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Miscellaneous Exercise · Q1

Q.Write down a unit vector in XY-plane, making an angle of 30∘30^\circ with the positive direction of x-axis.

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✓ Free question

A unit vector in the XY-plane at 30∘30^\circ from the positive x-axis has components (cos⁡30∘,sin⁡30∘)=(32,12)(\cos 30^\circ, \sin 30^\circ) = \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right). The answer is 32i^+12j^\boxed{\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}}.

Why Direction Vectors Work

Any vector in the XY-plane can be thought of as an arrow from the origin to a point (x,y)(x, y). For a unit vector (length = 1), the tip of the arrow lies exactly on the unit circle. The angle θ\theta measured counterclockwise from the positive x-axis gives the coordinates of that point as (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta). So the vector itself is cos⁡θ i^+sin⁡θ j^\cos\theta\,\hat{i} + \sin\theta\,\hat{j}.

This is the cleanest way to write a direction vector — no square roots, no Pythagoras needed after you know the angle.

Step-by-step

  1. Identify the angle. The problem says 30∘30^\circ with the positive x-axis. In standard position, that’s θ=30∘\theta = 30^\circ.

  2. Write the components using cosine and sine.

    For any angle θ\theta, a unit vector in the XY-plane is

u^=cos⁡θ i^+sin⁡θ j^.\hat{u} = \cos\theta\,\hat{i} + \sin\theta\,\hat{j}.

  1. Plug in θ=30∘\theta = 30^\circ.

cos⁡30∘=32,sin⁡30∘=12.\cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \sin 30^\circ = \frac{1}{2}.

  1. Assemble the vector.

u^=32 i^+12 j^.\hat{u} = \frac{\sqrt{3}}{2}\,\hat{i} + \frac{1}{2}\,\hat{j}.

Tip

If the angle were measured from the y-axis or given in some other reference, you’d swap sine and cosine. But “with the positive direction of x-axis” always means standard position — so cosine goes with i^\hat{i}, sine with j^\hat{j}.

Watch out

A common mistake is to write 12i^+32j^\frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j} — that’s the vector at 60∘60^\circ, not 30∘30^\circ. Always double-check which trig value belongs to which axis.

  1. Verify the length.

∣u^∣=(32)2+(12)2=34+14=1=1.\left|\hat{u}\right| = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1.

So it is indeed a unit vector.

✓Final answer

The required unit vector is 32i^+12j^\boxed{\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}}.

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