}}_1=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k}$, the angle is $\frac{\pi }{2}$. Since $\cos (\frac{\pi }{2})=0$, the dot product is zero.

We write:

This gives us $\frac{x}{\sqrt{2}}+\frac{z}{\sqrt{2}}=0$, or simply $x+z=0$. This is our first pillar.

For the second vector, ${\vec{p}}_2=\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{\sqrt{2}}\hat{k}$, the angle is $\frac{\pi }{3}$. Since $\cos (\frac{\pi }{3})=\frac{1}{2}$, we have:

This simplifies to $\frac{y}{\sqrt{2}}+\frac{z}{\sqrt{2}}=\frac{1}{2}$, which becomes $y+z=\frac{1}{\sqrt{2}}$.

Finally, for the third vector, ${\vec{p}}_3=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}$, the angle is $\frac{2\pi }{3}$. Since $\cos (\frac{2\pi }{3})=-\frac{1}{2}$, we have:

This simplifies to $\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=-\frac{1}{2}$, or $x+y=-\frac{1}{\sqrt{2}}$.

The Algebraic Symphony

We now have a system of three linear equations: 1) $x+z=0$ 2) $y+z=\frac{1}{\sqrt{2}}$ 3) $x+y=-\frac{1}{\sqrt{2}}$

From (1), we know $z=-x$. Substituting this into (2) yields $y-x=\frac{1}{\sqrt{2}}$. Now, look at (3) and this new equation side-by-side:

If we add these two equations, the $x$ terms cancel out perfectly! We get $2y=0$, which means $y=0$.

With $y=0$, equation (3) tells us $x=-\frac{1}{\sqrt{2}}$. Since $z=-x$, we find $z=\frac{1}{\sqrt{2}}$. We have found our vector:

The Final Act

The problem asks for $|\hat{u}-\vec{v}{|}^2$, where $\vec{v}=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{\sqrt{2}}\hat{k}$. Let's perform the subtraction component-wise:

Finally, we calculate the squared magnitude:

The final answer is $\frac{5}{2}$. Notice how the initial complexity of the vectors and angles melted away once we trusted the algebraic process. That is the power of a systematic approach.