CLASSICAL-MECHANICS · Interactive Practice

Same Vector, New Numbers: Components Under Rotated Axes

IKey Formulas

Formula Name Description
i^=i^cosθ+j^sinθ\hat{i}' = \hat{i}\cos\theta + \hat{j}\sin\theta Rotated unit vector New x-axis basis in old basis
j^=j^cosθi^sinθ\hat{j}' = \hat{j}\cos\theta - \hat{i}\sin\theta Rotated unit vector New y-axis basis in old basis
x=xcosθ+ysinθx' = x\cos\theta + y\sin\theta Component transform New x-component
y=xsinθ+ycosθy' = -x\sin\theta + y\cos\theta Component transform New y-component (note the minus sign)
R=x2+y2=x2+y2\lvert \vec{R}\rvert = \sqrt{x^2+y^2} = \sqrt{x'^2+y'^2} Magnitude invariance Length is the same in every frame

IIVisualization 1 — Components as Projections

Rotating the axes rewrites a fixed arrow's components — because xx' and yy' are its projections onto the tilted axes.

💡 Challenge: find the rotation θ\theta that drives xx' to 00 — the arrow then lies entirely along the new yy'-axis.

IIIVisualization 2 — Magnitude Is Invariant

Each component swings wildly as the axes turn, so what survives every rotation?

💡 Why it matters: length is a property of the vector itself, not of the frame — the same holds for the dot product of any two vectors.

IVQuiz Questions

Question 1

A vector has components x=3x = 3, y=4y = 4 in system SS. The axes are rotated by θ=90°\theta = 90° to form system SS'.

Using x=xcosθ+ysinθx' = x\cos\theta + y\sin\theta, what is xx'?

Correct! With θ = 90°, x' = 3(0) + 4(1) = 4.

Not quite. Remember cos 90° = 0 and sin 90° = 1, so the cosine term vanishes.

Show solution

Solution:

With θ=90°\theta = 90°: cos90°=0\cos 90° = 0 and sin90°=1\sin 90° = 1.

x=xcosθ+ysinθ=3(0)+4(1)=4x' = x\cos\theta + y\sin\theta = 3(0) + 4(1) = 4

After a 90°90° rotation, the new x-axis points along the old y-axis, so the new x-component picks up the old y-value.

Question 2

The most common error in this derivation involves a sign.

True or False: The correct transformation for the new y-component is y=xsinθ+ycosθy' = x\sin\theta + y\cos\theta.

Correct! The correct form is y' = −x sin θ + y cos θ — the minus sign is essential.

Not quite. Watch the sign: the x sin θ term must be negative.

Show solution

Solution:

The correct relation has a minus sign on the xsinθx\sin\theta term:

y=xsinθ+ycosθy' = -x\sin\theta + y\cos\theta

This minus sign appears because the new yy'-axis tilts away from the old xx-direction. The statement as written omits the minus sign, so it is False.

Question 3

A vector A\vec{A} has components Ax=5A_x = 5, Ay=3A_y = -3 in system SS. The axes are rotated by θ=60°\theta = 60° to form system SS', giving new components Ax0.10A_x' \approx -0.10 and Ay5.83A_y' \approx -5.83.

What is the magnitude A\lvert\vec{A}\rvert measured in system SS'?

Correct! Magnitude is invariant: √(25 + 9) = √34 in both frames.

Not quite. Length doesn't change under rotation — compute √(5² + (−3)²) in the original frame.

Show solution

Solution:

Magnitude is invariant under rotation, so compute it in either frame.

In SS: A=Ax2+Ay2=25+9=345.83\lvert\vec{A}\rvert = \sqrt{A_x^2 + A_y^2} = \sqrt{25 + 9} = \sqrt{34} \approx 5.83

In SS': A=(0.10)2+(5.83)2345.83\lvert\vec{A}\rvert = \sqrt{(-0.10)^2 + (-5.83)^2} \approx \sqrt{34} \approx 5.83

The components look completely different, but the length is identical in both frames.

Question 4

Consider the conceptual meaning of rotating the coordinate axes.

True or False: Rotating the axes by an angle +θ+\theta relabels the vector's components in exactly the same way as keeping the axes fixed and rotating the vector itself by θ-\theta.

Correct! A passive rotation of the axes by +θ matches an active rotation of the vector by −θ.

Not quite. Rotating the axes one way relabels components just like rotating the vector the opposite way.

Show solution

Solution:

This is the distinction between a passive rotation (rotating the axes) and an active rotation (rotating the vector).

Rotating the axes by +θ+\theta produces the same new component numbers as leaving the axes fixed and rotating the vector by θ-\theta. In this video nothing physical moved — only our description of the fixed arrow changed. So the statement is True.

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