Mikehepro's spesh thread

[mikehepro]

Hey guys:
Got a few questions on Circular functions.
Any help will be greatly appreciated
1.) Solve each of the following equations for x ()


2.) A curve on a light rail track is an arc of a circle of length 300m and the straight line joining the two ends of the curve is 270m long.
a. Show that, if the arc subtends an angle of at the centre of the circle is a soulution of the equation why is latex not working here? the equation is sin/theta=(\frac{\pi}{200})\theta
b.solve, correct to two decimal places, the equation for .
Thanks

[b^3]

1)

[mikehepro]

Hey guys, got a few questions on Vectors
Any help would be greatly appreciated.

1. A and B are defined by the position vectors  a=2i-2j-k and b=3i+4k
Find the unit vector which bisects the angle AOB. (Essential Ex 2D Q3 part b)
I thought that all i have to do is to find C which will be the unit vector of a+unit vector of b. Then find the unit vector of C to obtain the answer;however, i got a different answer to the answer in the book.


2. There's a few parts to this, i got a,b,c, but i have problem with part d.
A,B and C are points defined by the position vectors a=i+2j+k, b=2i+j-k and c=2i-3j+k. Find:

a, i, AB        ii, AC
b. the vector resolute of AB in the direction of AC
c. the shortest distance from B to line AC
and d.  the area of triangle ABC.
(Essential Ex 2D Q9 part d)
the answer is sqrt of 30 btw.

Thanks
Mike

[alchemy]

Hey guys, got a few questions on Vectors
Any help would be greatly appreciated.

1. A and B are defined by the position vectors  a=2i-2j-k and b=3i+4k
Find the unit vector which bisects the angle AOB. (Essential Ex 2D Q3 part b)
I thought that all i have to do is to find C which will be the unit vector of a+unit vector of b. Then find the unit vector of C to obtain the answer;however, i got a different answer to the answer in the book.

In general:
-Find the unit vectors of a and b.
-Find the sum of unit vectors a and b (this sum is actually the vector that bisects angle AOB)
-Find the unit vector of the sum vector of vectors a and b.

OR you could try this fantastic alternative method:

-Find the dot product of vectors a and b, and hence find the angle between vectors a and b. This is the angle AOB
-Let another vector (e.g. vector c) be the vector that bisects AOB. This vector would then make 1/2 of the angle AOB with the x-axis. So, divide the value you got in the first step to find the angle vector c makes with the x-axis.
-Let c = xi + yj +zk and it's magnitude equal to 1
-Find the dot product of a and c. Let this value equal to the angle vector c makes with the x-axis (refer to step 2), to form 'equation 1'.
-Find an equation for the magnitude of c and let this be equal to one (refer to step 3). Let this be equation 2.
-Now, the angle between vector b and c must be the same as the angle between vectors a and c. You can place that value for the angle in an equation that finds the dot product of vectors b and c. So, it would look like: b.c = |b||c| cos(angle between vectors a and c). Let this be equation 3.
-Solve the 3 equations to get your answer.

Note: I just 'copy pasted' this from an old post of mine on the spesh thread. Saves me from having to type it up again :/

[mikehepro]

Thanks for your help!!!
Personally i felt the second method requires more work.
I tried the first method and got a different answer to the one in the book.
so i worked out the sum of unit vectors a and b as
However when i took the unit vector of that, it didn't give me answer in the book which is , i got
I must have made a stupid mistake here.

[lzxnl]

Thanks for your help!!!
Personally i felt the second method requires more work.
I tried the first method and got a different answer to the one in the book.
so i worked out the sum of unit vectors a and b as
However when i took the unit vector of that, it didn't give me answer in the book which is , i got
I must have made a stupid mistake here.

You'll find that the magnitude of your answer isn't 1. 19^2 + 10^2 + 7^2 = 461 + 49 = 510
So you need to divide your original vector by sqrt(510)

[mikehepro]

You'll find that the magnitude of your answer isn't 1. 19^2 + 10^2 + 7^2 = 461 + 49 = 510
So you need to divide your original vector by sqrt(510)

ah i see what i did wrong here. Thanks.
Could anyone one please help me with the second question as well?

[lzxnl]

2. There's a few parts to this, i got a,b,c, but i have problem with part d.
A,B and C are points defined by the position vectors a=i+2j+k, b=2i+j-k and c=2i-3j+k. Find:

a, i, AB        ii, AC
b. the vector resolute of AB in the direction of AC
c. the shortest distance from B to line AC
and d.  the area of triangle ABC.
(Essential Ex 2D Q9 part d)
the answer is sqrt of 30 btw.

Thanks
Mike

I'm presuming 2a is fairly simple, so I'm not going to bother with those.
Hopefully you can do 2b; it's a standard question that you're expected to know in spesh.
Now let's look at c.
Draw a dot B and a line AC anywhere on a piece of paper, providing B isn't on AC. Now, draw a line from B to anywhere on AC. If you draw in the perpendicular from B to AC, you'll see that it's shorter than any other line connecting B to AC. This distance is what we mean by distance from a line to a point.
Now, if you look at your vector resolute parallel to AC, can you see that if you subtract this from AB, you get a vector perpendicular to AC by definition that goes from B to the line AC? The length of this is the distance you require. (do draw out a diagram for this)
As for 2d, the area is just half base height. You should be able to identify a base and a height now.

[mikehepro]

I'm presuming 2a is fairly simple, so I'm not going to bother with those.
Hopefully you can do 2b; it's a standard question that you're expected to know in spesh.
Now let's look at c.
Draw a dot B and a line AC anywhere on a piece of paper, providing B isn't on AC. Now, draw a line from B to anywhere on AC. If you draw in the perpendicular from B to AC, you'll see that it's shorter than any other line connecting B to AC. This distance is what we mean by distance from a line to a point.
Now, if you look at your vector resolute parallel to AC, can you see that if you subtract this from AB, you get a vector perpendicular to AC by definition that goes from B to the line AC? The length of this is the distance you require. (do draw out a diagram for this)
As for 2d, the area is just half base height. You should be able to identify a base and a height now.

Thanks for the help, I did a,b,c with out a problem, when i got to d, i knew that i had to use half base height. The base would be the answer in part C if i'm not wrong. I got lost when i needed the height, which i tried to use C^2=A^2+B^2 to solve witch gave me a wrong answer. Am i on the right track?
Thanks

[lzxnl]

Look at your triangle and see if it's a right angled triangle. Otherwise use normal trig, like 1/2 ab sin C

[mikehepro]

Look at your triangle and see if it's a right angled triangle. Otherwise use normal trig, like 1/2 ab sin C

Thanks mate.
I just drew a diagram now it become a lot clearer that all i have to do was 1/2bh. I mistaken answer C as the base instead of the height... Silly me