Why does a line have many different direction vectors rather than just one?

Because only the direction matters and not the magnitude, so any non-zero multiple of a direction vector points along the same line and works just as well. There is the same freedom in choosing the point, since any point on the line will do.

How do you check whether a given point lies on a line with a known vector equation?

Set the point's position vector equal to the line's equation and work out the value of the parameter from one component. That same value must then work in every other component, and if any one of them fails, the point is not on the line.

Define coincident lines.

Coincident lines are two lines that lie exactly on top of one another, so they are indistinguishable as sets of points. They are a special case of parallel lines: a pair of parallel lines either never meet at all, or are coincident.

What two things must you check to show that a pair of lines are coincident?

First that they are parallel , and then that they share any one point . The second step is enough on its own once the first holds, because two parallel lines with a single point in common must have every point in common.

When testing two lines for intersection, why is the third component equation the one that decides it?

Two equations in two unknowns will almost always have a solution, so finding values for the parameters proves nothing by itself. Only if those same values also satisfy the third equation do the lines genuinely meet; otherwise they pass by one another at different heights.

Once two lines are known to intersect, how do you find the point of intersection?

Substitute the value found for either parameter back into that parameter's own line equation. Repeating it with the other parameter in the other equation is a worthwhile check, since both must produce the same position vector.

Skew lines are lines that are neither parallel nor intersecting. A pair of them can be pictured in a room: one line running high along a wall and another running across the floor in a different direction, never meeting.

Why can no single plane contain a pair of skew lines?

Two lines lying in one plane form a two-dimensional problem, and there they are forced to be either parallel or crossing. Skew lines are neither of those, so no plane can hold them both.

True or False? Two lines in 3D that do not intersect must be parallel.

False. In three dimensions a pair of lines can miss each other while pointing in completely different directions. Such lines are skew , and ruling them out is exactly why being parallel has to be tested separately rather than assumed.

In what order do you test a pair of lines to classify them as parallel, coincident, intersecting or skew?

Test the directions first, because that single step separates the parallel cases from the non-parallel ones. Parallel lines then split into coincident or not, and non-parallel lines split into intersecting or skew, each by its own further test.

Why is the shortest distance from a point to a line always the perpendicular distance?

Any other route from the point to the line is the hypotenuse of a right-angled triangle whose shorter side is that perpendicular. A hypotenuse is always longer than either of the other two sides, so nothing can beat the perpendicular.

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

It is the point on the line closest to the given point, where the perpendicular dropped from that point meets the line. The shortest distance is then the length of the segment joining the two points.

What equation do you form to locate the foot of the perpendicular from a point to a line?

Write the foot's position vector in terms of the line's parameter, then form the displacement vector running from the given point to it. Setting the scalar product of that displacement with the line's direction vector equal to zero gives a single equation for the parameter.

How do you find the shortest distance between two parallel lines?

Take any point at all on one line and find its perpendicular distance to the other, exactly as for a point and a line. The answer does not depend on which point you start from, because parallel lines stay the same distance apart along their whole length.

What is special about the segment giving the shortest distance between two skew lines?

It is perpendicular to both lines at once, rather than to just one of them. That double requirement is what pins it down, since a segment perpendicular to only one of the lines could still be made shorter.

True or False? The same idea solves the point-to-line, the parallel-lines and the skew-lines problems.

True. Every one of them comes down to making a displacement vector perpendicular to a direction vector, which the scalar product turns into an equation to solve. Only the number of perpendicularity conditions changes from one case to the next.

A LevelFurther MathsEdexcelCore PureFlashcardsFurther VectorsVector Lines

Vector Lines (Edexcel A Level Further Maths: Core Pure): Flashcards

Exam code: 9FM0

1/40

0Still learning

Know0

FrontEquations of Lines in 3D
What do the two parts of the vector equation $r = a + λ d$ represent?

Was this flashcard helpful?

Cards in this collection (40)

What do the two parts of the vector equation $r = a + λ d$ represent?
$a$ is the position vector of one known point on the line, and $d$ is a direction vector telling you which way the line runs.
Comparing it with $y = m x + c$ , the point $a$ plays the part of the $+ c$ and the direction $d$ plays the part of the gradient $m$ .
A line passes through $A$ with position vector $4 i - 5 k$ and through $B$ with position vector $3 i - 3 k$ . Fill in a vector equation for the line:
$r = 4 i - 5 k + t (______)$
The completed equation is:
$r = 4 i - 5 k + t (- i + 2 k)$
A direction vector is $\vec{A B} = b - a$ , so subtracting the two position vectors gives $(3 - 4) i + (- 3 + 5) k$ .
Why does a line have many different direction vectors rather than just one?
Because only the direction matters and not the magnitude, so any non-zero multiple of a direction vector points along the same line and works just as well.
There is the same freedom in choosing the point, since any point on the line will do.
True or False?
Two vector equations that look completely different can still describe the same line.
True.
For instance $r = [\begin{matrix} 5 \\ 0 \end{matrix}] + t [\begin{matrix} 2 \\ 1 \end{matrix}]$ and $r = [\begin{matrix} 1 \\ - 2 \end{matrix}] + t [\begin{matrix} - 2 \\ - 1 \end{matrix}]$ describe exactly the same line.
Cartesian equations behave differently: two of those represent the same line only when one is a multiple of the other.
How do you check whether a given point lies on a line with a known vector equation?
Set the point's position vector equal to the line's equation and work out the value of the parameter from one component.
That same value must then work in every other component, and if any one of them fails, the point is not on the line.
A line has vector equation $r = a + λ d$ . Fill in its three parametric equations:
$x = a_{1} + λ______, y = a_{2} + λ______, z = a_{3} + λ______$
The completed equations are:
$x = a_{1} + λ d_{1}, y = a_{2} + λ d_{2}, z = a_{3} + λ d_{3}$
Each coordinate takes its own component from the point and its own component from the direction vector, and all three share the one parameter.
Write parametric equations for the line through $(- 2, 1, 0)$ with direction vector $[\begin{matrix} 3 \\ 1 \\ - 4 \end{matrix}]$ .
Pairing each coordinate with its own components gives:
$x = - 2 + 3 λ, y = 1 + λ, z = - 4 λ$
The $z$ equation has no constant term because the point's $z$ -coordinate is $0$ .
How do you get from the parametric equations of a line to its cartesian equation?
Make the parameter the subject of each of the three equations, then set the three expressions equal to one another.
That eliminates the parameter, leaving $\frac{x - a_{1}}{d_{1}} = \frac{y - a_{2}}{d_{2}} = \frac{z - a_{3}}{d_{3}}$ .
Find the cartesian equation of the line $r = [\begin{matrix} 1 \\ 0 \\ 2 \end{matrix}] + λ [\begin{matrix} 4 \\ - 2 \\ 1 \end{matrix}]$ .
The cartesian equation is $\frac{x - 1}{4} = \frac{y}{- 2} = z - 2$ .
The last part has no denominator written because the $z$ -component of the direction vector is $1$ .
Given a line in cartesian form, how do you read off a point on it and a direction vector?
Read them straight from the fractions: the numbers subtracted from $x$ , $y$ and $z$ give the coordinates of a point, and the three denominators give the components of a direction vector.
So $\frac{x + 3}{2} = \frac{y - 1}{5} = \frac{z}{- 1}$ passes through $(- 3, 1, 0)$ with direction $[\begin{matrix} 2 \\ 5 \\ - 1 \end{matrix}]$ .
What happens to the cartesian equation of a line when one component of its direction vector is zero?
That coordinate cannot be put over a zero denominator, so it is written as a separate equation instead: if the $y$ -component is $0$ then $y = a_{2}$ everywhere on the line.
The other two coordinates still form an ordinary pair, giving something like $\frac{x - a_{1}}{d_{1}} = \frac{z - a_{3}}{d_{3}}$ written alongside $y = a_{2}$ .
True or False?
Two lines with direction vectors $[\begin{matrix} 2 \\ 0 \\ - 8 \end{matrix}]$ and $[\begin{matrix} - 1 \\ 0 \\ 4 \end{matrix}]$ are parallel.
True.
One direction vector is $- 2$ times the other, and a scalar multiple runs along the same direction even when the scalar is negative.
Direction vectors do not have to be identical, or even to point the same way, for two lines to count as parallel.
Define coincident lines.
Coincident lines are two lines that lie exactly on top of one another, so they are indistinguishable as sets of points.
They are a special case of parallel lines: a pair of parallel lines either never meet at all, or are coincident.
What two things must you check to show that a pair of lines are coincident?
First that they are parallel, and then that they share any one point.
The second step is enough on its own once the first holds, because two parallel lines with a single point in common must have every point in common.
How do you test whether two non-parallel lines in 3D intersect?
Set the two vector equations equal to each other, using different parameters such as $λ$ and $μ$ , and write out the three component equations.
Solve two of them for the parameters, then check whether that same pair of values also satisfies the third.
When testing two lines for intersection, why is the third component equation the one that decides it?
Two equations in two unknowns will almost always have a solution, so finding values for the parameters proves nothing by itself.
Only if those same values also satisfy the third equation do the lines genuinely meet; otherwise they pass by one another at different heights.
Once two lines are known to intersect, how do you find the point of intersection?
Substitute the value found for either parameter back into that parameter's own line equation.
Repeating it with the other parameter in the other equation is a worthwhile check, since both must produce the same position vector.
Define skew lines.
Skew lines are lines that are neither parallel nor intersecting.
A pair of them can be pictured in a room: one line running high along a wall and another running across the floor in a different direction, never meeting.
Why can no single plane contain a pair of skew lines?
Two lines lying in one plane form a two-dimensional problem, and there they are forced to be either parallel or crossing.
Skew lines are neither of those, so no plane can hold them both.
True or False?
Two lines in 3D that do not intersect must be parallel.
False.
In three dimensions a pair of lines can miss each other while pointing in completely different directions.
Such lines are skew, and ruling them out is exactly why being parallel has to be tested separately rather than assumed.
In what order do you test a pair of lines to classify them as parallel, coincident, intersecting or skew?
Test the directions first, because that single step separates the parallel cases from the non-parallel ones.
Parallel lines then split into coincident or not, and non-parallel lines split into intersecting or skew, each by its own further test.
For $a = a_{1} i + a_{2} j + a_{3} k$ and $b = b_{1} i + b_{2} j + b_{3} k$ , fill in the scalar product:
$a \cdot b =______+______+______$
The completed scalar product is:
$a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$
Corresponding components are multiplied together and the three products are then added.
Why is the scalar product given that name?
Because the result is a scalar, an ordinary real number, rather than another vector.
It is also called the dot product, after the symbol used to write it in $a \cdot b$ .
Fill in the connection between the scalar product and the angle $θ$ between two vectors:
$a \cdot b = | a | | b |______$
The completed formula is:
$a \cdot b = | a | | b | \cos θ$
Here $θ$ is the angle between the two vectors when they are placed base to base, that is, starting from the same point.
What is the scalar product of a vector with itself?
It is the square of that vector's magnitude, so $a \cdot a = | a |^{2}$ .
That follows straight from the component form, since every component ends up multiplied by itself.
What does the sign of a scalar product tell you about the angle between two vectors?
A positive scalar product means an acute angle, a negative one means an obtuse angle, and zero means the two vectors are perpendicular.
Magnitudes are never negative, so the sign is carried entirely by $\cos θ$ .
True or False?
The scalar product of two vectors depends on the order in which they are written.
False.
The scalar product is commutative, so $a \cdot b$ and $b \cdot a$ are equal.
In this it behaves like ordinary multiplication of numbers, and brackets can be expanded in the usual way as well.
Find $t$ so that $[\begin{matrix} 2 \\ t \\ 5 \end{matrix}]$ and $(t - 1) i - j + k$ are perpendicular.
Perpendicular vectors have a scalar product of $0$ , so $2 (t - 1) + t (- 1) + 5 (1) = 0$ .
That simplifies to $t + 3 = 0$ , giving $t = - 3$ .
Fill in the missing term of this expansion, which works just as it does for ordinary numbers:
$(a - b) \cdot (a - b) = a \cdot a -______+ b \cdot b$
The completed expansion is:
$(a - b) \cdot (a - b) = a \cdot a - 2 a \cdot b + b \cdot b$
The middle term is doubled because the two cross terms of the expansion are equal to each other.
How do you find the angle between two lines given in vector form?
Find the angle between their direction vectors, since where the lines happen to pass through makes no difference to the angle.
For $r = a_{1} + s d_{1}$ and $r = a_{2} + t d_{2}$ that gives $θ = \arccos (\frac{d_{1} \cdot d_{2}}{| d_{1} | | d_{2} |})$ .
Calculate the angle between $[\begin{matrix} - 1 \\ 3 \\ 2 \end{matrix}]$ and $3 i + 4 j - k$ .
The scalar product is $- 3 + 12 - 2 = 7$ , and the magnitudes are $\sqrt{14}$ and $\sqrt{26}$ .
So $\cos θ = \frac{7}{\sqrt{14} \sqrt{26}} = 0.3668 \dots$ , giving $θ = 68.5 °$ to three significant figures.
Why is the shortest distance from a point to a line always the perpendicular distance?
Any other route from the point to the line is the hypotenuse of a right-angled triangle whose shorter side is that perpendicular.
A hypotenuse is always longer than either of the other two sides, so nothing can beat the perpendicular.
Define the foot of the perpendicular from a point to a line.
It is the point on the line closest to the given point, where the perpendicular dropped from that point meets the line.
The shortest distance is then the length of the segment joining the two points.
What equation do you form to locate the foot of the perpendicular from a point to a line?
Write the foot's position vector in terms of the line's parameter, then form the displacement vector running from the given point to it.
Setting the scalar product of that displacement with the line's direction vector equal to zero gives a single equation for the parameter.
Find the shortest distance from $A (1, 2, 0)$ to $r = [\begin{matrix} 2 \\ 0 \\ 6 \end{matrix}] + λ [\begin{matrix} 0 \\ 1 \\ 2 \end{matrix}]$ .
The displacement from $A$ to a general point on the line is $[\begin{matrix} 1 \\ λ - 2 \\ 6 + 2 λ \end{matrix}]$ , and setting its scalar product with the direction vector to zero gives $5 λ + 10 = 0$ .
With $λ = - 2$ the displacement becomes $[\begin{matrix} 1 \\ - 4 \\ 2 \end{matrix}]$ , so the shortest distance is $\sqrt{21}$ .
How do you find the shortest distance between two parallel lines?
Take any point at all on one line and find its perpendicular distance to the other, exactly as for a point and a line.
The answer does not depend on which point you start from, because parallel lines stay the same distance apart along their whole length.
What is special about the segment giving the shortest distance between two skew lines?
It is perpendicular to both lines at once, rather than to just one of them.
That double requirement is what pins it down, since a segment perpendicular to only one of the lines could still be made shorter.
Points $P$ and $Q$ on two skew lines give the shortest distance between them. Fill in the two conditions this imposes:
$(q - p) \cdot______= 0, (q - p) \cdot______= 0$
The completed conditions are:
$(q - p) \cdot d_{1} = 0, (q - p) \cdot d_{2} = 0$
Two equations in the two parameters is exactly what is needed to fix the positions of both $P$ and $Q$ .
Once both parameters are known, how do you finish finding the distance between two skew lines?
Substitute the two values into the displacement vector joining the points and take its magnitude.
The distance is $| q - p |$ , a single number rather than a vector.
True or False?
The same idea solves the point-to-line, the parallel-lines and the skew-lines problems.
True.
Every one of them comes down to making a displacement vector perpendicular to a direction vector, which the scalar product turns into an equation to solve.
Only the number of perpendicularity conditions changes from one case to the next.

Sign up to unlock flashcards

By signing up you agree to our Terms and Privacy Policy

Previous:Methods in CalculusNext:Vector Planes