Q.Find the values of $x$, $y$ and $z$ so that the vectors $\vec{a}=x\hat{i}+2\hat{j}+z\hat{k}$ and $\vec{b}=2\hat{i}+y\hat{j}+\hat{k}$ are equal.

Concept: Vector Equality — two vectors are equal if and only if their corresponding components are equal.

For $\vec{a} = x\hat{i} + 2\hat{j} + z\hat{k}$ and $\vec{b} = 2\hat{i} + y\hat{j} + \hat{k}$ to be equal:

1. Equate $\hat{i}$-components: $x = 2$
2. Equate $\hat{j}$-components: $2 = y$
3. Equate $\hat{k}$-components: $z = 1$

No other conditions are needed — component-wise equality is both necessary and sufficient.

The idea of vector equality is simple but powerful. Two vectors are the same mathematical object only if they have the same magnitude and the same direction. In component form — when we write a vector as a sum of unit vectors along the coordinate axes — this boils down to a very concrete condition: each corresponding component must match exactly.

Think of it like coordinates of a point in space. The point $(x, y, z)$ is the same as $(2, y, 1)$ only if $x=2$ and $z=1$. Vectors work the same way: the $\hat{i}$-component of $\vec{a}$ must equal the $\hat{i}$-component of $\vec{b}$, and so on for $\hat{j}$ and $\hat{k}$. There is no shortcut around this — it's the definition itself.

Let's apply it step by step.

1. Equate the $\hat{i}$-components. The $\hat{i}$ coefficient in $\vec{a}$ is $x$. In $\vec{b}$, it is $2$. For the vectors to be equal:

2. Equate the $\hat{j}$-components. The $\hat{j}$ coefficient in $\vec{a}$ is $2$. In $\vec{b}$, it is $y$. So:

3. Equate the $\hat{k}$-components. The $\hat{k}$ coefficient in $\vec{a}$ is $z$. In $\vec{b}$, it is $1$. Hence:

That's all there is to it. No extra equations, no hidden conditions. The three component equations are independent and each gives one value directly.

Method: Finding unknowns by equating vectors component-wise

Use this whenever two vectors are declared equal and you must solve for unknown components.

Steps

Step 1: Invoke the equality condition.

Two vectors are equal iff their corresponding components match exactly. For $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ and $\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$:

Step 2: Split the one vector equation into scalar equations.

Equate the $\hat{i}$-coefficients, the $\hat{j}$-coefficients, and the $\hat{k}$-coefficients separately — one vector equation becomes three independent scalar equations.

Step 3: Read off each unknown.

Each component equation yields one unknown directly; no magnitudes, dot products, or extra conditions are involved. Component-wise matching is both necessary and sufficient.

Common Mistakes

Mistake 1: Equating magnitudes (or forming $|\vec{a}|=|\vec{b}|$) instead of components.

Why it's wrong: vector equality is stricter than equal length — it requires each corresponding component to match, so a magnitude equation gives wrong values for $x,y,z$. Correct approach: equate the $\hat{i}$, $\hat{j}$, $\hat{k}$ coefficients separately: $x=2$, $2=y$, $z=1$.

Mistake 2: Cross-matching components (e.g. pairing $x$ with the $\hat{j}$-term).

Why it's wrong: only like components are equated — the $\hat{i}$-part of one with the $\hat{i}$-part of the other. Correct approach: line the vectors up axis-by-axis before setting coefficients equal.