2021-01-11
What is the vector equation for a straight line??
$$ \pmb{r} =\pmb{a} + \lambda \pmb{b} $$
$$\pmb{r} =\pmb{a} + \lambda \pmb{b}$$ What is the variable $\pmb{a}$ called??
The position vector.
$$\pmb{r} =\pmb{a} + \lambda \pmb{b}$$ What is the variable $\pmb{b}$ called??
The direction vector.
$$\pmb{r} =\pmb{a} + \lambda \pmb{b}$$ What is the variable $\pmb{r}$ called??
The general position vector for a point on the line.
$$\pmb{r} =\pmb{a} + \lambda \pmb{b}$$ What is the informal name for $\lambda$ called here??
The sliding factor.
Is $\lambda$ a scalar or vector parameter??
A scalar.
What are used to represent scalar parameters in vector equations??
Greek letters such as $\lambda$ or $\mu$.
$$ \left(\begin{matrix} 2 \ -1 \end{matrix}\right) + \lambda \left(\begin{matrix} 1 \ 2 \end{matrix}\right) $$ Which is the position vector??
$$ \left(\begin{matrix} 2 \\ -1 \end{matrix}\right) $$
$$ \left(\begin{matrix} 2 \ -1 \end{matrix}\right) + \lambda \left(\begin{matrix} 1 \ 2 \end{matrix}\right) $$ Which is the direction vector??
$$ \left(\begin{matrix} 1 \\ 2 \end{matrix}\right) $$
$$ \left(\begin{matrix} 1 \ 3 \end{matrix}\right) $$ What is the gradient of this vector??
$$ 3 $$
$$ \left(\begin{matrix} 7 \ 4 \end{matrix}\right) $$ What is the gradient of this vector??
$$ \frac{4}{7} $$
$$ \left(\begin{matrix} 0 \ -5 \end{matrix}\right) + \lambda \left(\begin{matrix} -1 \ -2 \end{matrix}\right) = \left(\begin{matrix} 2 \ -1 \end{matrix}\right) + \lambda \left(\begin{matrix} 2 \ 4 \end{matrix}\right) $$ What are the two reasons this is true??
- The two position vectors goes to a point on the same line
- The two direction vectors have the same gradient
$$ \left(\begin{matrix} 1 \ 2 \end{matrix}\right) $$ What is the direction vector that is perpindicular??
$$ \left(\begin{matrix} -2 \\ 1 \end{matrix}\right) $$
$$ \left(\begin{matrix} -1 \ -2 \end{matrix}\right) $$ What is the direction vector that is perpindicular??
$$ \left(\begin{matrix} 2 \\ -1 \end{matrix}\right) $$
$$ \left(\begin{matrix} a \ b \end{matrix}\right) $$ What is the direction vector that is perpindicular??
$$ \left(\begin{matrix} -b \\ a \end{matrix}\right) $$
$$ r = \left(\begin{matrix} -1 \ -4 \end{matrix}\right) + \lambda \left(\begin{matrix} 1 \ 3 \end{matrix}\right) $$ How could you rewrite this to find the point of intersection with another line??
$$ \left(\begin{matrix} x \\ y \end{matrix}\right) = \left(\begin{matrix} -1 \\ -4 \end{matrix}\right) + \lambda \left(\begin{matrix} 1 \\ 3 \end{matrix}\right) $$
$$\left(\begin{matrix} -1 \ -4 \end{matrix}\right) + \lambda \left(\begin{matrix} 1 \ 3 \end{matrix}\right)$$ How could you rewrite this as one vector??
$$ \left(\begin{matrix} -1 + \lambda \\ -4 + 3\lambda \end{matrix}\right) $$
$$\left(\begin{matrix} 2 \ 3 \end{matrix}\right) + \mu \left(\begin{matrix} 1 \ -1 \end{matrix}\right)$$ How can you rewrite this as one vector??
$$ \left(\begin{matrix} 2 +\mu \\ 3 - \mu \end{matrix}\right) $$
If $$\left(\begin{matrix} x \ y \end{matrix}\right) = \left(\begin{matrix} -1 + \lambda \ -4 + 3\lambda \end{matrix}\right)$$ and $$\left(\begin{matrix} x \ y \end{matrix}\right) = \left(\begin{matrix} 2 +\mu \ 3 - \mu \end{matrix}\right)$$, what must be true??
$$ \left(\begin{matrix} -1 + \lambda \\ -4 + 3\lambda \end{matrix}\right) = \left(\begin{matrix} 2 +\mu \\ 3 - \mu \end{matrix}\right) $$
If you solve the simulataneous equations $$\left(\begin{matrix} -1 + \lambda \ -4 + 3\lambda \end{matrix}\right) = \left(\begin{matrix} 2 +\mu \ 3 - \mu \end{matrix}\right)$$ and get values for $\lambda$ and $\mu$, what must you remember to do next??
Substitute into the line equation to get a position vector.
2021-01-12
What does the general 3d position vector look like??
$$ \left(\begin{matrix} x \\ y \\ z \end{matrix}\right) $$
$$ \left(\begin{matrix} 0 \ 0 \end{matrix}\right) + \lambda \left(\begin{matrix} 1 \ 3 \end{matrix}\right) $$ What would the Cartesian equation be??
$$ y = 3x $$
$$ \left(\begin{matrix} -1 \ -4 \end{matrix}\right) + \lambda \left(\begin{matrix} 1 \ 3 \end{matrix}\right) $$ When working out the Cartesian equation, what point do you know the line passes through??
$$ \left(\begin{matrix} -1 \\ -4 \end{matrix}\right) $$
If $$ \pmb{a} = \left(\begin{matrix} a_1 \ a_2 \ a_3 \end{matrix}\right) $$ and $$ \pmb{b} = \left(\begin{matrix} a_1 \ a_2 \ a_3 \end{matrix}\right) $$ for $$ r = a + \lambda b $$, how can you write the line in Cartesian form??
$$ \frac{x-a_1}{b_1} = \frac{x-a_2}{b_2} = \frac{x-a_3}{b_3} $$
$$x = -3 + 2\lambda$$ What is $\lambda$ equal to??
$$ \lambda = \frac{x+3}{2} $$
$$y = 3\lambda$$ What is $\lambda$ equal to??
$$ \lambda = \frac{y}{3} $$
$$z = -4 + 2\lambda$$ What is $\lambda$ equal to??
$$ \lambda = \frac{z-4}{2} $$
$$\lambda = \frac{x+3}{2} \ \lambda = \frac{y}{3} \ \lambda = \frac{z-4}{2}$$ How could you write the Cartesian equation of the line??
$$ \frac{x+3}{2} = \frac{y}{3} = \frac{z-4}{2} $$
$$\frac{x+3}{2} = \frac{y}{3} = \frac{z-4}{2}$$ What variable is every term equal to??
$$ \lambda $$
$$\lambda = \frac{x + 3}{2}$$ How could you rewrite this to convert back into a vector??
$$ x = -3 + 2\lambda $$
$$x = -3 + 2\lambda \ y = 3\lambda \ z = -4 + \lambda$$ What is this in vector form??
$$ \left(\begin{matrix} x \\ y \\ z \end{matrix}\right) = \left(\begin{matrix} -3+2\lambda \\ 3\lambda \\ -4+\lambda \end{matrix}\right) $$
$$ \left(\begin{matrix} 10 \ 6 \ -2 \end{matrix}\right) $$ How could you simplify this direction vector??
$$ \left(\begin{matrix} 5 \\ 3 \\ -1 \end{matrix}\right) $$
$$ \left(\begin{matrix} -1 \ -2 \ -3 \end{matrix}\right) $$ How could you simplify this direction vector??
$$ \left(\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}\right) $$
What is the formula for the line if it passes through point $\pmb{a}$ and is parallel to vector $\pmb{b}$??
$$ r = \pmb{a} + \lambda\pmb{b} $$
If a line passes through two points $\pmb{a}$ and $\pmb{b}$, what is the direction vector??
$$ \pmb{b} - \pmb{a} $$
If a line passes through two points $\pmb{a}$ and $\pmb{b}$, why is the direction vector $\pmb{b} - \pmb{a}$??
Because that is the vector between the two points which is where we need to go.
If $\overrightarrow{OA} = \pmb{a}$ and $\overrightarrow{OB} = \pmb{b}$, what is the vector $\overrightarrow{AB}$??
$$ \pmb{b} - \pmb{a} $$
If $\overrightarrow{OA} = \pmb{a}$ and $\overrightarrow{OB} = \pmb{b}$, what is the vector $\overrightarrow{BA}$??
$$ \pmb{a} - \pmb{b} $$
$$ \pmb{r} = \left(\begin{matrix} 1+\lambda \ 2+\lambda \ 3+\lambda \end{matrix}\right) $$ If $$ \left(\begin{matrix} 0 \ a \ b \end{matrix}\right) $$ lies on the line, what must the vector be equal to??
$$ \left(\begin{matrix} 0 \\ 1 \\ 2 \end{matrix}\right) $$
What are the two steps to show two lines are the same??
- Show direction vectors are multiples of each other.
- Show both points are on the same line.
$$ a = \left(\begin{matrix} 1 \ 2 \ 3 \end{matrix}\right) \ b = \left(\begin{matrix} 2 \ 4 \ 6 \end{matrix}\right) $$ How could you write the two direction vectors to show they’re the same??
$$ \left(\begin{matrix} 2 \\ 4 \\ 6 \end{matrix}\right) = 2 \left(\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}\right) $$
$$\pmb{r} = \left(\begin{matrix} -2+\lambda \ 1-2\lambda \ 4+\lambda \end{matrix}\right)$$ How could you show that the point $$ \left(\begin{matrix} 2 \ 1 \ 3 \end{matrix}\right) $$ isn’t on the line??
For the top part to be correct, $\lambda = 4$ but this makes the other points wrong.
2021-01-19
What are non-intersecting, non-parallel lines in 3d called??
Skew lines.
What are non-intersecting, non-parallel lines in 2d called??
Stupid because they don’t exist.
2021-01-25
What is true about the shortest distance between a point and a line??
It is perpindicular to the line and passing through the point.
What is the formula for $\overrightarrow{RP}$, the line passing through a general point on a line $R$ and a point $P$??
$$ \pmb{p} - \pmb{r} $$
What must $\overrightarrow{RP}$, the line passing through a general point on a line $R$ and a point $P$, be perpindicular to in order to make the shortest distance between $P$ and the line??
The direction vector of the line.
$$\overrightarrow{PA} = \left(\begin{matrix} 2\lambda-6 \ 1 \ 2+\lambda \end{matrix}\right) \ \pmb{r} = \left(\begin{matrix} 0 \ 4 \ 1 \end{matrix}\right) + \lambda \left(\begin{matrix} 2 \ 0 \ 1 \end{matrix}\right) $$ What equation could you solve to find a value of $\lambda$ for which $\overrightarrow{PA}$ is perpindicular to the line??
$$ \left(\begin{matrix} 2\lambda-6 \\ 1 \\ 2+\lambda \end{matrix}\right) \cdot \left(\begin{matrix} 2 \\ 0 \\ 1 \end{matrix}\right) $$
2021-01-25
What is true about the shortest distance between two lines??
The line segment joining them is parallel to both.
What’s the first step of finding the shortest distance between two parallel lines??
Rewriting both lines so they have the same position vector.
$$r = \pmb{a} + \lambda \pmb{b} \ \pmb{r} = \pmb{c} + \lambda \pmb{d}$$ If you’re subtracting two lines from one another with the same direction vector, what substitution can you make so things are easier??
$$ t = \lambda - \mu $$
The line segment $\overrightarrow{AB}$ is a combination of two points on parallel lines, $A$ and $B$. For $\overright{AB}$ to be perpindicular to the direction vector $\pmb{b}$ of the lines and therefore the shortest distance between them, what must be true??
$$ \overrightarrow{AB} \cdot \pmb{b} = 0 $$
What is the 5 step process for finding the distance between two parallel lines with direction vector $\pmb{b}$ and general points $A$ and $B$??
- Make direction vectors the same.
- Find an expression for $\overrightarrow{AB}$.
- Make an equation by setting $\overrightarrow{AB} \cdot \pmb{b} = 0$.
- Substitute solution back into $\overrightarrow{AB}$ in order to get the direction vector between the two points.
- Use Pythagoras to work out the length of the line.
Why is it easier to find the perpindicular distance between two parallel lines rather than two skew lines??
Because if the direction vectors are the same you only have to solve one equation rather than two.
What is the 6 step process for finding the perpindicular distance between two skew lines with direction vectors $\pmb{b}$ and $\pmb{d}$ for general points $A$ and $B$??
- Find an expression for $\overrightarrow{AB}$
- Make one equation for $\overrightarrow{AB} \cdot \pmb{b} = 0$.
- Make another equation for $\overrightarrow{AB} \cdot \pmb{d} = 0$.
- Solve the two equations simultaneously to get values for $\lambda$ and $\mu$.
- Substitute back into $\overrightarrow{AB}$ in order to get the direction vector between the two points.
- Use Pythagoras to work out the length of the line.
$$ \overrightarrow{AB} = \left(\begin{matrix} 1+2\lambda+3\mu \ 4+5\lambda+6\mu \ 7+8\lambda+9\mu \end{matrix}\right) $$ If the two direction vectors of two lines are $$ \left(\begin{matrix} 1 \ 2 \ 3 \end{matrix}\right) $$ and $$ \left(\begin{matrix} 4 \ 5 \ 6 \end{matrix}\right) $$, what two equations can you write in order to find the values of $\mu$ and $\lambda$ where the distance between them is the shortest??
$$ \left(\begin{matrix} 1+2\lambda+3\mu \\ 4+5\lambda+6\mu \\ 7+8\lambda+9\mu \end{matrix}\right) \cdot \left(\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}\right) = 0 \\ \left(\begin{matrix} 1+2\lambda+3\mu \\ 4+5\lambda+6\mu \\ 7+8\lambda+9\mu \end{matrix}\right) \cdot \left(\begin{matrix} 4 \\ 5 \\ 6 \end{matrix}\right) = 0 $$
2021-05-13
$$ \pmb{r} = \left(\begin{matrix} 5 \\ 8 \\ -2 \end{matrix}\right) + \lamba \left(\begin{matrix} 4 \\ 4 \\ -2 \end{matrix}\right) $$ What equal would you solve in order to find the value of $\lambda$ for which the line is closest to the origin??
$$ \left(\begin{matrix} 5+\lambda4 \\ 8+\lambda4 \\ -2-\lambda2 \end{matrix}\right) \cdot \left(\begin{matrix} 4 \\ 4 \\ -2 \end{matrix}\right) = 0 $$
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date: 2021-01-11 10:55
tags:
- '@?further-maths'
- '@?vectors'
- '@?latex-block-alt'
- '@?public'
title: Further Maths - Vector Equation of a Line