The scalar product in coordinates

Space geometry and vectors, grade 12 (4 units) Lesson 16 of 20Learning

The scalar product in coordinates

The scalar product from the coordinates

The unit vectors are perpendicular to each other and each is 1 long, so $\displaystyle {\vec{i}^2}={\vec{j}^2}={\vec{k}^2}={1}$ , and $\displaystyle {\vec{i}\cdot\vec{j}}={\vec{i}\cdot\vec{k}}={\vec{j}\cdot\vec{k}}={0}$ . Multiplying out by the distributive law leaves only the products of matching coordinates:

\displaystyle (x_1,y_1,z_1)\cdot(x_2,y_2,z_2)

\displaystyle =

\displaystyle x_1x_2+y_1y_2+z_1z_2

\displaystyle |\vec{u}|

\displaystyle =

\displaystyle \sqrt{x^2+y^2+z^2}

\displaystyle \cos\alpha

\displaystyle =

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

Example: a triangle’s angle and area

The points $\displaystyle A(-1,1,0)$ , $\displaystyle B(0,1,1)$ , $\displaystyle C(-4,3,-1)$ are given. Find the angle $\displaystyle BAC$ and the area of the triangle $\displaystyle ABC$ .

$\displaystyle {\overrightarrow{AB}}={(1,0,1)}$ , $\displaystyle {\overrightarrow{AC}}={(-3,2,-1)}$ .
$\displaystyle {\overrightarrow{AB}\cdot\overrightarrow{AC}}={-3+0-1}={-4}$ .
$\displaystyle {|\overrightarrow{AB}|}={\sqrt{2}}$ , $\displaystyle {|\overrightarrow{AC}|}={\sqrt{9+4+1}}={\sqrt{14}}$ .
$\displaystyle {\cos\angle BAC}={\frac{-4}{\sqrt{2}\cdot\sqrt{14}}}$ , so $\displaystyle \angle BAC\approx139.11^{\circ}$ – an obtuse angle, since the product is negative.
$\displaystyle {S}={\frac{1}{2}\cdot\sqrt{2}\cdot\sqrt{14}\cdot\sin139.11^{\circ}\approx1.73}$ .

Exercise 1

The vectors $\displaystyle {\vec{u}}={(2,-1,3)}$ and $\displaystyle {\vec{v}}={(1,4,2)}$ are given.

a

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

/ – 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

b

\displaystyle |\vec{u}|

, exactly

/ – 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

c

The angle between

\displaystyle \vec{u}

and

\displaystyle \vec{v}

, to two decimal places

/ – 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

Exercise 2

The points $\displaystyle A(3,-4,1)$ , $\displaystyle B(4,1,8)$ , $\displaystyle C(2,0,7)$ are given.

a

\displaystyle \overrightarrow{AB}\cdot\overrightarrow{AC}

/ – 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

b

The angle

\displaystyle BAC

, to two decimal places

/ – 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

c

The area of the triangle

\displaystyle ABC

, to two decimal places

/ – 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

Exercise 3

The points $\displaystyle A(3,-5,9)$ , $\displaystyle B(6,1,0)$ , $\displaystyle C(0,4,-9)$ , $\displaystyle D(-3,-2,0)$ are given.

a

Prove that the quadrilateral $\displaystyle ABCD$ is a parallelogram.

b

The parallelogram’s acute angle, to two decimal places

/ – 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

c

The parallelogram’s area, to two decimal places

/ – 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

Exercise 4

The points $\displaystyle A(2,1,0)$ , $\displaystyle B(4,3,1)$ , $\displaystyle C(1,3,-2)$ are given.

a

\displaystyle \overrightarrow{AB}\cdot\overrightarrow{AC}

/ – 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

b

What follows about the angle $\displaystyle BAC$ ? What kind of triangle is $\displaystyle ABC$ ?

c

The area of the triangle

\displaystyle ABC

/ – 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

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

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
$\displaystyle \vec{u}\perp\vec{v}\iff\vec{u}\cdot\vec{v}=0$