Animated Solution for Mathematics - Vector Algebra: Let $\vec{a}=\hat{i}-\hat{j}$, $\vec{b}=\hat{j}-\hat{k}$, $\vec{c}=\hat{k}-\hat{i}$. If $\vec{d}$ is a unit vector such that $\vec{a}\cdot \vec{d}=0=[\vec{b}\vec{c}\vec{d}]$, then $\vec{d}$ equals

A Vector Detective Story

Welcome, fellow explorer of the JEE Advanced landscape. Today, we are not just solving a problem; we are performing a geometric investigation.

We have an unknown vector $\vec{d}$, and it is hiding behind a veil of constraints. Our job is to strip away that veil using the power of vector algebra.

The Foundation of the Unknown

We begin by defining our target. Since we know nothing about $\vec{d}$, we assume the most general form: $\vec{d}=x\hat{i}+y\hat{j}+z\hat{k}$.

But wait—the problem gives us a gift. It tells us $\vec{d}$ is a unit vector. This is not just a label; it is a mathematical anchor.

It means the magnitude is unity, leading us to our first foundational equation:

Keep this equation close. It is the gatekeeper that will allow us to solve for the specific values of $x$, $y$, and $z$ once we find their relationships.

The Orthogonality Filter

Next, we look at the condition $\vec{a}\cdot \vec{d}=0$. In the world of vectors, a dot product of zero is a loud signal: it screams perpendicularity.

We are told $\vec{a}=\hat{i}-\hat{j}$. When we compute the dot product with our unknown $\vec{d}$, we get:

This simplifies beautifully to $x-y=0$, or $x=y$. Just like that, we have reduced our degrees of freedom. We no longer have three unknowns; we have two.

The Coplanarity Key

Now, we face the most intimidating part of the problem: $[\vec{b}\vec{c}\vec{d}]=0$. Do not let the notation scare you.

The scalar triple product is simply the volume of a parallelepiped. If the volume is zero, the vectors are trapped in a flat, two-dimensional plane. This means $\vec{d}$ is coplanar with $\vec{b}$ and $\vec{c}$.

If $\vec{d}$ lives in the plane defined by $\vec{b}$ and $\vec{c}$, it must be perpendicular to the normal of that plane. How do we find that normal? We use the cross product: $\vec{n}=\vec{b}\times \vec{c}$.

Calculating this determinant is a rite of passage:

Since $\vec{d}$ is perpendicular to this normal $\vec{n}$, their dot product must also be zero:

The Synthesis

We are in the endgame. We have $x=y$ from our first condition, and $x+y+z=0$ from our second.

Substituting $y=x$ into the second equation gives us $2x+z=0$, or $z=-2x$. Now, every component of $\vec{d}$ is expressed in terms of $x$:

Finally, we return to our anchor—the unit vector condition $x^2+y^2+z^2=1$. Substituting our expressions for $y$ and $z$:

With $x$ determined, the vector $\vec{d}$ reveals itself. By substituting $x$ back into our expression, we arrive at the final result:

Look at that elegance. We navigated through orthogonality, coplanarity, and normalization to find the exact orientation of our vector. You have successfully decoded the geometry. Well done.