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TAEuler
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#1
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The position vectors of points A and B are (7, 1, -3) and (-2,8,-1)

a) Calculate vector OA . vector OB

Very simple, (7, 1, -3) . (-2, 8, -1) = -14 + 8 + 3 = -3

b) Find the size, in degrees, of the acute angle between the vectors.

lal = root (49 + 1 + 9) = root 59

lbl = root (4 + 64 + 1) = root 69

a.b = lal lbl cos x
so, -3 = root 59 x root 69 (cos x)
so -3 = root (4071) cos x
so cos x = -3/root (4071)

x = arccos (-3/root (4071)
= 92.7 degrees,

but it says the acute angle, so the angle = 180-x = 87.3

But the textbook says the answer is 87.1? Have I done anything wrong?
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mqb2766
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#2
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(Original post by TAEuler)
The position vectors of points A and B are (7, 1, -3) and (-2,8,-1)

a) Calculate vector OA . vector OB

Very simple, (7, 1, -3) . (-2, 8, -1) = -14 + 8 + 3 = -3

b) Find the size, in degrees, of the acute angle between the vectors.

lal = root (49 + 1 + 9) = root 59

lbl = root (4 + 64 + 1) = root 69

a.b = lal lbl cos x
so, -3 = root 59 x root 69 (cos x)
so -3 = root (4071) cos x
so cos x = -3/root (4071)

x = arccos (-3/root (4071)
= 92.7 degrees,

but it says the acute angle, so the angle = 180-x = 87.3

But the textbook says the answer is 87.1? Have I done anything wrong?
Everything is right apart from it asks for the acute angle.
Just draw two lines (vectors). You could measure the angle between them as either 93 or 87 degrees depending on the direction of the vector.

Flip the sign of OA, you now get 3 for the dot product and the arccos() is 87.

Edit - misread your question.
Last edited by mqb2766; 3 weeks ago
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ghostwalker
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#3
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(Original post by TAEuler)
= 92.7 degrees,

but it says the acute angle, so the angle = 180-x = 87.3

But the textbook says the answer is 87.1? Have I done anything wrong?
Agree with 87.3 (3dec.pl.). Error in book.
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TAEuler
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#4
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#4
I need help with another question as I disagree with their answer, this one i'm less confident with my workings:

Is the point (-2,4,7) closer to the line with equation r = 5i + 2k + λ(i - 2j + 4k) or the line parallel to the vector 2i + j that passes through the origin?:

we know Line 1: r = (5, 0, 2) + λ (1, -2, 4)
and line 2: r = (0,0,0) + s (2, 1, 0)

let point B be on line 1 so that the vector AB is perpendicular to line 1.

so vector AB . (1, -2, 4) = 0

Vector OA = (-2, 4, 7) and Vector OB = (5 + λ, -2λ, 2 + 4λ)

so vector AB = (7 + λ, -2λ - 4, 4λ - 5)

so (7 + λ, -2λ - 4, 4λ - 5) . (1, -2, 4) = 0
7 + λ - 2(-2λ - 4) + 4 (4λ - 5) = 0
7 + λ + 4 λ + 8 + 16λ - 20 = 0
so λ = 5/21

so Vector AB = (152/21, -94/21, -85/21)

lABl = root (1865/21) = 9.424 (3sf)

Let point C be on line 2 so that vector AC is perpendicular to line 2

So vector OC = (2s, s, 0)

vector AC = (2s, s, 0) - (-2, 4, 7)
= (2s + 2, s - 4, -7)

So Vector AC . (2, 1, 0) = 0

(2s + 2, s - 4, -7) . (2, 1, 0) = 0

4s + 4 + s -4 = 0
5s = 0
so s = 0

so Vector AC = (2, -4, -7)

lACl = root 69

therefore, the point is closer to the second line.

However, the book says it's closer to the first?

Have I gone wrong or is it another textbook mistake?
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ghostwalker
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#5
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#5
(Original post by TAEuler)
I need help with another question as I disagree with their answer, this one i'm less confident with my workings:

Is the point (-2,4,7) closer to the line with equation r = 5i + 2k + λ(i - 2j + 4k) or the line parallel to the vector 2i + j that passes through the origin?:

we know Line 1: r = (5, 0, 2) + λ (1, -2, 4)
and line 2: r = (0,0,0) + s (2, 1, 0)

let point B be on line 1 so that the vector AB is perpendicular to line 1.

so vector AB . (1, -2, 4) = 0

Vector OA = (-2, 4, 7) and Vector OB = (5 + λ, -2λ, 2 + 4λ)

so vector AB = (7 + λ, -2λ - 4, 4λ - 5)

so (7 + λ, -2λ - 4, 4λ - 5) . (1, -2, 4) = 0
7 + λ - 2(-2λ - 4) + 4 (4λ - 5) = 0
7 + λ + 4 λ + 8 + 16λ - 20 = 0
so λ = 5/21

so Vector AB = (152/21, -94/21, -85/21)

lABl = root (1865/21) = 9.424 (3sf)

Let point C be on line 2 so that vector AC is perpendicular to line 2

So vector OC = (2s, s, 0)

vector AC = (2s, s, 0) - (-2, 4, 7)
= (2s + 2, s - 4, -7)

So Vector AC . (2, 1, 0) = 0

(2s + 2, s - 4, -7) . (2, 1, 0) = 0

4s + 4 + s -4 = 0
5s = 0
so s = 0

so Vector AC = (2, -4, -7)

lACl = root 69

therefore, the point is closer to the second line.

However, the book says it's closer to the first?

Have I gone wrong or is it another textbook mistake?

I did it by minimising the square of the distances - a different method - and got the exact same results as you. So, book's in error.
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TAEuler
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Another question, this one in regards to me not being able to visualise something as it's in 3-D:

the line L1 is given by r = (6, -3, 1) + λ (-1, 1, -2)

The line L2 is given by r = (4, 2, -5) + s(-5, 2, -8).

The point A is the intersection of the lines L1 and L2 and the point B has coordinates (1, -2, 7). Calculate the area of the triangle OAB:

To find A...
(6, -3, 1) + λ (-1, 1, -2) = (4, 2, -5) + s(-5, 2, -8)

x: 6 - λ = 4 - 5s
y: λ - 3 = 2 + 2s
z: 1 - 2λ = -5 -8s

through simultaneous equations, I got the A as (-1, 4, -13)

This is where I struggle to visualise it, I would try and use dot product but I can't figure out how the angle calculated plays into it, it could potentially be 180 - the angle from dot product and then simply 1/2 ab sin C to find the area. So I'm not too sure on what to do, or how to visualise it. I also tried a sketch but I couldn't seem to display the vectors.

I also worked out lOAl as root 186
and lOBl as 3 root 6.
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RDKGames
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#7
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(Original post by TAEuler)
This is where I struggle to visualise it, I would try and use dot product but I can't figure out how the angle calculated plays into it, it could potentially be 180 - the angle from dot product and then simply 1/2 ab sin C to find the area.
Well the fact that \sin x \equiv \sin(180-x) it will not make a difference whether you use one angle or the other.

So I'm not too sure on what to do, or how to visualise it. I also tried a sketch but I couldn't seem to display the vectors.

I also worked out lOAl as root 186
and lOBl as 3 root 6.
You know A. So to use \dfrac{1}{2}ab\sin C you should calculate a=|AO| and b=|AB| with C being the angle between AO and AB.

Any sort of diagram like this is sufficient to understand what you need to calculate:

[EDIT: It's actually less work if you just consider the vectors OA and OB instead]

Last edited by RDKGames; 3 weeks ago
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TAEuler
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#8
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A simpler question:

This is for my mate but I wasn't able to explain it for him:

Points A and B have position vectors OA = (2, 2, 3) and OB = (-1, 7, 2). Find the angle between AB and OA

The textbook specifies to use the scalar product of two vectors to find the angle between them, you need both vectors to be directed away from the angle, otherwise the angle you find will be 180 - the angle.

His workings:

Let the angle between the position vectors AB and AO be x.

Vector AB = (-1, 7, 2) - (2, 2, 3) = (-3, 5, -1)
lABl = root 35
A
Vector AO = (-2, -2, -3). lAOl = root 17

AB . AO = -1

x = arccos (-1/root 595) = 92.3 degrees

However, the answer given is 87.7 degrees.

This has confused both me and him, have they calculated the wrong angle? And if not why is it 180 - x
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RDKGames
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#9
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(Original post by TAEuler)
A simpler question:

This is for my mate but I wasn't able to explain it for him:

Points A and B have position vectors OA = (2, 2, 3) and OB = (-1, 7, 2). Find the angle between AB and OA

The textbook specifies to use the scalar product of two vectors to find the angle between them, you need both vectors to be directed away from the angle, otherwise the angle you find will be 180 - the angle.

His workings:

Let the angle between the position vectors AB and AO be x.

Vector AB = (-1, 7, 2) - (2, 2, 3) = (-3, 5, -1)
lABl = root 35
A
Vector AO = (-2, -2, -3). lAOl = root 17

AB . AO = -1

x = arccos (-1/root 595) = 92.3 degrees

However, the answer given is 87.7 degrees.

This has confused both me and him, have they calculated the wrong angle? And if not why is it 180 - x
Nothing in the question suggests you seek the acute/obtuse angle, so no need to aim for one or the other.

Also, since the questions says between AB and OA then those are the vectors you need to dot.
You dotted AB with AO instead.

The dot product formula gives the angle between the vectors that are ‘head-to-head’ or ‘tail-to-tail’. It may not seem that you can take the dot product between AB and OA to get anything meaningful given this, but notice that if (on a diagram) you extend the vector OA through A to A’ so that it is now OA’ = 2*OA, you should realise that OA = AA’ and AA’ has its tail to AB. So that’s the angle the between the tails that the question effectively wants.
Last edited by RDKGames; 3 weeks ago
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TAEuler
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#10
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(Original post by RDKGames)
Nothing in the question suggests you seek the acute/obtuse angle, so no need to aim for one or the other.

Also, since the questions says between AB and OA then those are the vectors you need to dot.
You dotted AB with AO instead.

The dot product formula gives the angle between the vectors that are ‘head-to-head’ or ‘tail-to-tail’. It may not seem that you can take the dot product between AB and OA to get anything meaningful given this, but notice that if (on a diagram) you extend the vector OA through A to A’ so that it is now OA’ = 2*OA, you should realise that OA = AA’ and AA’ has its tail to AB. So that’s the angle the between the tails that the question effectively wants.
I don't get it. It says between vectors OA and AB, but the angle between these two is OAB, the vector OA ends at A so why should it be the acute angle because the two and not the obtuse?
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RDKGames
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#11
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(Original post by TAEuler)
I don't get it. It says between vectors OA and AB, but the angle between these two is OAB, the vector OA ends at A so why should it be the acute angle because the two and not the obtuse?
It’s not necessarily *that* angle tho...

The vector OA is just that, a vector. On your diagram this vector can be anywhere you want it to be as long as it has the same magnitude and direction. It need not start start at O, which is what you are relying on too much. I explained in my last post how AA’ is an equivalent vector to OA, so “the angle between OA and AB” is the same as saying “the angle between AA’ and AB” and if you look at the vector that starts at A and goes to A’ you would see that it has its tail to the vector going from A to B. So at no point do we get the angle you mistakenly look for.
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TAEuler
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#12
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"Show that the lines with equations r = (5i - 2j + k) + t(2i - j - k) and r = (i + j - k) + s(-2 + i + j + 5k) are skew."

Is this an error in question, since the -2 is left on it's own and doesn't have an i, j or k?
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RDKGames
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(Original post by TAEuler)
"Show that the lines with equations r = (5i - 2j + k) + t(2i - j - k) and r = (i + j - k) + s(-2 + i + j + 5k) are skew."

Is this an error in question, since the -2 is left on it's own and doesn't have an i, j or k?
[Without working it out] I think it's supposed to be -2i instead. Typo.
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