Using the scalar product: proofs, lengths and angles

The scalar product adds two tools to proofs with vectors:

• perpendicularity: prove that $\displaystyle {\underline{u}\cdot\underline{v}}={0}$;
• lengths and angles: work out $\displaystyle \underline{w}^2$ or $\displaystyle {\cos\alpha}={\frac{\underline{u}\cdot\underline{v}}{|\underline{u}|\cdot|\underline{v}|}}$.

What you need to know about the base vectors: their lengths and the angles between them – that is, all the products $\displaystyle \underline{u}^2$, $\displaystyle \underline{v}^2$, $\displaystyle \underline{u}\cdot\underline{v}$.

We prove: the diagonals of a rhombus are perpendicular to each other.

The proof below multiplies the diagonals in a scalar product, and shows the product is 0.

• $\displaystyle {\overrightarrow{AB}}={\underline{u}}$, $\displaystyle {\overrightarrow{AD}}={\underline{v}}$, and in a rhombus $\displaystyle {|\underline{u}|}={|\underline{v}|}$.
• $\displaystyle {\overrightarrow{AC}}={\underline{u}+\underline{v}}$, $\displaystyle {\overrightarrow{BD}}={\underline{v}-\underline{u}}$.
• $\displaystyle {\overrightarrow{AC}\cdot\overrightarrow{BD}}={(\underline{v}+\underline{u})\cdot(\underline{v}-\underline{u})}={\underline{v}^2-\underline{u}^2}={0}$.
• So $\displaystyle AC\perp BD$.

We prove: if a line is perpendicular to two intersecting lines in a plane through its foot, it is perpendicular to every line in the plane through the foot.

The proof below writes any line in the plane as a combination of the two lines.

• $\displaystyle \underline{n}$ is a vector along the line, and $\displaystyle \underline{u}$, $\displaystyle \underline{v}$ vectors along the two intersecting lines: $\displaystyle {\underline{n}\cdot\underline{u}}={0}$, $\displaystyle {\underline{n}\cdot\underline{v}}={0}$.
• $\displaystyle \underline{u}$ and $\displaystyle \underline{v}$ are independent, so every vector $\displaystyle \underline{w}$ in the plane is $\displaystyle s\underline{u}+t\underline{v}$.
• $\displaystyle {\underline{n}\cdot\underline{w}}={s(\underline{n}\cdot\underline{u})+t(\underline{n}\cdot\underline{v})}={0}$.

The three perpendiculars theorem is proved the same way: $\displaystyle {\overrightarrow{PA}}={\overrightarrow{PP'}+\overrightarrow{P'A}}$, and for every vector $\displaystyle \underline{d}$ in the plane $\displaystyle {\underline{d}\cdot\overrightarrow{PP'}}={0}$. So $\displaystyle {\underline{d}\cdot\overrightarrow{PA}}={\underline{d}\cdot\overrightarrow{P'A}}$: one is 0 if and only if the other is.

In the box $\displaystyle ABCDA'B'C'D'$: $\displaystyle {\overrightarrow{AB}}={\underline{u}}$, $\displaystyle {\overrightarrow{AD}}={\underline{v}}$, $\displaystyle {\overrightarrow{AA'}}={\underline{w}}$, and $\displaystyle {|\underline{u}|}={4}$, $\displaystyle {|\underline{v}|}={3}$, $\displaystyle {|\underline{w}|}={2}$.
Find the angle between the diagonals $\displaystyle AC'$ and $\displaystyle BD'$.

The solution below uses the fact that the box’s edges are perpendicular: the product of any two different vectors among $\displaystyle \underline{u}$, $\displaystyle \underline{v}$, $\displaystyle \underline{w}$ is 0.

• $\displaystyle {\overrightarrow{AC'}}={\underline{u}+\underline{v}+\underline{w}}$, $\displaystyle {\overrightarrow{BD'}}={-\underline{u}+\underline{v}+\underline{w}}$.
• $\displaystyle {\overrightarrow{AC'}\cdot\overrightarrow{BD'}}={-\underline{u}^2+\underline{v}^2+\underline{w}^2}={-16+9+4}={-3}$.
• $\displaystyle {|\overrightarrow{AC'}|^2}={16+9+4}={29}$, and also $\displaystyle {|\overrightarrow{BD'}|^2}={29}$.
• $\displaystyle {\cos\alpha}={\frac{-3}{29}}$, so $\displaystyle \alpha\approx95.94^{\circ}$. The angle between the lines is the acute one: $\displaystyle {180^{\circ}-95.94^{\circ}}={84.06^{\circ}}$.

$\displaystyle AB$ is a diameter of a circle with centre $\displaystyle O$, and $\displaystyle C$ is a point on the circle. $\displaystyle {\overrightarrow{OB}}={\underline{a}}$, $\displaystyle {\overrightarrow{OC}}={\underline{c}}$.

Prove: an inscribed angle on a diameter is a right angle.

In the parallelogram $\displaystyle ABCD$: $\displaystyle {\overrightarrow{AB}}={\underline{u}}$, $\displaystyle {\overrightarrow{AD}}={\underline{v}}$.
Prove: the sum of the squares of the diagonals equals the sum of the squares of the sides. Open $\displaystyle \overrightarrow{AC}^2+\overrightarrow{BD}^2$.

In the triangle $\displaystyle ABC$, $\displaystyle CD$ is the height to the side $\displaystyle AB$ (or to its extension).
$\displaystyle {\overrightarrow{CB}}={\underline{v}}$, $\displaystyle {\overrightarrow{CA}}={\underline{u}}$, $\displaystyle {\overrightarrow{AD}}={t\cdot\overrightarrow{AB}}$.
$\displaystyle {|\underline{v}|}={2}$, $\displaystyle {|\underline{u}|}={1}$, $\displaystyle {\cos\angle ACB}={\frac{3}{4}}$.

$\displaystyle {\overrightarrow{CD}\cdot\overrightarrow{AB}}={0}$. Open it, put in $\displaystyle \underline{u}^2$, $\displaystyle \underline{v}^2$, $\displaystyle \underline{u}\cdot\underline{v}$, and find $\displaystyle t$.

What does the sign of $\displaystyle t$ tell you about where $\displaystyle D$ is?

In a cube $\displaystyle ABCDA'B'C'D'$ of edge 2: $\displaystyle {\overrightarrow{AB}}={\underline{u}}$, $\displaystyle {\overrightarrow{AD}}={\underline{v}}$, $\displaystyle {\overrightarrow{AA'}}={\underline{w}}$.
Find the angle between the diagonal $\displaystyle AC'$ and the edge $\displaystyle AB$. Round to two decimal places.