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# C4 Vectors Watch

1. (Original post by BabyMaths)
OK = 4 i -11j +4k + lambda(i + 4j + k)

The dot product of this with i+4j+k (the direction of AB) is 0.
Why is OM equal to this ? Only M can be a point not the whole equation (OM). Then it can't give 0 with cross product.
2. (Original post by zaback21)
Why is OM equal to this ? Only M can be a point not the whole equation (OM). Then it can't give 0 with cross product.
Do the first part of the equation and you get
r=(4,-11,4)+lambda(1,4,1)

M is some point on this line, between A and B. Remember the direction of AB is (1,4,1). So when we are at A we must move a certain multiple of (1,4,1) vector to get to this point M. We let this multiple=lambda. To break this down, the journey from O to M, which gives the position vector of M, is like this

1. Go to A from O, so vector (4, -11,4)
2. Move a certain multiple of (1,4,1) to go from A to M
Overall, you went from O to M

A sketch really really helps a lot to visualise things .
3. (Original post by krisshP)
Do the first part of the equation and you get
r=(4,-11,4)+lambda(1,4,1)

M is some point on this line, between A and B. Remember the direction of AB is (1,4,1). So when we are at A we must move a certain multiple of (1,4,1) vector to get to this point M. We let this multiple=lambda. To break this down, the journey from O to M, which gives the position vector of M, is like this

1. Go to A from O, so vector (4, -11,4)
2. Move a certain multiple of (1,4,1) to go from A to M
Overall, you went from O to M

A sketch really really helps a lot to visualise things .
I know what O to M is. I know what OM is. But the equation of line OM is not what he wrote. It is only equal at the point M and O has nothing to do with it. Its the equation AB which should be (4,-11,4)+lambda(1,4,1) not OM=(4,-11,4)+lambda(1,4,1).

Equation of OM is = t(ai+bj+ck)
where O=(0,0,0) M=(a,b,c) and t is a parameter.
4. (Original post by zaback21)
I know what O to M is. I know what OM is. But the equation of line OM is not what he wrote. It is only equal at the point M and O has nothing to do with it. Its the equation AB which should be (4,-11,4)+lambda(1,4,1) not OM=(4,-11,4)+lambda(1,4,1).

Equation of OM is = t(ai+bj+ck)
where O=(0,0,0) M=(a,b,c) and t is a parameter.
You are correct. If I was not being lazy I would have said for some value of .

Sorry for any confusion caused.

I must be less sloppy.
I must be less sloppy.
Only another 98 lines to do..
5. (Original post by BabyMaths)
You are correct. If I was not being lazy I would have said for some value of .

Sorry for any confusion caused.

I must be less sloppy.
I must be less sloppy.
Only another 98 lines to do..
Ha ha lol !
6. (Original post by BabyMaths)
X
a=3i+5j+7k find the angle a makes with the x-axis?

With |a|=√83

Cos∅=3/√83
∅=71° <-- correct

Tan∅=5/3
∅=59° <-- wrong, but WHY?

Thank you very much
7. (Original post by krisshP)
a=3i+5j+7k find the angle a makes with the x-axis?

With |a|=√83

Cos∅=3/√83
∅=71° <-- correct

Tan∅=5/3
∅=59° <-- wrong, but WHY?

Thank you very much
Well, that triangle is all wrong. You can't even make a triangle with those side lengths.

I might have a go at a diagram in a minute. But mainly I should say, stick to using the dot product in these questions, even though it's good to try out your own ideas too.
8. Thanks a lot. The diagram did help me to visualise the situation far better.
9. (Original post by krisshP)
As previously said, here's the correct working out
With |a|=√83

Cos∅=3/√83
∅=71° <-- correct

But the thing is the working out implies that 3 is the length of the adjacent side. But the diagram shows that 3 is NOT the length of a side of the triangle. i hope i am making some sense here. But i agree though that the triangle's hypotenuse is sqrt34.

From the diagram only,

3^2 + 5^2=a^2
a=sqrt34

Tanø=7/sqrt34
Ø=47.7

Sorry, but the diagram did help me to visualise the situation far better, but I now question the "correct" working out.
The angle you found, , is the angle between the vector and the xy plane. To find the angle between the vector and the x axis you could use the cosine rule but you should just use the dot product.
10. (Original post by BabyMaths)
The angle you found, , is the angle between the vector and the xy plane. To find the angle between the vector and the x axis you could use the cosine rule but you should just use the dot product.
I get it now, what I Previously said doesn't give the required angles that is needed. I looked at everything again and then realised this, so I edited my post. Yes I understand why you suggest using the dot product rather than basic trig. Looking at that angle makes it be difficult to define what really even is the opposite/adjacent/hypotenuse side. So it's better to just stick to the dot product and keep things simple. I guess a "fresh eye" helps a lot in Maths if you are stuck on something

Thank you
11. (Original post by BabyMaths)
X

I'm stuck on the second part and was wondering during if you can help me out please.

Part a.
AB=(4,-8)

Line 1
r= (3,4) + lambda(4,-8)
r=(3,4) + lambda(1,-2)

Part b.

Let the direction of line 2 be (x,y), so for line 2
r=(14,7) + mu(x,y)

Since line 2 is perpendicular to line 1,

(1,-2).(x,y)=0

x-2y=0

I suspect that simultaneous equations are involved to get x and y, but I'm unsure how to get the second equation. I feel like I have a brain freeze.

Thank you
12. (Original post by krisshP)

x-2y=0

I suspect that simultaneous equations are involved to get x and y, but I'm unsure how to get the second equation. I feel like I have a brain freeze.

Thank you
You don't need simultaneous equations.

Recall that the direction vector of a line is only unique up to scalar multiples, i.e. (1,2), (2,4) specify the same direction.

So you have 2y=x.

And your direction vector (x,y) becomes (2y,y). Can you take it from there?
13. (Original post by ghostwalker)
You don't need simultaneous equations.

Recall that the direction vector of a line is only unique up to scalar multiples, i.e. (1,2), (2,4) specify the same direction.

So you have 2y=x.

And your direction vector (x,y) becomes (2y,y). Can you take it from there?
r=(14,7) + mu(2,1)

This is right according to the textbook answer of
r=(14,7) + mu(8,4)

But where did the 8 and 4 one from? Surely that means that perhaps there's another method getting exactly these numbers?
14. (Original post by krisshP)
r=(14,7) + mu(2,1)

This is right according to the textbook answer of
r=(14,7) + mu(8,4)

But where did the 8 and 4 one from? Surely that means that perhaps there's another method getting exactly these numbers?
If you recall if two lines are perpendicular then the product of their gradients is -1.

The gradient of a line with direction vector (4,-8) is -8/4, So the gradient of a prependicular lines is 4/8, which gives a direction vector of (8,4).
15. (Original post by ghostwalker)
If you recall if two lines are perpendicular then the product of their gradients is -1.

The gradient of a line with direction vector (4,-8) is -8/4, So the gradient of a prependicular lines is 4/8, which gives a direction vector of (8,4).
Makes perfect sense

Thank you for helping me out
16. thank you for the help
17. can you explain it more clearly by drawing the square and point o in it.
18. (Original post by davros)
OD is just the vector from O to D. You can get there in a number of ways e.g. O to A then A to D; or O to B and then B to D; or O to C and then C to D.

Using the fact that ABCD is a square, you should be able to express vectors like AD, BD and CD in terms of a, b and c.
i can't t understand it clearly can you draw the square and show it.

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