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Vectors: magnitude, direction and products · Resolve parallel and perpendicular components

Choose the object the question asks for

A signed scalar, a length and a vector component are different results. Start with the same unit direction, then choose the required form.

  1. Normalise the target direction

    The target vector b must be nonzero. The signed scalar component s may be negative; an ordinary length cannot.

    u=b/∣b∣,s=a⋅u.\mathbf u=\mathbf b/|\mathbf b|,\qquad s=\mathbf a\cdot\mathbf u.

    Projection length · scalarTake the absolute value

    Use ∣s∣|s|. The signed scalar component itself is ss; preserve its sign when that is what the question asks for.

    Parallel component · vectorMultiply by the unit direction

    Use sus\mathbf u. A negative ss means this component points opposite to u\mathbf u; do not replace it by ∣s∣|s|.

    Perpendicular component · vectorSubtract the parallel component

    Use a−su\mathbf a-s\mathbf u. Its dot product with u\mathbf u is zero. If a perpendicular length is requested, take the magnitude of this vector.

    Check that the two vector components add back to a. Reversing the target direction changes the signed scalar’s sign, while the projection length and both vector components stay unchanged.