Exercise 9.4VectorsExtension to 3 DimensionsThe vector methods we have learn so far apply equally well in 3-Dimesions; so there really is no further learning to do in order to extend our ideas to deal with problems in 3-DInevitably, sketches and diagrams will no longer be practically accurate, but they can still be used to aid thinking
Vector NotationIn 3-D, the coordinates of a point may be given as A=(4, 3, 2)In vector notation, we write the coordinates of 'A' as 4i + 3j + 2kWhere 'i' stands for: along the x-axis
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If a vector is given by:┌ ┐ │ p │ PQ = │ p │ or PQ = pi + qj + rk │ q │ └ ┘ Then its length is given by: |PQ| = √ pｲ + qｲ + rｲ |
e.g. Find the length of the vector OA in the diagram:
┌ ┐ │ 4 │ OA = │ 4 │ |OA| = √ 4ｲ + 4ｲ + 0ｲ │ 0 │ └ ┘ = 5.66 units
e.g. Find the length of the vector AB in the diagram:
┌ ┐ │ 4 │ AB = │-2 │ |OA| = √ 4ｲ + (-2)ｲ + 3ｲ │ 3 │ └ ┘ = √29 units
Parallel Vectors
e.g. Find the relative vector CA:
CA = A - C ┌ ┐ ┌ ┐ ┌ ┐ │ 4 │ │ 0 │ │ 4 │ OB = │ 4 │ - │ 2 │ = │ 2 │ (or: 4i + 2j) │ 0 │ │ 0 │ │ 0 │ └ ┘ └ ┘ └ ┘
e.g. Since we now know the coordinates of 'B', find the relative vector DB:
DB = B - D ┌ ┐ ┌ ┐ ┌ ┐ │ 8 │ │ 6 │ │ 2 │ OB = │ 2 │ - │ 1 │ = │ 1 │ (or: 2i + 1j) │ 3 │ │ 3 │ │ 0 │ └ ┘ └ ┘ └ ┘
So, the coordinates of 'D' are: (15, 3)
e.g. Hence compare CA and DB:
CA DB ┌ ┐ ┌ ┐ │ 4 │ ◄隣2覧 │ 2 │ │ 2 │ ◄隣2覧 │ 1 │ So: 2DB = CA │ 0 │ ◄隣2覧 │ 0 │ └ ┘ └ ┘
Which tells us that they ARE parallel; but also that CA is twice as long as DB
The Vector Equation of a Line
e.g. Find the vector equation of the line AB in the diagram:
Since we know the vector position vector of 'A'
And we know the relative vector 'AB' (which is a vector along the line)
r = OA + λ AB ┌ ┐ ┌ ┐ │ 4 │ │ 4 │ r = │ 4 │ + λ │-2 │ or: r = (4i + 4j) + λ(4i - 2j + 3k) │ 0 │ │ 3 │ └ ┘ └ ┘ ╘═╦═╛ ╘═╦═╛ the position vector ╝ ╚ the dirn vector of a point on the line parallel to the line
So, this is the equation of the line though 4i +4j and parallel to the vector 4i - 2j + 3k
For example: If we put in λ=0 into the equation:
┌ ┐ ┌ ┐ ┌ ┐ │ 4 │ │ 4 │ │ 4 │ r = │ 4 │ + 0 │-2 │ = │ 4 │ or: 4i + 4j │ 0 │ │ 3 │ │ 0 │ └ ┘ └ ┘ └ ┘
Which is the position vector of the point 'A' on the line AB
If we put in λ=1 into the equation:
┌ ┐ ┌ ┐ ┌ ┐ │ 4 │ │ 4 │ │ 8 │ r = │ 4 │ + 1 │-2 │ = │ 2 │ or: 8i + 2j + 3k │ 0 │ │ 3 │ │ 3 │ └ ┘ └ ┘ └ ┘
Which is the position vector of the point 'B' on the line AB
If we put in λ=½ into the equation:
┌ ┐ ┌ ┐ ┌ ┐ │ 4 │ │ 4 │ │ 6 │ r = │ 4 │ + ｽ │-2 │ = │ 3 │ or: 6i + 3j + 1.5k │ 0 │ │ 3 │ │1.5│ └ ┘ └ ┘ └ ┘
Which is the position vector of 'M', the midpoint of AB
Actually; we could have found the equation of the line AB in a different way:
We could have started with the point 'B' and the vector AB:
r = OB + η AB ┌ ┐ ┌ ┐ │ 8 │ │ 4 │ r = │ 2 │ + η │-2 │ or: r = (8i+2j+3k) + η(4i-2j+3k) │ 3 │ │ 3 │ └ ┘ └ ┘
Note: Since I have written the equation a different way, I have used a different variable 'η' so we don't get mixed up...
e.g. Find the equation of the line CD:
Since we know the vector position vector of 'C'
And we know the relative vector 'CD' (from earlier)
r = OC + μ CD ┌ ┐ ┌ ┐ │ 0 │ │ 6 │ r = │ 2 │ + μ │-1 │ or: r = (2j) + μ(6i - j + 3k) │ 0 │ │ 3 │ └ ┘ └ ┘ ╘═╦═╛ ╘═╦═╛ the position vector ╝ ╚ the dirn vector of a point on the line parallel to the line
Finding where a line intercepts the Axes' Planes
In 2-D, most lines intersect both the 'x' and 'y' axes (which lines don't intersect on of the axes?)
But, in 3-D, it is rare for a line to intersect the axes. Instead, the equivalent is to think of a line intersecting the x-y plane (or the y-z plane, or the x-z plane):
1. To find where a line intercepts the x-y plane; we put 'z=0'
e.g. Find where r = (4i + 4j) + λ(4i - 2j + 3k) intercepts the x-y plane:
We put z=0, so we can call the point (?, ?, 0): ┌ ┐ ┌ ┐ ┌ ┐ │ 4 │ │ 4 │ │ ? │ 覧覧► 4 + 4λ = ? ◄════════════╗ │ 4 │ + λ │-2 │ = │ ? │ 覧覧► 4 - 2λ = ? ◄════════════╣ │ 0 │ │ 3 │ │ 0 │ 覧覧► 0 + 3λ = 0 覧覧► λ = 0 ═╝ └ ┘ └ ┘ └ ┘ Solving the bottom row gives: λ = 0 Substituting in the top and middle rows gives: x = 4, y = 4 So the point is (4, 4, 0)
2. To find where a line intercepts the y-z plane; we put 'x=0'
e.g. Find where r = (4i + 4j) + λ(4i - 2j + 3k) intercepts the y-z plane:
We put x=0, so we can call the point (0, ?, ?): ┌ ┐ ┌ ┐ ┌ ┐ │ 4 │ │ 4 │ │ 0 │ 覧覧► 4 + 4λ = 0 覧覧► λ = -1 ═╗ │ 4 │ + λ │-2 │ = │ ? │ 覧覧► 4 - 2λ = ? ◄════════════╣ │ 0 │ │ 3 │ │ ? │ 覧覧► 0 + 3λ = ? ◄════════════╝ └ ┘ └ ┘ └ ┘ Solving the top row gives: λ = -1 Substituting in the middle and bottom rows gives: y = 6, z = -3 So the point is (0, 6, -3)
3. To find where a line intercepts the x-z plane; we put 'y=0'
e.g. Find where r = (4i + 4j) + λ(4i - 2j + 3k) intercepts the x-z plane:
We put y=0, so we can call the point (?, 0, ?): ┌ ┐ ┌ ┐ ┌ ┐ │ 4 │ │ 4 │ │ ? │ 覧覧► 4 + 4λ = ? ◄════════════╗ │ 4 │ + λ │-2 │ = │ 0 │ 覧覧► 4 - 2λ = 0 覧覧► λ = 2 ═╣ │ 0 │ │ 3 │ │ ? │ 覧覧► 0 + 3λ = ? ◄════════════╝ └ ┘ └ ┘ └ ┘ Solving the middle row gives: λ = 2 Substituting in the top and bottom rows gives: x = 12, z = 6 So the point is (12, 0, 6)
Identifying Equivalent Equations
Before, we learnt that there are many different ways of writing the vector equation of the line AB.
"How do we tell that: r = (4i + 4j) + λ(4i - 2j + 3k) and r = (-4i + 8j - 6k) + κ(8i - 4j + 6k) represent the same line?"
1. The direction vectors either the same, or parallel:
d.v. of eqn 1 d.v. of eqn 2 ┌ ┐ ┌ ┐ │ 4 │ 覧ﾗ2・#9658; │ 8 │ │-2 │ 覧ﾗ2・#9658; │-4 │ │ 3 │ 覧ﾗ2・#9658; │ 6 │ └ ┘ └ ┘
2. We must prove that position vector given in the 2nd line is also a point that lies on the 1st line:
┌ ┐ ┌ ┐ │-4 │ │ 8 │ The second line is: │ 8 │ + η │-4 │ │-6 │ │ 6 │ └ ┘ └ ┘ ▼ ┌ ┐ So, we know that │-4 │ is the position vector of a point on line 2 │ 8 │ │-6 │ └ ┘ We have to check it this point also lies on the first line: ┌ ┐ ┌ ┐ ┌ ┐ │ 4 │ │ 4 │ │-4 │ 覧覧► 4 + 4λ = -4 覧覧► λ = -2 ╗ same values of λ │ 4 │ + λ │-2 │ = │ 8 │ 覧覧► 4 - 2λ = 8 覧覧► λ = -2 ║ so (-4,8,-6) does │ 0 │ │ 3 │ │-6 │ 覧覧► 0 + 3λ = -6 覧覧► λ = -2 ╝ lie on the line └ ┘ └ ┘ └ ┘
Which shows that both lines pass through the point (-4, 8, -6) and since we have already shown that the direction vectors are parallel, we can say that these two equations represent the SAME line...
Finding where two Lines Intersect
In 2-D, most lines intersect (unless they are parallel)
However, in 3-D lines can be parallel, intersecting (i.e. non-parallel and meet in one point) or skew (i.e. non-parallel but don't meet)
To find if and where two lines intersect:
Look at the direction vectors to see if they are parallel or not:
If they are parallel, then the lines might be "equivalent", or they might be "parallel". To decide:
Take the position vector given in the equation of the 2nd line and test to see if it is a point on the first line (just like we did in the example above)
If it is, then the lines are equivalent (i.e. different ways of writing the same equation). If not, the lines are different, but parallel (so they don't meet each other)
If they are not parallel then the lines might be "skew", or they might be "intersecting". To decide:
-
Make them equation to each other
-
Read the "TOP ROW" of this vector equation and call this equation (1)
-
Read the "MIDDLE ROW" of this vector equation and call this equation (2)
-
Solve simultaneously to find 'λ' and 'ｵ'
-
Now, we can decide if the lines are intersecting or skew:
Check that these values of 'λ' and 'ｵ' also work in the "BOTTOM ROW":Note: This check is very important: Solving the simultaneous equations doesn't tell if they actually DO meet... ║ ╠══► Verifying the values of λ and ｵ also work in ROW 3 confirms these lines are intersecting ╚══► But if the values of λ and ｵ don't work in ROW 3, then the lines are skew
-
If the lines are intersecting, then put the value of 'λ' back into its equation to find the point of intersection
-
e.g. Find where the lines AB [ r = (4i + 4j) + λ(4i - 2j + 3k) ] and CD [ r = (2j) + ｵ(6i - j + 3k) ] intersect:
┌ ┐ ┌ ┐ ┌ ┐ ┌ ┐ │ 4 │ │ 4 │ │ 0 │ │ 6 │ 覧覧► 4+4λ = 6μ 覧覧► 2+2λ = 3μ │ 4 │ + λ │-2 │ = │ 2 │ + μ │-1 │ 覧覧► 4-2λ = 2-μ 覧覧► 4-2λ = 2-μ + │ 0 │ │ 3 │ │ 0 │ │ 3 │ ════╗ ------------- └ ┘ └ ┘ └ ┘ └ ┘ ║ 8 = 2+2μ ║ => 2 = μ ║ => 2 = λ ║ ╔══╩═══════════════════════════════════╗ ║ USE THIS TO SEE IF ACTUALLY DO MEET: ║ ║ 0 + 3λ = 0 + 3μ ║ ║ 0 + 3(2) = 0 + 3(2) √ It works! ║ ║ SO THEY DO MEET ║ ╚══════════════════════════════════════╝ Now, either put λ = 2 into the first line, or put μ = 2 into the second line to find WHERE: Putting λ = 2 into line (1): ┌ ┐ ┌ ┐ ┌ ┐ │ 4 │ │ 4 │ │12 │ r = │ 4 │ + 2 │-2 │ = │ 0 │ = 12i + 6k │ 0 │ │ 3 │ │ 6 │ └ ┘ └ ┘ └ ┘
So, they meet at (12, 0, 6)
Note: It is not always the case that we will solve ROW 1 and ROW 2 simultaneously. If necessary, we can switch to solving ROW 2 and ROW 3 simultaneously (and them using ROW 1 to verify intersection)
Converting a Vector Equation to Cartesian Form
The method is as follows:
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Replace 'r' with (x, y, z)
-
Read the "TOP ROW" of this vector equation
-
Read the "MIDDLE ROW" of this vector equation
-
Read the "BOTTOM ROW" of this vector equation
-
Make 'λ' the subject of ALL equations and then make them equal to each other. This is the Cartesian equation
e.g. Find the Cartesian Equation for: r = (4i + 4j) + λ(4i - 2j + 3k):
┌ ┐ ┌ ┐ │ 4 │ │ 4 │ r = │ 4 │ + λ │-2 │ │ 0 │ │ 3 │ ║ └ ┘ └ ┘ ▼ ┌ ┐ ┌ ┐ ┌ ┐ │ x │ │ 4 │ │ 4 │ 覧・#9658; x = 4 + 4λ ═╗ │ y │ = │ 4 │ + λ │-2 │ 覧・#9658; y = 4 - 2λ ═╬═══════════════╗ │ z │ │ 0 │ │ 3 │ 覧・#9658; z = 3λ ═╬═══════════════╬═══════════════╗ └ ┘ └ ┘ └ ┘ ║ ║ ║ ▼ ▼ ▼ λ = x - 4 λ = y - 3 λ = z 5 12 3 => x - 4 = y - 3 = z 5 12 3
Now you can see why we prefer the vector methods: The equation of a line in 3-D does not look very pretty in Cartesian form...
In practice, it is not necessary to go through this, because you can 'see' how to get quickly from one form to the other
e.g. Find the Cartesian Equation for: r = (5i + 2j + 1k) + λ(7i + 3j + 4k):
=> x - 5 = y - 2 = z - 1 7 3 4
e.g. Find the Cartesian Equation for: r = (2i + 3j + 4k) + λ(5i + 6j + 7k):
=> x - 2 = y - 3 = z - 4 5 6 7
e.g. Find the Cartesian Equation for: r = (2i -6j + 1k) + λ(7i + 4k):
=> x - 2 = y + 6 = z - 1 7 0 4
But this one doesn't seem to make sense, because we can't "divide by 0" In fact, the only way it comes close to making sense is if y=-6 (then the top of the fraction would also be zero)
=> So the equation is: x - 2 = z - 1 and y = -6 7 4
Question 1, part (a): The Cartesian Equation for: r = (2i + 3j -1k) + λ(1i + 1j + 1k):
=> x - 2 = y - ・/font> = z + 1 1 ・/font> ・/font> Which simplifies to・
Question 1, part (b): The Cartesian Equation for: r = (0i + 4j + 0k) + λ(3i + 0j + 5k):
=> x - 0 = y - 4 = z - 0 3 0 5
Now the middle bit of this equation doesn't make sense and can only make sense if y = 4; so the equation becomes:
=> So the equation is: x = z and y = ... 3 5
Question 1, part (d): They've written the equation slightly differently, but it is easy to convert to the form that we are used to:
┌──────────┐ ┌ ┐ ┌─┴─┐ └─────────────────────►│ 2 - 1λ │ r = (2-λ)i + (3+2λ)j + λk => r = │ 3 + 3λ │◄────┐ └─┬──┘ └┬┘ ┌───────►│ 0 + 1λ │ │ │ └───────┘ └ ┘ │ └───────────────────────────────────────┘ ┌ ┐ ┌ ┐ │ 2 │ + λ │-1 │ r = │ 3 │ │ ・│ │ ・│ │ ・│ └ ┘ └ ┘
Question 2, part (a): It is fairly easy to reverse the process used in question 1:
┌ ┐ ┌ ┐ x - 3 = y - 1 = z - 7 │ 3 │ + λ │ 4 │ 4 2 6 => │ 1 │ │ ・│ │ ・│ │ ・│ └ ┘ └ ┘ └─┬─┘ This is the direction vecotr of the line ─────────┘ It can be cancelled down
Question 2, part (b): If we write the equation as:
x - 1 = y - ・/font> = z - ・/font> -3 ・/font> 1
Then it becomes easier to see how to convert it...
Question 2, part (c): If we make λ the subject of each equation first...
Question 2, part (d): At first, lets ignore the z = 4, and concentrate on the:
2x + 3y = 12 => 2x = 3y - 12 => 2(x-0) = 3(y-4) => x - 0 = y - 4 3 2 .
Now, dealing with the z = 4: If we write the equation as:
x - ・/font> = y - ・/font> = z - 4 ・/font> ・/font> 0 └──┬──┘ └─────────────── Similar to qu 1(b)
Then we can easily convcert it to vector form...
Question 3, part (a): They have kindly given us the position vector of a point on the line and a direction vector along the line:
r = OA + λ b
┌ ┐ ┌ ┐
│ 1 │ │ 5 │
r = │-3 │ + λ │ 4 │ or: r = (i - 3j + 2k) + λ(5i + 4j - k)
│ 2 │ │-1 │
└ ┘ └ ┘
╘═╦═╛ ╘═╦═╛
the position vector ╝ ╚ the dirn vector
of a point on the line parallel to the line
Question 4, part (a): To decide if they are parallel or not, we need to compare the direction vectors:
d.v. of eqn 1 d.v. of eqn 2 ┌ ┐ ┌ ┐ │ 2 │ ◄隣2覧 │ 1 │ │-3 │ ◄隣2覧 │ 1 │ │ 1 │ ◄隣2覧 │-1 │ └ ┘ └ ┘
Is there a multiplier that works for the TOP ROW, MIDDLE ROW and BOTTOM ROW? (If so, they are parallel; if not, they are non-parallel)
Question 4, part (b): It is easier to compare the direction vectors of the lines (to see which are parallel) if they are all written in vector form...
Line 1: Line 2: ┌ ┐ ┌ ┐ ┌ ┐ ┌ ┐ │ 1 │ │ ・│ │ 0 │ │ ・│ r = │ ・│ + λ │ 3 │ r = │-5 │ + μ │ ・│ │-1 │ │-4 │ │ ・│ │-8 │ └ ┘ └ ┘ └ ┘ └ ┘
Question 5: A very quick sketch is a good idea, as your mind can work more easily with a sketch to work from
I can produce a superb 3D sketch, because I'm doing it on a computer:
But you only need to produce a 'rough' sketch like this:
The direction vector of the line AB is found by finding the vector AB:
┌ ┐ │-2 │ AB = │-2 │ │-6 │ └ ┘
But the direction vector of the line AB can be any multiple of this vector; so it makes sense to cancel it down by multiplying by '-ｽ':
┌ ┐ ┌ ┐ │ ・│ │ 1 │ r = │ ・│ + λ │ 1 │ │ 4 │ │ 3 │ └ ┘ └ ┘ ╘═╦═╛ ╘═╦═╛ you can use either the ╝ ╚ the dirn vector position vector of the parallel to the line point 'A', or of the point 'B'
Similarly, the equation of the line BC is found using the position vector of 'B' (say) and the direction vector given by finding the relative vector CB・/p>
Question 5, part (b): We need to add 'D' to our diagram - placing 'D' in such a way that is would form a parallelogram (i.e. the opposite sides would be parallel)
On my super-dooper sketch, it would look like this:
But on your 'rough and ready' sketch, it might look like this:
Since it is a parallelogram, the the vector BA and the vector CD must be equivalent (I've labelled them as p in the diagram below):
We can easily find BA (the relative vector from A to B, by subtracting):
┌ ┐ │ 2 │ BA = │ ・│ = p │ ・│ └ ┘
Which means that we also know the vector CD:
┌ ┐ │ 2 │ CD = │ ・│ = p │ ・│ └ ┘
Now, since we know the coordinates of 'C' (which means we know how to get from 'O' to 'C') and we know the vector from 'C' to 'D' - we can find the coordinates of 'D' by adding vectors:
OD = OC + CD ┌ ┐ ┌ ┐ ┌ ┐ │ ・│ │ 2 │ │ ・│ OB = │ ・│ + │ ・│ = │ 4 │ │-1 │ │ ・│ │ ・│ └ ┘ └ ┘ └ ┘
So, they meet at (・ 4, ・
Question 6, part (a): It is easy to find the equation of the line (we need to find AB first - the direction vector of the line; well, you can divide it by '2' to make the direction vector simpler...)
To find where the line crosses the xy plane, we make z=0. Lets call the point (?, ?, 0):
┌ ┐ ┌ ┐ ┌ ┐ │ 3 │ │-1 │ │ ? │ 覧覧► 3 - λ = ? ◄═════════════╗ │ 1 │ + λ │ ・│ = │ ? │ 覧覧► 1 + ・#955; = ? ◄═════════════╣ │-4 │ │ 5 │ │ 0 │ 覧覧► -4 + 5λ = 0 覧覧► λ = 4/5 ═╝ └ ┘ └ ┘ └ ┘ Solving the bottom row gives: λ = 4/5 Substituting in the top and middle rows gives: x = ・ y = ・ So the point is (・ ・ 0)
Question 7: Write the equation of the line given in vector from, and notice that its direction vector is:
┌ ┐ │ 3 │ OA = │ 4 │ │ 5 │ └ ┘
Our line has the same direction vector (i.e. it is parallel), but the position vector of a point on the line is 5i - 2j - 4k...
Question 7, part (b): To find where y=0, we call the point
(?, 0 , ?) and make it
equal to the equation...
Solving (the MIDDLE ROW) to find 'λ'
And then using this value of 'λ' in the TOP ROW and BOTTOM
ROW to find 'x' and 'z'...
Question 8: Re-write it as:
x - 2 = y + ｽ = z - ・/span> 1 ｽ ⅓
Then it becomes easier to see how to convert it...
Don't forget that you can 'simplify' the direction vector of the
line by multiplying through by ...
Our line is parallel to this, but passes through (2, -1, -1)...
Question 9: Start each part by making sure the equations are written in vector form.
Step 1: Look at the direction vectors of each pair to see if they are parallel lines (or perhaps even equivalent equations for the same line)
Step 2: If not, make the equations equal to each other; solve the TOP and MIDDLE ROWS to find λ and μ and put these values into the BOTTOM ROW to see if they are intersecting (if those values work) or skew (if those values don't work in the BOTTOM ROW)
Step 3: If you've discovered they are intersecting, then put this value of 'λ' into the equation of the first line to find the point of intersection (you can also put the value of μ into the equation of the second line as a check)
Question 9, part (a): Both direction vectors simplify to ... So the lines are...
Question 9, part (b): Re-writing the lines in vector form:
Line 1: Line 2: ┌ ┐ ┌ ┐ ┌ ┐ ┌ ┐ │ 4 │ │ 1 │ │ 7 │ │ 6 │ r = │ 8 │ + λ │ 2 │ r = │ 6 │ + μ │ 4 │ │ 3 │ │ 1 │ │ 5 │ │ 5 │ └ ┘ └ ┘ └ ┘ └ ┘ └─┬─┘ └─┬─┘ └─────────────────not parallel──────────────────┘
Looking at the direction vectors, we can see instantly that these lines are non-parallel...
Since they are not parallel, we make them equal each other:
┌ ┐ ┌ ┐ ┌ ┐ ┌ ┐ │ 4 │ │ 1 │ │ 7 │ │ 6 │ 覧覧► 4+ λ = 7+6μ 覧覧► 4+λ = 7+6μ │ 8 │ + λ │ 2 │ = │ 6 │ + μ │ 4 │ 覧覧► 8+2λ = 6+4μ 覧覧► 4+λ = 3+2μ + │ 3 │ │ 1 │ │ 5 │ │ 5 │ ════╗ ------------ └ ┘ └ ┘ └ ┘ └ ┘ ║ 0 = 4+4μ ║ => ・= μ ║ => ・= λ ║ ╔══╩═══════════════════════════════════╗ ║ USE THIS TO SEE IF ACTUALLY DO MEET: ║ ║ 3 + λ = 5 + 5μ ║ ║ 3 + (・ = 5 + 5(・ √ It works! ║ ║ SO THEY DO MEET ║ ╚══════════════════════════════════════╝ Now, either put λ = ・/b> into the first line, or put μ = ・into the second line to find WHERE: Putting λ = ・/font> into line (1): ┌ ┐ ┌ ┐ ┌ ┐ │ 4 │ │ 1 │ │ 1 │ r = │ 8 │ + ・/b> │ 2 │ = │ ・│ = i + ・u>k │ 3 │ │ 1 │ │ ・│ └ ┘ └ ┘ └ ┘
Question 9, part (c): They are clearly non-parallel, so make the equations equal and get your simultaneous equations from the TOP and MIDDLE ROWS; solve them and check the values in the BOTTOM ROW to see if they actually do intersect...
Question 10: They've TOLD US that these lines do intersect...
So that means that whatever rows we use to get the values of λ and μ, we know that those answers will work in the remaining row...
So, for this question, since ROW 1 contains an unknown (疎・, let's start by ignoring that row and working with ROW 2 and ROW 3:
┌ ┐ ┌ ┐ ┌ ┐ ┌ ┐
┌──►│ 2 │ │ 1 │ │ a │ │-1 │ ┐
│ │ 9 │ + λ │ 2 │ = │ 7 │ + μ │ 2 │ 覧覧► 9 + 2λ = 7 + 2μ ╞╡ Solving these simultaneously
│ │ 13│ │ 3 │ │-2 │ │-3 │ 覧覧► 13 + 3λ = -2 - 3μ ╞╡ gives λ = ・ and μ = ・
│ └ ┘ └ ┘ └ ┘ └ ┘ ┘ └──────────────┬──────────────┘
│ ▼
│ ┌───────────────────────────────────────────────────────────┐ │
└──────◄──────┤Putting these values of λ and μ into ROW 1 lets us find 疎・#9500;─────◄────┘
└───────────────────────────────────────────────────────────┘
Question 12: Start by finding the equation of the line AB: We know A lies on the line and the relative vector AB = 3i + 6j + 9k is the direction vect0r of the line:
┌ ┐ ┌ ┐ │ 0 │ │ 1 │ r = │ 1 │ + λ │ 2 │ │ 1 │ │ 3 │ └ ┘ └ ┘ └─┬─┘ This is the direction vecotr of the line ─────┘ I cancelled it down
Next find the equation of the line CD...
Then solve simulaneously...
BUT THERE'S A SUBTLE DIFFERENCE: Our equation of the Line AB implies it it is infinitely long (i.e. it starts long before A and it continues way past B)...
But that's not true - the question says the line from A to B (which are λ=0 and λ=3). So, if you solve them simultaneoulsy and get an answer for λ that is NOT between 0 and 3, then - well they don't actually meet...