Perpendicularity: perpendicular vectors and a vector perpendicular to a plane

Space geometry and vectors, grade 12 (4 units) Lesson 17 of 20Learning

Perpendicularity: perpendicular vectors and a vector perpendicular to a plane

Perpendicular vectors

$\displaystyle {\cos90^{\circ}}={0}$ , and it is the only angle between $\displaystyle 0^{\circ}$ and $\displaystyle 180^{\circ}$ whose cosine is 0. So:

Two vectors are perpendicular if and only if their scalar product is 0: $\displaystyle {\vec{u}\perp\vec{v}\iff\vec{u}\cdot\vec{v}}={0}$ .

For example $\displaystyle {(1,2,3)\cdot(3,0,-1)}={3+0-3}={0}$ : the vectors are perpendicular.

A vector perpendicular to a plane

A vector is perpendicular to a plane when it is perpendicular to every vector in the plane.

The sufficient condition: if $\displaystyle \vec{n}$ is perpendicular to two non-collinear vectors $\displaystyle \vec{a}$ , $\displaystyle \vec{b}$ that determine the plane, it is perpendicular to the plane.

The reason: every vector in the plane is $\displaystyle s\vec{a}+t\vec{b}$ , and $\displaystyle {\vec{n}\cdot(s\vec{a}+t\vec{b})}={s\,\vec{n}\cdot\vec{a}+t\,\vec{n}\cdot\vec{b}}={0}$ .
This is how a segment is proved to be the height of a pyramid or a prism: it is perpendicular to two intersecting edges of the base.

Example: a right trapezoid and a pyramid

The points $\displaystyle A(3,0,1)$ , $\displaystyle B(3,1,-1)$ , $\displaystyle C(-2,-1,-7)$ , $\displaystyle D(-2,-4,-1)$ , $\displaystyle E(-1,4,3)$ are given.
Prove that $\displaystyle ABCD$ is a right trapezoid, and that the edge $\displaystyle AE$ of the pyramid $\displaystyle ABCDE$ is perpendicular to the plane $\displaystyle ABCD$ .

The proof below shows that $\displaystyle \overrightarrow{DC}$ is collinear with $\displaystyle \overrightarrow{AB}$ , and works out scalar products.

$\displaystyle {\overrightarrow{AB}}={(0,1,-2)}$ , $\displaystyle {\overrightarrow{DC}}={(0,3,-6)}={3\overrightarrow{AB}}$ : $\displaystyle AB\parallel DC$ , and $\displaystyle {DC}={3AB}$ . So $\displaystyle ABCD$ is a trapezoid.
$\displaystyle {\overrightarrow{AD}}={(-5,-4,-2)}$ , $\displaystyle {\overrightarrow{AB}\cdot\overrightarrow{AD}}={0-4+4}={0}$ : $\displaystyle AD\perp AB$ – the trapezoid is right-angled.
$\displaystyle {\overrightarrow{AE}}={(-4,4,2)}$ : $\displaystyle {\overrightarrow{AE}\cdot\overrightarrow{AB}}={0+4-4}={0}$ and $\displaystyle {\overrightarrow{AE}\cdot\overrightarrow{AD}}={20-16-4}={0}$ .
$\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AD}$ are not collinear, so $\displaystyle AE$ is perpendicular to the plane $\displaystyle ABCD$ – it is the pyramid’s height.

Exercise 1

Find the scalar product, and say whether the vectors are perpendicular.

\displaystyle (2,-1,4)

and

\displaystyle (1,2,1)

/ – 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 (3,-2,5)

and

\displaystyle (4,1,-2)

/ – 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 quadrilateral pyramid $\displaystyle SABCD$ the base’s vertices are $\displaystyle A(0,0,0)$ , $\displaystyle B(4,0,0)$ , $\displaystyle C(4,4,0)$ , $\displaystyle D(0,4,0)$ .
The apex is $\displaystyle S(2,2,5)$ .

Prove that the pyramid’s base is a square.

Why is the pyramid’s base in the plane $\displaystyle xy$ ?

\displaystyle |\overrightarrow{SA}|

, exactly

/ – 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{SC}|

, exactly

/ – 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

Find $\displaystyle SB$ and $\displaystyle SD$ too, and explain why the pyramid is right.

Exercise 3

In the parallelepiped $\displaystyle ABCDA'B'C'D'$ the base $\displaystyle ABCD$ is a rhombus of side 1.
$\displaystyle {\angle BAD}={\angle A'AB}={\angle A'AD}={60^{\circ}}$ , $\displaystyle {AA'}={1.2}$ .
$\displaystyle {\overrightarrow{AB}}={\vec{u}}$ , $\displaystyle {\overrightarrow{AD}}={\vec{v}}$ , $\displaystyle {\overrightarrow{AA'}}={\vec{w}}$ .
The point $\displaystyle E$ is on the diagonal $\displaystyle AC$ , and $\displaystyle {AE:EC}={2:3}$ .

\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 \vec{u}\cdot\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 \overrightarrow{A'E}

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 \overrightarrow{A'E}\cdot\vec{u}

/ – 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'E}\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

Explain why $\displaystyle A'E$ is perpendicular to the base plane.

Exercise 4

In the pyramid $\displaystyle SABCD$ the vertices are $\displaystyle A(3,-5,9)$ , $\displaystyle B(6,1,0)$ , $\displaystyle C(0,4,-9)$ , $\displaystyle D(-3,-2,0)$ , $\displaystyle S(6,-19,-10)$ .
The point $\displaystyle K$ satisfies $\displaystyle {\overrightarrow{DK}}={\frac{1}{3}\overrightarrow{DB}}$ .

Explain why $\displaystyle K$ is on the diagonal $\displaystyle DB$ , and find its coordinates.

\displaystyle \overrightarrow{SK}\cdot\overrightarrow{DB}

/ – 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{SK}\cdot\overrightarrow{DA}

/ – 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

What follows about $\displaystyle SK$ ? Explain.

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