Define collinear points.

Points that all lie on the same straight line . Showing it needs two vectors that are parallel and share a common point, since parallel vectors on their own only give parallel lines.

In a shape such as a cuboid or a parallelogram, what is true of the vectors along opposite sides?

They are equal : the same magnitude and the same direction. That is what lets you find a missing vertex. Travel to the unknown corner along a route made of vectors you already know, adding them as you go.

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Vectors in 3D (Edexcel A Level Maths: Pure): Flashcards

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FrontVectors in 3 Dimensions
Define unit vector.

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Define unit vector.
A vector with a magnitude of 1.
$i$ , $j$ and $k$ are the unit vectors in the directions of the $x$ , $y$ and $z$ axes, and they are mutually perpendicular.
What are the two ways of writing a 3D vector?
As a column vector, or in $i$ , $j$ , $k$ form. The three components are the same numbers either way:
$(\begin{matrix} 3 \\ 7 \\ - 2 \end{matrix}) = 3 i + 7 j - 2 k$
Each component is the distance moved in the direction of one axis.
Complete the formula for the magnitude of a 3D vector:
$| x i + y j + z k | = \sqrt{______}$
The completed formula is:
$| x i + y j + z k | = \sqrt{x^{2} + y^{2} + z^{2}}$
It is Pythagoras' theorem carried into three dimensions. This one is on the list of formulae you are expected to know, so it is not given in the formulae booklet.
The magnitude is also called the modulus of the vector.
True or False?
The magnitude of $- 2 i + 3 j - 6 k$ is negative, because two of its components are negative.
False.
A magnitude is a length, so it is never negative. Squaring the components removes the signs:
$| - 2 i + 3 j - 6 k | = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$
How do you find the unit vector in the direction of a vector $a$ ?
Divide the vector by its own magnitude, giving $\frac{a}{| a |}$ . It is written $\hat{a}$ .
So, for example, if $a = 3 i - 4 j + 5 k$ then $| a | = \sqrt{50} = 5 \sqrt{2}$ , so:
$\hat{a} = \frac{1}{5 \sqrt{2}} (3 i - 4 j + 5 k)$
How can you tell whether two vectors are parallel?
One is a scalar multiple of the other, so every component is multiplied by the same number.
So, for example:
$(\begin{matrix} 6 \\ - 4 \\ 2 \end{matrix}) = \frac{2}{3} (\begin{matrix} 9 \\ - 6 \\ 3 \end{matrix})$
so those two vectors are parallel.
Two points are given by their position vectors. How do you find the distance between them?
Subtract one position vector from the other to get the vector joining the points, then find that vector's magnitude.
So, for example, for $A (5, 3, - 7)$ and $B (3, - 6, 5)$ :
$\sqrt{(5 - 3)^{2} + (3 - (- 6))^{2} + (- 7 - 5)^{2}} = \sqrt{229} = 15.1$
It does not matter which way round you subtract, because each difference is squared.
Define collinear points.
Points that all lie on the same straight line.
Showing it needs two vectors that are parallel and share a common point, since parallel vectors on their own only give parallel lines.
How do you show that three points are collinear?
Use their position vectors to find two vectors that start at the same point, such as $\vec{A B}$ and $\vec{A C}$ , then show one is a multiple of the other.
So, for example, for $A (1, 2, 3)$ , $B (3, 8, 1)$ and $C (7, 20, - 3)$ , $\vec{A B} = 2 i + 6 j - 2 k$ and $\vec{A C} = 6 i + 18 j - 6 k = 3 \vec{A B}$ , and both start at $A$ .
The vector $a = x i + y j + z k$ makes an angle $θ_{x}$ with the $x$ -axis. Complete the formula:
$\cos θ_{x} = \frac{______}{______}$
The completed formula is:
$\cos θ_{x} = \frac{x}{| a |}$
It is just cosine = adjacent divided by hypotenuse, in the right-angled triangle formed by the vector and the axis: the side along the axis has length $x$ , and the vector itself has length $| a |$ .
The $y$ and $z$ versions work the same way, and none of them is in the formulae booklet.
How do you find the angle between two vectors in three dimensions?
Make them two sides of a triangle and find the vector for the third side, then find the length of all three sides using the magnitude formula.
The angle then comes from the cosine rule:
$\cos θ = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$
True or False?
To show that a triangle with sides given as vectors is isosceles, you show that two of those vectors are equal.
False.
You compare their magnitudes. Two side vectors of a triangle can never be equal, since equal vectors point the same way and the sides of a triangle do not.
So, for example, sides $5 i + 6 j - 2 k$ and $7 i + 4 k$ are different vectors, but both have magnitude $\sqrt{65}$ .
Why must you find an angle before you can find the area of a triangle whose sides are given as vectors?
Because the area formula $Area = \frac{1}{2} a b \sin θ$ needs the angle between the two sides, and the vectors do not give it to you directly.
That is why an area question of this kind always begins by finding all three side lengths: the angle has to be worked out from those first.
In a shape such as a cuboid or a parallelogram, what is true of the vectors along opposite sides?
They are equal: the same magnitude and the same direction.
That is what lets you find a missing vertex. Travel to the unknown corner along a route made of vectors you already know, adding them as you go.

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