Q.Write all the unit vectors in XY-plane.

The key idea is that any vector in the XY-plane has zero $z$-component, and a unit vector has magnitude 1.

Step 1: A general vector in the XY-plane is $\vec{v}=a\hat{i}+b\hat{j}$, with magnitude $\sqrt{a^2+b^2}$.

Step 2: For a unit vector we need $\sqrt{a^2+b^2}=1$, i.e. $a^2+b^2=1$ — the unit circle.

Step 3: Parameterise by the angle $\theta$ the vector makes with the positive $x$-axis: $a=\cos\theta$, $b=\sin\theta$. Then

and $|\vec{v}|=\sqrt{\cos^2\theta+\sin^2\theta}=1$, so every such vector is indeed a unit vector.

Setting up the condition

A unit vector is any vector with magnitude exactly 1. In the XY-plane every vector can be written as $a\hat{i}+b\hat{j}$ with $a,b$ real (the $z$-component is 0). The unit-vector condition is

So the task reduces to finding all ordered pairs $(a,b)$ satisfying $a^2+b^2=1$ — the equation of the unit circle. Every point on that circle gives exactly one unit vector.

Step-by-step

1. Parameterise the circle. The standard parameterisation of $a^2+b^2=1$ is

where $\theta$ is the angle measured anticlockwise from the positive $x$-axis. As $\theta$ runs over $[0,2\pi)$ we cover every point of the circle exactly once.

2. Write the vector form. Substituting into $a\hat{i}+b\hat{j}$ gives

3. Check the magnitude.

so every such vector is a unit vector.

4. Are there any others? No. If $a^2+b^2=1$ then $(a,b)$ lies on the unit circle, so some $\theta$ gives $a=\cos\theta$, $b=\sin\theta$. The parameterisation captures every possibility.

Method: Describing all unit vectors in a plane

To list every unit vector lying in a coordinate plane, translate "length $1$" into an equation, recognise it as the unit circle, and parametrise it with an angle.

Steps

Step 1: Write the general vector in the plane

A vector in the XY-plane has zero out-of-plane component: $\vec{v} = x\hat{i} + y\hat{j}$ (its $\hat{k}$ component is $0$).

Step 2: Impose the unit-length condition

Length $1$ means $\sqrt{x^2+y^2} = 1$, i.e. $x^2+y^2 = 1$ — the equation of the unit circle. The problem is now "all points on the unit circle."

Step 3: Parametrise with an angle

Every point of $x^2+y^2=1$ is $x = \cos\theta$, $y = \sin\theta$. So the complete family is

Because $\cos^2\theta+\sin^2\theta = 1$ automatically, every such vector is a unit vector, and every planar unit vector arises this way — so the description is exhaustive, not just a few examples.

Common Mistakes

Mistake 1: Listing only $\pm\hat{i}$ and $\pm\hat{j}$ as "the unit vectors."

Why it's wrong: those are just four of infinitely many; any direction in the plane has a unit vector, e.g. $\tfrac{1}{\sqrt{2}}(\hat{i}+\hat{j})$. Correct approach: give the whole continuous family $\cos\theta\,\hat{i}+\sin\theta\,\hat{j}$.

Mistake 2: Forgetting the zero $\hat{k}$ component (staying "in the XY-plane").

Why it's wrong: including any $\hat{k}$ part takes the vector out of the plane the question restricts to. Correct approach: enforce $z = 0$, leaving only $x\hat{i}+y\hat{j}$ with $x^2+y^2=1$.