Lengths, angles and areas by the scalar product

Space geometry and vectors, grade 12 (4 units) Lesson 15 of 20Learning

Lengths, angles and areas by the scalar product

The method

To find a length or an angle in a solid with vectors:

choose base vectors – three edges or segments whose lengths and the angles between them are known, not in one plane;
work out the table of their products: each vector squared, and each pair. A vector perpendicular to a plane – its product with every vector of that plane is 0;
express the segments asked for as combinations of the base vectors;
a length: $\displaystyle {|\vec{a}|}={\sqrt{\vec{a}^2}}$ , expanded by the short products;
an angle: $\displaystyle {\cos\alpha}={\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\cdot|\vec{b}|}}$ – with both segments leaving the same point;
a triangle’s area from two sides and the angle between them (grade 11):

\displaystyle S

\displaystyle =

\displaystyle \frac{1}{2}ab\sin\gamma

Example: a right pyramid

In the right pyramid $\displaystyle SABCD$ the base $\displaystyle ABCD$ is a rectangle, and $\displaystyle SO$ is the height.
$\displaystyle {AB}={4}$ , $\displaystyle {BC}={3}$ , $\displaystyle {SO}={5}$ .
$\displaystyle {\overrightarrow{AB}}={\vec{u}}$ , $\displaystyle {\overrightarrow{BC}}={\vec{v}}$ , $\displaystyle {\overrightarrow{OS}}={\vec{w}}$ .
Find the angle $\displaystyle BSC$ .

The table of products: $\displaystyle {\vec{u}^2}={16}$ , $\displaystyle {\vec{v}^2}={9}$ , $\displaystyle {\vec{w}^2}={25}$ . $\displaystyle {\vec{u}\cdot\vec{v}}={0}$ (a rectangle), $\displaystyle {\vec{u}\cdot\vec{w}}={\vec{v}\cdot\vec{w}}={0}$ (the height is perpendicular to the base).
$\displaystyle O$ is the diagonals’ midpoint: $\displaystyle {\overrightarrow{OB}}={\frac{1}{2}\overrightarrow{DB}}={\frac{1}{2}\vec{u}-\frac{1}{2}\vec{v}}$ , and $\displaystyle {\overrightarrow{OC}}={\frac{1}{2}\vec{u}+\frac{1}{2}\vec{v}}$ .
$\displaystyle {\overrightarrow{SB}}={-\vec{w}+\frac{1}{2}\vec{u}-\frac{1}{2}\vec{v}}$ , $\displaystyle {\overrightarrow{SC}}={-\vec{w}+\frac{1}{2}\vec{u}+\frac{1}{2}\vec{v}}$ .
$\displaystyle {|\overrightarrow{SB}|^2}={25+\frac{1}{4}\cdot16+\frac{1}{4}\cdot9}={31.25}$ , and $\displaystyle {|\overrightarrow{SC}|^2}={31.25}$ as well. (All the mixed products are 0.)
$\displaystyle {\overrightarrow{SB}\cdot\overrightarrow{SC}}={\vec{w}^2+\frac{1}{4}\vec{u}^2-\frac{1}{4}\vec{v}^2}={25+4-2.25}={26.75}$ .
$\displaystyle {\cos\angle BSC}={\frac{26.75}{31.25}}$ , so $\displaystyle \angle BSC\approx31.13^{\circ}$ .

Exercise 1

In the right prism $\displaystyle ABCA'B'C'$ the bases are equilateral triangles of side 3.
The side edge is 4 long.
$\displaystyle {\overrightarrow{AB}}={\vec{u}}$ , $\displaystyle {\overrightarrow{AC}}={\vec{v}}$ , $\displaystyle {\overrightarrow{AA'}}={\vec{w}}$ .

\displaystyle \overrightarrow{A'C}

in terms of

\displaystyle \vec{u}

\displaystyle \vec{v}

\displaystyle \vec{w}

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle |\overrightarrow{A'C}|

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle \vec{u}\cdot\vec{v}

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle \overrightarrow{A'B}\cdot\overrightarrow{A'C}

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

The angle

\displaystyle BA'C

, to two decimal places

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

The area of the triangle

\displaystyle BA'C

, to two decimal places

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 2

In the right prism $\displaystyle ABCDA'B'C'D'$ the base $\displaystyle ABCD$ is a parallelogram.
$\displaystyle E$ is the midpoint of the edge $\displaystyle D'C'$ .
$\displaystyle {\overrightarrow{AB}}={\vec{u}}$ , $\displaystyle {\overrightarrow{BC}}={\vec{v}}$ , $\displaystyle {\overrightarrow{AA'}}={\vec{w}}$ .
$\displaystyle {|\vec{u}|}={|\vec{v}|}={8}$ , $\displaystyle {|\vec{w}|}={5}$ , $\displaystyle {|\overrightarrow{AE}|}={11}$ .

\displaystyle \overrightarrow{AE}

in terms of

\displaystyle \vec{u}

\displaystyle \vec{v}

\displaystyle \vec{w}

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle \vec{u}\cdot\vec{v}

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

The angle

\displaystyle BAD

, to two decimal places

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

The area of the base

\displaystyle ABCD

, to two decimal places

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 3

The edge of the cube $\displaystyle ABCDA'B'C'D'$ is 2 long, and $\displaystyle M$ is the midpoint of the edge $\displaystyle CC'$ .
$\displaystyle {\overrightarrow{AB}}={\vec{u}}$ , $\displaystyle {\overrightarrow{AD}}={\vec{v}}$ , $\displaystyle {\overrightarrow{AA'}}={\vec{w}}$ .

\displaystyle |\overrightarrow{AM}|

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle \overrightarrow{AM}\cdot\overrightarrow{AC'}

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

The angle

\displaystyle MAC'

, to two decimal places

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Formulas: space geometry and vectors, grade 12 (4 units)

Open as a page

The right-angled triangle and a triangle's area

Pythagoras’ theorem – $\displaystyle c$ the hypotenuse
$\displaystyle a^2+b^2$ $\displaystyle =$ $\displaystyle c^2$
$\displaystyle a$ the leg opposite the angle $\displaystyle \alpha$ , $\displaystyle c$ the hypotenuse
$\displaystyle \sin\alpha$ $\displaystyle =$ $\displaystyle \frac{a}{c}$
$\displaystyle b$ the leg next to the angle $\displaystyle \alpha$
$\displaystyle \cos\alpha$ $\displaystyle =$ $\displaystyle \frac{b}{c}$
The leg opposite the angle over the leg next to it
$\displaystyle \tan\alpha$ $\displaystyle =$ $\displaystyle \frac{a}{b}$
The cosine of an obtuse angle
$\displaystyle \cos(180^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle -\cos\alpha$
A triangle’s area – two sides and the angle between them
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{1}{2}ab\sin\gamma$

The prism

The surface area – the lateral surface and the two bases; $\displaystyle B$ the base’s area
$\displaystyle S$ $\displaystyle =$ $\displaystyle M+2B$
The lateral area of a right prism – $\displaystyle P$ the base’s perimeter, $\displaystyle h$ the height
$\displaystyle M$ $\displaystyle =$ $\displaystyle P\cdot h$
The volume of a prism – $\displaystyle h$ the height, the distance between the bases
$\displaystyle V$ $\displaystyle =$ $\displaystyle B\cdot h$

The pyramid

The surface area – the lateral surface and the base
$\displaystyle S$ $\displaystyle =$ $\displaystyle M+B$
The volume of a pyramid – a third of the prism with the same base and height
$\displaystyle V$ $\displaystyle =$ $\displaystyle \frac{B\cdot h}{3}$

Vectors, geometrically

The triangle rule
$\displaystyle \overrightarrow{AB}+\overrightarrow{BC}$ $\displaystyle =$ $\displaystyle \overrightarrow{AC}$
The difference of two vectors leaving the same point
$\displaystyle \overrightarrow{AD}-\overrightarrow{AB}$ $\displaystyle =$ $\displaystyle \overrightarrow{BD}$
Collinear vectors
$\displaystyle \vec{v}=t\cdot\vec{u}$
A linear combination of $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$
$\displaystyle \vec{w}=s\vec{u}+t\vec{v}$

Vectors in coordinates

The vector between two points – the end minus the start
$\displaystyle \overrightarrow{AB}$ $\displaystyle =$ $\displaystyle (x_B-x_A,\ y_B-y_A,\ z_B-z_A)$
A vector’s length
$\displaystyle |\vec{u}|$ $\displaystyle =$ $\displaystyle \sqrt{x^2+y^2+z^2}$
The midpoint of the segment $\displaystyle AB$
$\displaystyle M=\left(\frac{x_A+x_B}{2},\frac{y_A+y_B}{2},\frac{z_A+z_B}{2}\right)$

The scalar product

The definition – $\displaystyle \alpha$ the angle between the vectors
$\displaystyle \vec{u}\cdot\vec{v}$ $\displaystyle =$ $\displaystyle |\vec{u}|\cdot|\vec{v}|\cdot\cos\alpha$
In coordinates
$\displaystyle \vec{u}\cdot\vec{v}$ $\displaystyle =$ $\displaystyle x_1x_2+y_1y_2+z_1z_2$
The angle between two vectors
$\displaystyle \cos\alpha$ $\displaystyle =$ $\displaystyle \frac{\vec{u}\cdot\vec{v}}{|\vec{u}|\cdot|\vec{v}|}$
A vector’s length by the scalar product
$\displaystyle |\vec{u}|$ $\displaystyle =$ $\displaystyle \sqrt{\vec{u}^2}$
A short scalar product
$\displaystyle (\vec{u}+\vec{v})^2$ $\displaystyle =$ $\displaystyle \vec{u}^2+2\vec{u}\cdot\vec{v}+\vec{v}^2$
A short scalar product
$\displaystyle (\vec{u}+\vec{v})\cdot(\vec{u}-\vec{v})$ $\displaystyle =$ $\displaystyle \vec{u}^2-\vec{v}^2$
Perpendicular vectors
$\displaystyle \vec{u}\perp\vec{v}\iff\vec{u}\cdot\vec{v}=0$