Heya all you smart (and helpful/bored) people of rhp 😀
This not that important, but it's really bugging me, so if anyone has some time on their hands:
We had homework in maths for tomorrow. She already told us what the answer was, and we were supposed to see if we could figure it out by ourselves. Problem is, I have a different answer than she does, and I can't find my mistake.
Here's the question:
In the triangle ABC with the center (I dunno the english word... you know, when you connect each corner with the middle of the opposing side, that point where the lines meet) S (0|0|4); A (1|1|0) and B (3|2|0), find point C.
The answer she gave was C (-4|-3|12).
My answer was C (-8|-7|12).
Any opinions on this?
Thanks,
Angie
Originally posted by angie88I think you got the wrong answer as C (-4|-3|12) does not equal
Heya all you smart (and helpful/bored) people of rhp 😀
This not that important, but it's really bugging me, so if anyone has some time on their hands:
We had homework in maths for tomorrow. She already told us what the answer was, and we were supposed to see if we could figure it out by ourselves. Problem is, I have a different answer than she does, and I ...[text shortened]... gave was C (-4|-3|12).
My answer was C (-4|-3|12).
Any opinions on this?
Thanks,
Angie
C (-4|-3|12).
Originally posted by angie88i don't understand the way you've wrote the question, could you explain in a different way or add a diagram.
Heya all you smart (and helpful/bored) people of rhp 😀
This not that important, but it's really bugging me, so if anyone has some time on their hands:
We had homework in maths for tomorrow. She already told us what the answer was, and we were supposed to see if we could figure it out by ourselves. Problem is, I have a different answer than she does, and I ...[text shortened]... gave was C (-4|-3|12).
My answer was C (-8|-7|12).
Any opinions on this?
Thanks,
Angie
Originally posted by trevor33That's all we were given... maybe you don't understand the display of (x1|x2|x3)... in a 3-D coordinate plane, x1 is the value on the axis pointing towards you, x2 is on the regular x-axis and x3 is the y-axis.
i don't understand the way you've wrote the question, could you explain in a different way or add a diagram.
Did that make it any clearer?
Originally posted by angie88I am not sure I understand the dimentions of the ABC triangle..
Heya all you smart (and helpful/bored) people of rhp 😀
This not that important, but it's really bugging me, so if anyone has some time on their hands:
We had homework in maths for tomorrow. She already told us what the answer was, and we were supposed to see if we could figure it out by ourselves. Problem is, I have a different answer than she does, and I ...[text shortened]... gave was C (-4|-3|12).
My answer was C (-8|-7|12).
Any opinions on this?
Thanks,
Angie
Is S the centre point? and it is where the say the imaginery lines drawn from the middle of each line towards each other? you are trying to find where C is?
If I can understand the lengths of A and B, and what angle they are to each other, then that would be of help....
1 edit
Originally posted by Peachyyes, S is the centre point. The lines are drawn from eg. point A to the middle of the line between B and C. From point B to the middle of the line between C and A. From point C to the middle of the line between A and B.
I am not sure I understand the dimentions of the ABC triangle..
Is S the centre point? and it is where the say the imaginery lines drawn from the middle of each line towards each other? you are trying to find where C is?
If I can understand the lengths of A and B, and what angle they are to each other, then that would be of help....
Yes, we're trying to find C.
For explanations of the axis, see above post.
The angles aren't given... we've just started with this 3-D vector stuff, so I'm not very good at explaining it.
Hope that makes it a bit clearer.
Originally posted by angie88yes i know what you mean, i've done something similar. were you given the distance of a+b or just the axis.
That's all we were given... maybe you don't understand the display of (x1|x2|x3)... in a 3-D coordinate plane, x1 is the value on the axis pointing towards you, x2 is on the regular x-axis and x3 is the y-axis.
Did that make it any clearer?
2 edits
Originally posted by angie88It does.
yes, S is the centre point. The lines are drawn from eg. point A to the middle of the line between B and C. From point B to the middle of the line between C and A. From point C to the middle of the line between A and B.
Yes, we're trying to find C.
For explanations of the axis, see above post.
The angles aren't given... we've just started with this ...[text shortened]... -D vector stuff, so I'm not very good at explaining it.
Hope that makes it a bit clearer.
Now is S = (0|0|4)
OR: 0, 0, -4 ?
EDIT: The answer given has -ve values. One of the points you gave must have -ve value too.
Originally posted by angie88out of the blue, I can suggest you a method for finding the 3D coordinates of the baricenter of the triangle (that is, where the medians intersects).
yes, S is the centre point. The lines are drawn from eg. point A to the middle of the line between B and C. From point B to the middle of the line between C and A. From point C to the middle of the line between A and B.
Yes, we're trying to find C.
For explanations of the axis, see above post.
The angles aren't given... we've just started with this ...[text shortened]... -D vector stuff, so I'm not very good at explaining it.
Hope that makes it a bit clearer.
First, project the 3 vertices of the triangle over any plane. You can do it easily cos you know the components of the vectors (the sides of the triangle).
Now, you have the projection of the the triangle on a plane (any of the the 3 ones).
Now, find the C point tracing the median segments. 2 will suffice.
And finally, return to the 3D space using the trig values you have used
while projecting the triangle pn a plane.
This should work.
Originally posted by CrazyLilTingto be honest... I have no idea what you're talking about 🙁
out of the blue, I can suggest you a method for finding the 3D coordinates of the baricenter of the triangle (that is, where the medians intersects).
First, project the 3 vertices of the triangle over any plane. You can do it easily cos you know the components of the vectors (the sides of the triangle).
Now, you have the projection of the the triangle ...[text shortened]... ng the trig values you have used
while projecting the triangle pn a plane.
This should work.
Thanks though.
Peachy-it's 4. the C has to be the negative thing... draw a sketch and you'll see.