Further Maths - Vectors
See also:
- [[Further Maths - Vector Equation of a Line]]A
- [[Further Maths - Vector Equation of a Plane]]A
- [[Further Maths - Dot Product]]A
What is the formula for the distance to a three dimensional point $(a,b,c)$?
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\[\left(\begin{matrix} a \\ b \\ c \end{matrix}\right)\]What is the formula for the length of the vector??
\[\sqrt{a^2 + b^2 + c^2}\]#####
\[\pmb{a} = \left(\begin{matrix} a _ 1 \\ a _ 2 \\ a _ 3 \end{matrix}\right) \\ \pmb{b} = \left(\begin{matrix} b _ 1 \\ b _ 2 \\ b _ 3 \end{matrix}\right)\]What is the formula for the distance between the two vectors??
\[\sqrt{(b_1 - a_1)^2 + (b_2 - a_2)^2 + (b_3 - a_3)^2}\]What is the vector parallel to the $x$-axis?
What is the vector parallel to the $y$-axis?
What is the vector parallel to the $z$-axis?
2021-01-20
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\[\left(\begin{matrix} x \\ y \\ z \end{matrix}\right)\]If this is a direction vector, how could you eliminate one of the unknowns??
\[\left(\begin{matrix} 1 \\\\ y \\\\ z \end{matrix}\right)\]#####
\[\left(\begin{matrix} x \\ y \\ z \end{matrix}\right) \to \left(\begin{matrix} 1 \\\\ y \\\\ z \end{matrix}\right)\]When does this trick not work?? When the value of $x$ is actually $0$.
If you have two vectors $\pmb{a}$ and $\pmb{b}$ and you wish to find a vector $\pmb{c}$ that is perpindicular to both, what must be true?
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\[x + y + z = 0\\5x - 2y + 3z = 4\]Why can’t you solve these two equations?? Because there are 3 unknowns but only 2 equations.
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\[\left(\begin{matrix} 2 \\ 1 \\ 3 \end{matrix}\right)\]How could you write this vector for $x$ equal to $1$??
\[\left(\begin{matrix} 1 \\\\ \frac{1}{2} \\\\ \frac{3}{2} \end{matrix}\right)\]What is the technique for finding a vector perpindicular vector to two other vectors?
Use the fact the dot product must be equal to zero to find and solve two simulatenous equations.
2021-05-17
What does $\pmb{\hat{X}}$ mean?
Unit/normalised vector; in the same direction as $\pmb{X}$ but has magnitude $1$.
What is the formula for $\pmb{\hat{X}}$?
2021-09-20
Suppose you have
\[Ax + By + Cz = D \\ \alpha x + \beta y + \gamma z = \delta\]These planes intersect at a sheaf. What’s the general formula for a new plane that also passes through this sheaf??
\[(Ax + By + Cz - D) + t(\alpha x + \beta y + \gamma z - \delta) = 0\]Suppose you have
\[Ax + By + Cz = D \\ \alpha x + \beta y + \gamma z = \delta\]These planes intersect at a sheaf. What is the first step in finding the equation of the sheaf?? Making the substitution $z = \lambda$.
Suppose you have
\[Ax + By + Cz = D \\ \alpha x + \beta y + \gamma z = \delta\]and you have made the substitution $z = \lambda$ to get
\[Ax + By = D - C\lambda \\ \alpha x + \beta y = \delta - \gamma \lambda\]What is the next step?? Solving these equations in general to come up with
\[\]Explain the general process for finding the equation of a sheaf?
If you’ve made the substitution $z = \lambda$ in order to find the equation of a sheaf, what will always be the value of $z$ in the final vector for the parametric form of the line?
2022-01-19
Why must you be careful using the
\[\frac{ \vert a\alpha + b\beta + c\gamma - d \vert }{\sqrt{a^2 + b^2 + c^2}}\]formula?? Because you subtract $d$ which can be confusing if $d$ is negative.
Why must you be careful using the “let $z = \lambda$” approach to finding the line of intersection between two planes?
Because it’s very easy to write the final vector in the wrong order.
What’s the thought process for finding the reflection of a point in a plane?
Find a line that passes through the point and has the normal vector of the plane, and double the parameter for the point of intersection with the plane.
What’s the thought process for finding the reflection of a line in a plane?
Create a new line out of the point of intersection with the plane and the reflection of one point in the line.
2022-05-17
How could you show two lines lie in the same plane?
Show they intersect.
How can you ensure you get acute angles when using the scalar product to find angles between vectors?
Take the modulus of what you’re putting into $\arccos$.
Do you need to do any subtracting to work out the angle between two planes after you’ve worked out the angle between their normals?
No.
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\[\cos \theta = \left \vert \frac{\pmb{a}\cdot\pmb{b}}{ \vert \pmb{a} \vert \vert \pmb{b} \vert } \right \vert\]Why are the modulus signs around this important?? It ensures you only get acute angles.
2022-05-17
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\[8x + 4y - 2z = 20\]How do they always want you to write your answers for direction vectors and planes in an exam?? As simple as possible, divide through by 2