Dot Product of Like Vectors/Mistake

Source Work

1951: B. Hague: An Introduction to Vector Analysis (5th ed.):

Chapter $\text {II}$: The Products of Vectors

$2$. The Scalar Product: $(2.3)$

Mistake

When two vectors are perpendicular, therefore,

$\mathbf A \cdot \mathbf B = 0$, $\qquad (2.2)$

and when they are parallel,

$\mathbf A \cdot \mathbf B = A B$. $\qquad (2.3)$

Correction

Being parallel is insufficient.

Parallel vector quantities may have opposite directions, that is, such that $\theta = 180 \degrees$, in which case $\mathbf A \cdot \mathbf B = -A B$.

For $\mathbf A \cdot \mathbf B = AB$, it is necessary for $\mathbf A$ and $\mathbf B$ to be like, that is, for their directions to be equal.

Sources

• 1951: B. Hague: An Introduction to Vector Analysis (5th ed.) ... (previous) ... (next): Chapter $\text {II}$: The Products of Vectors: $2$. The Scalar Product: $(2.3)$