- #36
CaliforniaRoll88
- 35
- 6
How do you disregard the ##1/3##?PeroK said:Note that the factor of ##\frac 1 3## is an unnecessary complication. Instead, we can look for:
$$\vec B=B\{2, -2, 1\}$$
How do you disregard the ##1/3##?PeroK said:Note that the factor of ##\frac 1 3## is an unnecessary complication. Instead, we can look for:
$$\vec B=B\{2, -2, 1\}$$
Why do you include it?CaliforniaRoll88 said:How do you disregard the ##1/3##?
Let's look at two questions. Let A be any point:CaliforniaRoll88 said:How do you disregard the ##1/3##?
It can be done much more easily using the Post #27 and @PeroK's suggestions!CaliforniaRoll88 said:Finally got the answer with this method.
Let ##B## be any point not ##A##:PeroK said:a) What are the coordinates of a point a distance ##c## units from ##A## in the direction of the ##\vec b##?
That's not right at all. The point of my question was that the answers to a) and b) are the same. A direction is a direction.CaliforniaRoll88 said:Let ##B## be any point not ##A##:
Then the coordinates of ##B## away from ##A## by ##c## units is ##B(cA_x,cA_y,cA_z)##
Is this right?
I suggest that you do some 2D problems so that you can draw a diagram. I think you are struggling with 3D geometry because you are struggling with the basic concepts.CaliforniaRoll88 said:Let ##B## be any point not ##A##:
Then the coordinates of ##B## away from ##A## by ##c## units is ##B(cA_x,cA_y,cA_z)##
Is this right?
Hi @CaliforniaRoll88. That's not right. As suggested by @PeroK, maybe it would help you to do a 2D problem first - a diagram is then easy to draw so you can see what's happening. How about trying this:CaliforniaRoll88 said:Let ##B## be any point not ##A##:
Then the coordinates of ##B## away from ##A## by ##c## units is ##B(cA_x,cA_y,cA_z)##
Is this right?
##\vec P=P\left<1,2\right>##Steve4Physics said:Hi @CaliforniaRoll88. That's not right. As suggested by @PeroK, maybe it would help you to do a 2D problem first - a diagram is then easy to draw so you can see what's happening. How about trying this:
"The vector from the origin to point ##P## is given as ##<1, 2>##, and the unit vector directed from the origin towards point ##Q## is ##{\frac 15}{<3,4>}## . If points ##P## and ##Q## are ten units apart, find the coordinates of point ##Q##.
And you might want to re-read Post #27!
Edited.
Looks good!CaliforniaRoll88 said:##\vec P=P\left<1,2\right>##
##\vec Q=Q\frac{1}{5}\left<3,4\right>##
Let ##k=\frac{1}{5}Q##
##\vec Q=k\left<3,4\right>=\left<3k,4k\right>##
##\vec P\cdot\vec Q=1\cdot3k+2\cdot4k=11k##
##\vec R## is the resultant of ##\vec P## and ##\vec Q##
Law of Cosines:
##R^2=P^2+Q^2-2\vec P\cdot\vec Q##
##10^2=5+25k-22k##
Am I headed in the right direction?
A couple of points (pun intended).CaliforniaRoll88 said:##\vec R## is the resultant of ##\vec P## and ##\vec Q##
I missed that you had ##25k## instead of ##25k^2##. You need to find some way of reducing the number of simple algebraic errors. And, you need a way of spotting them yourself.CaliforniaRoll88 said:##10^2=5+25k-22k##
##95=3k##
##k=\frac {95}{3}##
##\vec Q=k\left<3,4\right>=\left<3k,4k\right>=\left<3\frac {95}{3},4\frac {95}{3}\right>=\left<95,126\frac {2}{3}\right>##
Is this right @Steve4Physics, @PeroK?
Edit: I am sure it's wrong.
As a completely separate correction, it’s probably worth noting that the above is wrong. ##\vec P = \left< 1, 2 \right>##. There is no "##P##" on the right-hand side.CaliforniaRoll88 said:##\vec P=P\left<1,2\right>##
Hi @CaliforniaRoll88. Yes, it's wrong!CaliforniaRoll88 said:##10^2=5+25k-22k##
##95=3k##
##k=\frac {95}{3}##
##\vec Q=k\left<3,4\right>=\left<3k,4k\right>=\left<3\frac {95}{3},4\frac {95}{3}\right>=\left<95,126\frac {2}{3}\right>##
Is this right @Steve4Physics, @PeroK?
Edit: I am sure it's wrong.
##d^2=(x_Q-x_P)^2+(y_Q-y_P)^2##Steve4Physics said:You have 2 points P(1,2) and Q(3k, 4k). You require the distance between them to be 10. So what equation can you now write down and solve?
You have repeated a line, but no matter.CaliforniaRoll88 said:##d^2=(x_Q-x_P)^2+(y_Q-y_P)^2##
##100=(3k-1)^2+(4k-2)^2##
##100=(3k-1)^2+(4k-2)^2##
No. You have used b=-1. But b=1 because the minus sign is already taken into account in the term ##a-b##.CaliforniaRoll88 said:##(a-b)^2=a^2-2ab+b^2##
##(3k-1)^2=(3k)^2-2(3k)(-1)+(-1)^2=9k^2+6k+1##
No. Same mistake as described above.CaliforniaRoll88 said:##(4k-2)^2=(4k)^2-2(4k)(-2)+(-2)^2=16k^2+16k+4##
This expression (when corrected) must be equal to 100, This gives you a quadratic equation you can solve to find ##k##. Then you know Q##(3k,4k)## and the problem is done.CaliforniaRoll88 said:##(3k-1)^2+(4k-2)^2=25k^2+22k+5##
You do not need the law of cosines if you use the above method.CaliforniaRoll88 said:@Steve4Physics, I can see that I am using the quadratic formula method. How do I employ the Law of Cosines here?