Further Maths - Vector Equation of a Line
2021-01-11
What is the vector equation for a straight line?
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\[\pmb{r} =\pmb{a} + \lambda \pmb{b}\]What is the variable $\pmb{a}$ called?? The position vector.
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\[\pmb{r} =\pmb{a} + \lambda \pmb{b}\]What is the variable $\pmb{b}$ called?? The direction vector.
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\[\pmb{r} =\pmb{a} + \lambda \pmb{b}\]What is the variable $\pmb{r}$ called?? The general position vector for a point on the line.
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\[\pmb{r} =\pmb{a} + \lambda \pmb{b}\]What is the informal name for $\lambda$ called here?? The sliding factor.
\[\pmb{r} =\pmb{a} + \lambda \pmb{b}\]
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.
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\[\left(\begin{matrix} 2 \\ -1 \end{matrix}\right) + \lambda \left(\begin{matrix} 1 \\ 2 \end{matrix}\right)\]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
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\[\left(\begin{matrix} 1 \\ 2 \end{matrix}\right)\]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
\[\pmb{r} =\pmb{a} + \lambda \pmb{b}\]
What is the variable $\pmb{r}$ called?
The general position vector for a point on the line.
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\[\left(\begin{matrix} 0 \\ 0 \end{matrix}\right) + \lambda \left(\begin{matrix} 1 \\ 3 \end{matrix}\right)\]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??
\[\left(\begin{matrix} 1 \\\\ 2 \\\\ 3 \end{matrix}\right)\]\[\pmb{r} =\pmb{a} + \lambda \pmb{b}\]
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?
\[\left(\begin{matrix} 2 \\ -1 \end{matrix}\right) + \lambda \left(\begin{matrix} 1 \\ 2 \end{matrix}\right)\]
Which is the direction vector?
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\[\pmb{r} = \left(\begin{matrix} 1+\lambda \\ 2+\lambda \\ 3+\lambda \end{matrix}\right)\]If
\[\left(\begin{matrix} 0 \\ a \\ b \end{matrix}\right)\]lies on the line, what must the vector be equal to??
\[\left(\begin{matrix} 0 \\\\ 1 \\\\ 2 \end{matrix}\right)\]\[\left(\begin{matrix} 1 \\ 3 \end{matrix}\right)\]
What is the gradient of this vector?
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\[a = \left(\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}\right) \\ b = \left(\begin{matrix} 2 \\ 4 \\ 6 \end{matrix}\right)\]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
\[\left(\begin{matrix} 7 \\ 4 \end{matrix}\right)\]
What is the gradient of this vector?
\[\left(\begin{matrix} 0 \\ -5 \end{matrix}\right) + \lambda \left(\begin{matrix} -1 \\ -2 \end{matrix}\right) = \left(\begin{matrix} 2 \\ -1 \end{matrix}\right) + \lambda \left(\begin{matrix} 2 \\ 4 \end{matrix}\right)\]
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
2021-01-25
\[\left(\begin{matrix} 1 \\ 2 \end{matrix}\right)\]
What is the direction vector that is perpindicular?
\[\left(\begin{matrix} -1 \\ -2 \end{matrix}\right)\]
What is the direction vector that is perpindicular?
\[\left(\begin{matrix} a \\ b \end{matrix}\right)\]
What is the direction vector that is perpindicular?
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\[\overrightarrow{PA} = \left(\begin{matrix} 2\lambda-6 \\ 1 \\ 2+\lambda \end{matrix}\right) \\ \pmb{r} = \left(\begin{matrix} 0 \\ 4 \\ 1 \end{matrix}\right) + \lambda \left(\begin{matrix} 2 \\ 0 \\ 1 \end{matrix}\right)\]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
\[r = \left(\begin{matrix} -1 \\ -4 \end{matrix}\right) + \lambda \left(\begin{matrix} 1 \\ 3 \end{matrix}\right)\]
How could you rewrite this to find the point of intersection with another line?
\[\left(\begin{matrix} -1 \\ -4 \end{matrix}\right) + \lambda \left(\begin{matrix} 1 \\ 3 \end{matrix}\right)\]
How could you rewrite this as one vector?
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\[r = \pmb{a} + \lambda \pmb{b} \\ \pmb{r} = \pmb{c} + \lambda \pmb{d}\]If you’re subtracting two lines from one another with the same direction vector, what substitution can you make so things are easier??
\[t = \lambda - \mu\]\[\left(\begin{matrix} 2 \\ 3 \end{matrix}\right) + \mu \left(\begin{matrix} 1 \\ -1 \end{matrix}\right)\]
How can you rewrite this as one vector?
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?
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.
What does the general 3d position vector look like?
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\[\overrightarrow{AB} = \left(\begin{matrix} 1+2\lambda+3\mu \\ 4+5\lambda+6\mu \\ 7+8\lambda+9\mu \end{matrix}\right)\]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
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\[\pmb{r} = \left(\begin{matrix} 5 \\\\ 8 \\\\ -2 \end{matrix}\right) + \lamba \left(\begin{matrix} 4 \\\\ 4 \\\\ -2 \end{matrix}\right)\]What equal would you solve in order to find the value of $\lambda$ for which the line is closest to the origin??
\[\left(\begin{matrix} 5+\lambda4 \\\\ 8+\lambda4 \\\\ -2-\lambda2 \end{matrix}\right) \cdot \left(\begin{matrix} 4 \\\\ 4 \\\\ -2 \end{matrix}\right) = 0\]