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Where do the standard basis vectors land during a 90-degree rotation around the y-axis in 3D space?

During a 90-degree rotation around the y-axis, i^\hat{i} moves to the negative z-axis at (0,0,−1)(0, 0, -1), j^\hat{j} remains stationary on the y-axis at (0,1,0)(0, 1, 0), and k^\hat{k} swings to the positive x-axis at (1,0,0)(1, 0, 0).

Conditions

  • Rotation angle is exactly 90 degrees
  • Axis of rotation is the y-axis
  • Right-handed coordinate system convention assumed for sign determination

Reasoning, step by step

  1. Visualize the rotation axis: the y-axis remains fixed.
  2. Track i^\hat{i} (initially on positive x): it rotates 90 degrees towards the negative z-direction.
  3. Track j^\hat{j} (initially on positive y): since it lies on the axis of rotation, its position does not change.
  4. Track k^\hat{k} (initially on positive z): it rotates 90 degrees towards the positive x-direction.
  5. Compile the final coordinates into column vectors to form the rotation matrix.

Example

The script describes: 'the i^\hat{i} vector moves down to the negative z-axis at coordinates (0,0,−1)(0, 0, -1). The j^\hat{j} vector remains stationary... at (0,1,0)(0, 1, 0)... Meanwhile, the k^\hat{k} vector swings over to the positive x-axis at (1,0,0)(1, 0, 0).'

Common misconceptions

  • Assuming j^\hat{j} changes direction because the whole space is rotating; it stays fixed because it is on the axis.
  • Mixing up the signs for i^\hat{i} and k^\hat{k} depending on whether the rotation is clockwise or counter-clockwise relative to the viewer.

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