Revise Scalar and vector products; planes for IB Mathematics: Analysis and Approaches (3.9)

The scalar product multiplies two vectors and returns a single number. It has two equivalent forms — an algebraic one you compute from components, and a geometric one built from magnitudes and the angle between the vectors. Setting the two equal is what lets you extract an angle, and the special case where the number is zero is what tests for a right angle.

For $\mathbf{a}=\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$ and $\mathbf{b}=\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}$: $\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta = a_1b_1+a_2b_2+a_3b_3.$

The vector product is available only in 3D. It takes two vectors and returns a third vector that is perpendicular to both of them — which is exactly the ingredient you need to describe a plane. Its magnitude carries geometric meaning too: it equals the area of the parallelogram the two vectors span.

$\mathbf{a}\times\mathbf{b}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\a_1&a_2&a_3\\b_1&b_2&b_3\end{vmatrix}=\begin{pmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{pmatrix},$ $\quad |\mathbf{a}\times\mathbf{b}|=|\mathbf{a}||\mathbf{b}|\sin\theta.$

A plane is a flat surface stretching out forever in 3D. To fix it uniquely you need just two things: a normal vector $\mathbf{n}$ (the direction 'straight out' of the surface) and one point known to lie on it. Every form of the plane's equation is really the same statement — that the vector from the known point to any point on the plane is perpendicular to $\mathbf{n}$.

With normal $\mathbf{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}$ and a point with position vector $\mathbf{a}_0$ on the plane: vector form: $\mathbf{r}\cdot\mathbf{n}=\mathbf{a}_0\cdot\mathbf{n}$ Cartesian form: $ax+by+cz=d, \quad d=\mathbf{a}_0\cdot\mathbf{n}$ normal form: $\mathbf{r}\cdot\hat{\mathbf{n}}=p, \quad \hat{\mathbf{n}}=\dfrac{\mathbf{n}}{|\mathbf{n}|}.$

Both angle calculations run through the dot product, but you must feed it the right vectors and pick the right trig ratio. For two planes, dot their normals and use cosine, because the angle between the normals equals the angle between the planes. For a line and a plane, dot the line's direction with the plane's normal — but that gives the angle to the normal, so the tilt from the surface itself comes out through sine.

| Line (direction $\mathbf{d}$) and plane (normal $\mathbf{n}$): $\sin\theta=\dfrac{ | \mathbf{d}\cdot\mathbf{n} | }{ | \mathbf{d} | | \mathbf{n} | }$. | | --- | --- | --- | --- | --- | --- | | Two planes (normals $\mathbf{n}_1,\mathbf{n}_2$): $\cos\theta=\dfrac{ | \mathbf{n}_1\cdot\mathbf{n}_2 | }{ | \mathbf{n}_1 | | \mathbf{n}_2 | }$. |

On Paper 1 the marks are analytic: each is tied to a specific line of working — a method mark ($M$) for the correct approach, or an accuracy mark ($A$) for the right value, where an $A$ mark is dependent on the $M$ mark before it. The engine accepts exact forms and any correctly-rounded value, and awards degrees or radians alike. Watch how each of the four marks below is earned by a specific line.