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Vectors

Hi guys,

Finding it a little troublesome doing these three questions, I'm sure the first one's relatively straightforward but I seem to be having a bit of trouble :s-smilie:. Can anyone give me a couple hints? Cheers.

1) Simplify a×ba.b\frac{\vec{a} \times \vec{b}}{\vec a . \vec b}

2) Show that the foot of the perpendicular from the point p\vec{p} to the line r=a+λu\vec{r} = \vec{a} + \lambda{\vec u} has position vector:

a+u.(pa)u2u\vec{a} + \frac{\vec{u}.(\vec{p} - \vec{a})}{|\vec{u}|^2} \vec{u}

3) A high energy particle beam is fired at a sphere of matter. The beam is initiated from point a\vec{a} in a direction d\vec{d} at a sphere, centred on the origin, with radius pp. What must the radius of the sphere be in terms of a\vec{a} and d\vec{d} in order for there to be a collision with the energy beam?
(edited 9 years ago)
Original post by r3l3ntl3ss

1) Simplify a×ba.b\frac{\vec{a} \times \vec{b}}{\vec a . \vec b}



Looking at the first one, I'm not sure what they're trying to achieve. If we go to the definitions of the dot and cross products this reduces to (tanθ)n(\tan\theta) \vec{n}.

But the question arises as to how to specify n\vec{n} and tanθ\tan\theta and the simplest method as far as I can see is with the original formula.


2) Show that the foot...


Hint: What do you know about perpendicular vectors?


3) A high energy particle...


Diagram to start.
Question comes down to:

Spoiler

(edited 9 years ago)
Original post by r3l3ntl3ss


2) Show that the foot of the perpendicular from the point p\vec{p} to the line r=a+λu\vec{r} = \vec{a} + \lambda{\vec u} has position vector:

a+u.(pa)u2u\vec{a} + \frac{\vec{u}.(\vec{p} - \vec{a})}{|\vec{u}|^2} \vec{u}


1. Sketch the line r=a+λu\vec{r} = \vec{a} + \lambda{\vec u}, showing vectors a,u\vec{a}, \vec{u}

2. Draw a position vector p\vec{p} somewhere in the plane.

3. Draw the perpendicular vector s\vec{s}, say, from p\vec{p} to the line. Call the point where they meet F. Note that the position vector of F is rF=a+λpu\vec{r}_F = \vec{a}+\lambda_p \vec{u} for some scalar λp\lambda_p which depends on p\vec{p}.

Your job is to find an expression for λp\lambda_p in terms of quantities you already know.

4. Let the vector along the line from a\vec{a} to F be t\vec{t}. Write down an equation relating s,a,p,t\vec{s}, \vec{a}, \vec{p}, \vec{t}. (Hint: draw a suitable triangle, one of whose sides is t\vec{t}).

5. Note that we can write t=λpu\vec{t} = \lambda_p \vec{u}

6. Write down in vector notation the condition that s\vec{s} is perpendicular to the line.

This is enough info to find an expression for λp\lambda_p
Reply 3
Original post by atsruser
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Original post by ghostwalker
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Thanks for the help guys! Yeah the first question was a bit odd, I've managed to do the second question, and now attempting the third :smile:

Repped!
Reply 4
I'm still quite confused about question 3 :/
Original post by r3l3ntl3ss
I'm still quite confused about question 3 :/


Using the result of question 2, what's the position vector for the foot of the perpendicular from the origin to the line?

Then just find its length. This will be the minimum radius to intersect the trajectory. I.e. the path will be a tangent to the sphere, the perpendicular being a radius.
Reply 6
Original post by ghostwalker
Using the result of question 2, what's the position vector for the foot of the perpendicular from the origin to the line?

Then just find its length. This will be the minimum radius to intersect the trajectory. I.e. the path will be a tangent to the sphere, the perpendicular being a radius.


Sorry I'm just really baffled by this question, it seems so straightforward but I can't put my mind to it :frown:

Am I right in saying the position vector is aa.dd2d\vec{a} - \frac{\vec{a}.\vec{d}}{|\vec{d}|^2}\vec{d}?
(edited 9 years ago)
Original post by r3l3ntl3ss
Sorry I'm just really baffled by this question, it seems so straightforward but I can't put my mind to it :frown:

Am I right in saying the position vector is aa.dd2d\vec{a} - \frac{\vec{a}.\vec{d}}{|\vec{d}|^2}\vec{d}?


Looks good.

If you've not done so already, do draw a diagram, and make the trajectory a tangent to the sphere.
Reply 8
Original post by ghostwalker
Looks good.

If you've not done so already, do draw a diagram, and make the trajectory a tangent to the sphere.


ah, so all it is, is

p=aa.dd2dp = |\vec{a} - \frac{\vec{a}.\vec{d}}{|\vec{d}|^2}\vec{d}|?
Original post by r3l3ntl3ss
ah, so all it is, is

p=aa.dd2dp = |\vec{a} - \frac{\vec{a}.\vec{d}}{|\vec{d}|^2}\vec{d}|?


Yep.

If you're allowed to have a unit vector in the direction of d then you can simplify that second term a bit, otherwise I'd leave it as that.
(edited 9 years ago)

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