Two lines that are not parallel and which do not intersect . They can only occur in three dimensions; in two dimensions, lines that are not parallel always meet somewhere.

How do you tell whether two lines in 3D are parallel?

Check whether their direction vectors are scalar multiples of each other. The points the lines happen to pass through are irrelevant to this; only the directions count.

Two lines have been shown to be parallel. What are the only two possibilities?

They either never meet at all, or they are the same line written two different ways. Which of the two it is still has to be checked, since two very different-looking equations can describe one line.

How do you check whether two parallel lines are actually identical?

Take any point on one line and test whether it also lies on the other. A single point in common is enough, because parallel lines cannot cross and then separate again.

Two lines are known not to be parallel. How do you find out whether they intersect?

Equate the two general points and write down three equations , one for each component. Solve any two of them for the two parameters, then test whether those values also satisfy the third.

What does the third component equation tell you?

Whether the two lines actually meet. If the parameter values satisfy it as well, the lines intersect ; if they do not, the lines are skew .

True or False? The scalar product of two vectors is another vector.

False. The result is a real number , which is exactly why it is called the scalar product. That is worth holding on to, since most other operations on vectors hand you back a vector.

How do you find the angle between two lines in 3D?

Find the angle between their direction vectors . Where the lines happen to sit makes no difference to the angle between them, so the position vectors play no part at all.

Define the foot of the perpendicular from a point to a line.

The point on the line that is closest to the given point. The segment joining the two is perpendicular to the line, which is where the name comes from.

International A Level (IAL)MathsEdexcelPure 4FlashcardsVectorsVector Equations of Lines & The Scalar Product

Vector Equations of Lines & The Scalar Product (Edexcel International A Level (IAL) Maths: Pure 4): Flashcards

Exam code: YMA01

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FrontEquation of a Line in Vector Form
Complete the vector equation of the line through the points with position vectors $a$ and $b$ :
$r = a + t (______)$

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Cards in this collection (26)

Complete the vector equation of the line through the points with position vectors $a$ and $b$ :
$r = a + t (______)$
The completed equation is:
$r = a + t (b - a)$
The bracket is a direction vector: the step that takes you from one of the points to the other.
What two pieces of information fix a line in vector form?
The position vector of one point on it, and a direction vector along it.
Those give $r = a + t d$ , where $t$ runs through every real value as you move along the line.
How does $r = a + t d$ compare with $y = m x + c$ ?
$a$ plays the part of the $+ c$ , fixing where the line sits.
$d$ plays the part of the $m$ , fixing which way it goes, and unlike a gradient it works just as well in three dimensions.
Is the point $(2, 0, - 1)$ on the line $r = 3 i + 2 j - k + t (i + 2 j)$ , and how can you tell?
Yes, it is, because $t = - 1$ makes all three components match at once.
A point lies on the line only if a single value of $t$ works for every component; if no one value does, it does not.
True or False?
Two vector equations that look different must represent different lines.
False.
Any point on the line will serve as $a$ , and any multiple of the direction will serve as $d$ , so a single line has infinitely many equations.
That is unlike a Cartesian equation, where two equations describe the same line only if one is a multiple of the other.
Can a vector equation use column vectors instead of $i$ , $j$ and $k$ ?
Yes, and the two forms say exactly the same thing.
So $r = 3 i + j - 7 k + t (i - 2 j)$ and $r = (\begin{matrix} 3 \\ 1 \\ - 7 \end{matrix}) + t (\begin{matrix} 1 \\ - 2 \\ 0 \end{matrix})$ are one and the same equation.
Define skew lines.
Two lines that are not parallel and which do not intersect.
They can only occur in three dimensions; in two dimensions, lines that are not parallel always meet somewhere.
How do you tell whether two lines in 3D are parallel?
Check whether their direction vectors are scalar multiples of each other.
The points the lines happen to pass through are irrelevant to this; only the directions count.
Two lines have been shown to be parallel. What are the only two possibilities?
They either never meet at all, or they are the same line written two different ways.
Which of the two it is still has to be checked, since two very different-looking equations can describe one line.
How do you check whether two parallel lines are actually identical?
Take any point on one line and test whether it also lies on the other.
A single point in common is enough, because parallel lines cannot cross and then separate again.
Two lines are known not to be parallel. How do you find out whether they intersect?
Equate the two general points and write down three equations, one for each component.
Solve any two of them for the two parameters, then test whether those values also satisfy the third.
What does the third component equation tell you?
Whether the two lines actually meet.
If the parameter values satisfy it as well, the lines intersect; if they do not, the lines are skew.
True or False?
The same letter can be used for the parameter in both lines' equations.
False.
The two lines reach any common point at different parameter values in general, so one letter cannot stand for both at once.
Using $s$ for one line and $t$ for the other keeps them properly apart.
Complete the scalar product of $a = a_{1} i + a_{2} j + a_{3} k$ and $b = b_{1} i + b_{2} j + b_{3} k$ :
$a \cdot b = a_{1} b_{1} +______+______$
The completed definition is:
$a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$
Corresponding components are multiplied together and the results are then added.
True or False?
The scalar product of two vectors is another vector.
False.
The result is a real number, which is exactly why it is called the scalar product.
That is worth holding on to, since most other operations on vectors hand you back a vector.
What is the scalar product in terms of the angle between the vectors?
$a \cdot b = |a| |b| \cos θ$ .
Here $θ$ is the angle between them when they are placed base to base, that is starting from the same point.
What is $a \cdot a$ equal to?
${|a|}^{2}$ , the square of the vector's own magnitude.
It follows from the angle formula, since a vector makes an angle of zero with itself and $\cos 0 ° = 1$ .
Which two familiar algebraic rules does the scalar product obey?
The order does not matter, $a \cdot b = b \cdot a$ , and brackets multiply out, $a \cdot (b + c) = a \cdot b + a \cdot c$ .
Together they let you expand something like $(a - b) \cdot (a - b)$ exactly as you would in ordinary algebra.
Work out $(3 i - k) \cdot (2 i + 9 j + k)$ .
The answer is $5$ .
The absent $j$ in the first vector counts as $0$ , so the sum is $3 \times 2 + 0 \times 9 + (- 1) \times 1 = 5$ .
Complete the angle between two vectors:
$θ =______(\frac{a \cdot b}{|a| |b|})$
The completed formula is:
$θ = \cos^{- 1} (\frac{a \cdot b}{|a| |b|})$
The scalar product divided by the two magnitudes gives $\cos θ$ , so an inverse cosine is what finally produces the angle itself.
How do you find the angle between two lines in 3D?
Find the angle between their direction vectors.
Where the lines happen to sit makes no difference to the angle between them, so the position vectors play no part at all.
True or False?
Two non-zero vectors are perpendicular exactly when their scalar product is zero.
True.
If they are perpendicular then $\cos 90 ° = 0$ makes the product zero; and if the product is zero then neither magnitude can be, so $\cos θ$ must be.
The condition holds in both directions, which is what makes it such a quick test.
Are $2 i - 3 j + 5 k$ and $- 4 i - j + k$ perpendicular?
Yes, they are perpendicular.
Their scalar product works out as $2 \times (- 4) + (- 3) \times (- 1) + 5 \times 1 = - 8 + 3 + 5 = 0$ .
Define the foot of the perpendicular from a point to a line.
The point on the line that is closest to the given point.
The segment joining the two is perpendicular to the line, which is where the name comes from.
A point $P$ does not lie on the line $r = a + t d$ . How do you locate the closest point on the line?
Write that point as $a + t_{0} d$ , form the vector running from it to $P$ , and set the scalar product of that vector with $d$ equal to zero.
Solving the resulting equation gives $t_{0}$ , and substituting it back gives the point itself.
Once you have found the closest point on the line, how do you get the shortest distance?
Take the magnitude of the vector joining that point to $P$ .
That length is sometimes called the length of the perpendicular, and it is smaller than the distance to any other point of the line.

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