The Vector That Would Not Normalise
A graphics programmer is rotating a triangle in the plane. The vertices are A(1, 2), B(4, 6) and C(-1, 5). She computes the edge vectors AB = (3, 4) and AC = (-2, 3), notes that AB has length 5 and AC has length sqrt(13), and builds the 2x2 matrix M whose columns are AB and AC. She claims that because the columns are not parallel, M is invertible, and that M maps the unit square to a parallelogram of area equal to |det M|. A colleague warns that the triangle's area is not |det M| and that the rotation matrix she plans to multiply by must be orthogonal, which M is not. She must sort out which claims hold and compute the triangle's area correctly.
1. Compute det M, decide whether M is invertible, and determine the area of the parallelogram spanned by AB and AC. Then state the triangle's area and explain the factor relating the two.
Read the worked answer
Determinant. With columns AB = (3, 4) and AC = (-2, 3),
det M = 3*3 - (-2)*4 = 9 + 8 = 17.
Since det M = 17 is nonzero, M is invertible; the columns are linearly independent (not parallel), confirming the programmer's invertibility claim.
Parallelogram area = |det M| = 17.
Triangle area = half the parallelogram area = 17/2 = 8.5. The factor of 1/2 arises because the triangle with vertices A, B, C is exactly half of the parallelogram with vertices A, B, C, B + AC. So the colleague is right that |det M| is not the triangle's area; it is twice it.
2. The programmer wants a rotation matrix R (orthogonal, det = 1) to rotate the triangle. Explain what properties R must satisfy, why M fails them, and compute the image of AB under a 90-degree anticlockwise rotation.
Read the worked answer
Orthogonality requirements. A rotation matrix R must satisfy R^T R = I (orthonormal columns) and det R = 1 (orientation-preserving). Equivalently R = [[cos t, -sin t],[sin t, cos t]].
Why M fails. Columns of M are (3,4) and (-2,3). Their dot product is 3(-2) + 4(3) = -6 + 12 = 6, not 0, so the columns are not orthogonal, and their lengths are 5 and sqrt(13), not 1. Hence M^T M is not the identity; M is invertible but not orthogonal, so it is not a rotation (it also scales and shears).
Image of AB under 90 degrees anticlockwise. R = [[0, -1],[1, 0]]. Then R(3, 4) = (03 - 14, 13 + 04) = (-4, 3). Its length is 5, same as AB, as a rotation must preserve length. The programmer should use R, not M, for the rotation step.