AHL 3.12—Vector definitions

A scalar is a mathematical quantity described by a single real number (its magnitude), such as temperature or speed. A vector requires both a magnitude (size) and a direction to be fully specified.

The definition directly matches: scalar → magnitude only, vector → magnitude and direction. This is captured in the second option.

The correct answer is "A scalar has magnitude only; a vector has both magnitude and direction." This is the standard IB definition used throughout the course.

The question asks which option correctly distinguishes a scalar from a vector based on their defining properties.

"A scalar has both magnitude and direction; a vector has magnitude only" reverses the definitions entirely — this is the opposite of correct.

"Both scalars and vectors have magnitude and direction, but vectors are always positive" is false on both counts — scalars have no direction, and vectors are not restricted to positive values.

"A vector is defined only in three-dimensional space" is incorrect; vectors can exist in 2D, 3D, or any number of dimensions.

The remaining option, "A scalar has magnitude only; a vector has both magnitude and direction", is the correct and standard definition.

The position vectors are $\mathbf{p}=\left(\begin{array}{c}2\\ -3\\ 5\end{array}\right)$ and $\mathbf{q}=\left(\begin{array}{c}6\\ 1\\ -1\end{array}\right)$.

$\overrightarrow{PQ}=\mathbf{q}-\mathbf{p}=\left(\begin{array}{c}6-2\\ 1-(-3)\\ -1-5\end{array}\right)=\left(\begin{array}{c}4\\ 4\\ -6\end{array}\right)$

The displacement vector $\overrightarrow{PQ}=\left(\begin{array}{c}4\\ 4\\ -6\end{array}\right)$, which is the first option.

We need $\overrightarrow{PQ}=\mathbf{q}-\mathbf{p}$, which is destination minus start (i.e., $Q$ minus $P$).

$\left(\begin{array}{c}-4\\ -4\\ 6\end{array}\right)$ is $\mathbf{p}-\mathbf{q}=\overrightarrow{QP}$, the reverse direction — this is $-\overrightarrow{PQ}$, not $\overrightarrow{PQ}$.

$\left(\begin{array}{c}8\\ -2\\ 4\end{array}\right)$ appears to come from adding components incorrectly; it does not equal $\mathbf{q}-\mathbf{p}$.

$\left(\begin{array}{c}4\\ -2\\ -6\end{array}\right)$ uses $1-3=-2$ for the $y$-component, ignoring the negative sign on $-3$; the correct $y$-component is $1-(-3)=4$.

Only $\left(\begin{array}{c}4\\ 4\\ -6\end{array}\right)$ correctly applies $\mathbf{q}-\mathbf{p}$ component-wise.