Vector Projections Qs Help Plz UGRENT (1 Viewer)

csi

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1. Given a=(3,4) and b=(2,1), find the area of the triangle formed by a, b and the projection of b onto a.
ANS: 1 (note: bolded letters are all vectors)

2. Given a=(3,4) and b=(2,1), find the area of the triangle formed by a, b and the projection of a onto b.
ANS: 5

Can someone please explain the difference in approaching questions 1 and 2 please.

3. Points P and A are on the number plane. The vector OA is (3,1). Point B is chosen so that the area of triangle PAB is 10u^2 and |OB| = 4root5. Find all possible vectors for OP.

Thanks!
 
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CM_Tutor

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Can someone please explain the difference in approaching questions 1 and 2 please.
The approach is the same, but the answers are differ as the vector produced by projecting onto is different from that resulting from the projection of onto .

So, for









Question 1: The area of the triangle formed by , , and . We need two vectors that are perpendicular to each other, and



Question 2: The area of the triangle formed by , , and . We need two vectors that are perpendicular to each other, and

 
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csi

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The approach is the same, but the answers are differ as the vector produced by projecting onto is different from that resulting from the projection of onto .

So, for









Question 1: The area of the triangle formed by , , and . We need two vectors that are perpendicular to each other, and



Question 2: The area of the triangle formed by , , and . We need two vectors that are perpendicular to each other, and

thanks!!
 

CM_Tutor

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3. Points P and A are on the number plane. The vector OA is (3,1). Point B is chosen so that the area of triangle PAB is 10u^2 and |OB| = 4root5. Find all possible vectors for OP.

Thanks!
Csi, are you sure that there is not something missing from this question? As far as I can see, unless I've made a mistake in reasoning, there are infinitely many possible vectors .

We know that is a fixed point represented by and that, since then must lie on the circle with centre and radius

Now, consider the simple case where , , and are collinear with located between and . The distance from to is given by



and I can form a triangle of area 10 u2 by locating at a position where the perpendicular distance () from to the line segment is



In other words, can be located anywhere on either of the two line segments that are parallel to and a distance of units from and with one end point of each line segment on the line through that is perpendicular to and the other end point of each line segment on the line through that is perpendicular to .

Or, consider a rectangle with and and being the midpoints of and , respectively. So long as , then with at any point on or , then has base and height and thus an area of



There will also be two line segments parallel to along which can be located for every other possible position of on the circle . The greater is the distance from to then the longer are the two line segments but the shorter is the perpendicular distance between those line segments and .

Unless my reasoning is flawed, there is no reasonable way to determine every possible vector to answer the question, which is why I think there is a problem.
 
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CM_Tutor

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On further reflection, there need not be limitations on the line segments parallel to because the area of the triangle and position of are restricted by the length of and the perpendicular distance from to produced, which restricts to the parallel lines but not to any segment of them.

Another proof of infinite possibilities for . Take as any point outside the circle and join , letting produced meet the circle at and . We know from above that must be greater than whichever is the smaller of and and it follows from above that



and thus that , the perpendicular distance from to for , must satisfy



Since , the radius of the circle, it follows that there must exist at least one point on the line segment (and so within the circle ) where a line perpendicular to at meets the circle at a distance from M, and this is a point to complete the required triangle with the required area.

We thus can show that can be any point satisfying . I think it can also be any point on the circle, and a great many points within the circle. It is not all points as (i) clearly being at is impossible and (ii) if is sufficiently close to (say we set ) then constructing a triangle with the required area mandates that the distance is too large for it to be possible for to lie on the required circle. Determining the complete region within which can lie is an interesting problem, and one I don't know how to solve...
 
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Qeru

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Csi, are you sure that there is not something missing from this question? As far as I can see, unless I've made a mistake in reasoning, there are infinitely many possible vectors .

We know that is a fixed point represented by and that, since then must lie on the circle with centre and radius

Now, consider the simple case where , , and are collinear with located between and . The distance from to is given by



and I can form a triangle of area 10 u2 by locating at a position where the perpendicular distance () from to the line segment is



In other words, can be located anywhere on either of the two line segments that are parallel to and a distance of units from and with one end point of each line segment on the line through that is perpendicular to and the other end point of each line segment on the line through that is perpendicular to .

Or, consider a rectangle with and and being the midpoints of and , respectively. So long as , then with at any point on or , then has base and height and thus an area of



There will also be two line segments parallel to along which can be located for every other possible position of on the circle . The greater is the distance from to then the longer are the two line segments but the shorter is the perpendicular distance between those line segments and .

Unless my reasoning is flawed, there is no reasonable way to determine every possible vector to answer the question, which is why I think there is a problem.
Yeah looks good I dont see any issues with your working. @csi can you send a screenshot of the original question?
 

CM_Tutor

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Yeah looks good I dont see any issues with your working. @csi can you send a screenshot of the original question?
Thanks, Qeru. I was thinking more broadly of the problem of the region within the circle in which can lie.

Can be located anywhere on the plane so long as



where is the largest of the perpendicular distances from to a tangent to the circle that is parallel to ?

If so, what does the region where cannot lie look like?
 

idkkdi

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Yo, is everyone having a vectors test or what?

@CM_Tutor chuck a doc with juicy vector questions.
 

csi

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Csi, are you sure that there is not something missing from this question? As far as I can see, unless I've made a mistake in reasoning, there are infinitely many possible vectors .

We know that is a fixed point represented by and that, since then must lie on the circle with centre and radius

Now, consider the simple case where , , and are collinear with located between and . The distance from to is given by



and I can form a triangle of area 10 u2 by locating at a position where the perpendicular distance () from to the line segment is



In other words, can be located anywhere on either of the two line segments that are parallel to and a distance of units from and with one end point of each line segment on the line through that is perpendicular to and the other end point of each line segment on the line through that is perpendicular to .

Or, consider a rectangle with and and being the midpoints of and , respectively. So long as , then with at any point on or , then has base and height and thus an area of



There will also be two line segments parallel to along which can be located for every other possible position of on the circle . The greater is the distance from to then the longer are the two line segments but the shorter is the perpendicular distance between those line segments and .

Unless my reasoning is flawed, there is no reasonable way to determine every possible vector to answer the question, which is why I think there is a problem.
I’ve written exactly what the question says, please check out thread 9 for a photo of the original question if that would help. Thanks!
 

CM_Tutor

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I’ve written exactly what the question says, please check out thread 9 for a photo of the original question if that would help. Thanks!
Thanks. I thought that there might be more, but I see there isn't. So, either I've made a mistake in interpretation or the question is not what whoever wrote it intended. Do you have an answer or solution?
 

csi

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Thanks. I thought that there might be more, but I see there isn't. So, either I've made a mistake in interpretation or the question is not what whoever wrote it intended. Do you have an answer or solution?
Unfortunately no, sorry about that.
 

CM_Tutor

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Unfortunately no, sorry about that.
That's ok. it's what I guessed. It will be interesting if you get any further info because I don't see how this is a reasonable question for MX2, let alone MX1.
 

idkkdi

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That's ok. it's what I guessed. It will be interesting if you get any further info because I don't see how this is a reasonable question for MX2, let alone MX1.
anything inside the region bounded by OB, except the point A looks like a possibility for P.

no idea how to prove tho haha.
 

Qeru

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Another Idea is to maybe find the vector projection which allows u to find the perpendicular and find the area of the triangle that way. Not bothered to Latex it tho.
 

CM_Tutor

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anything inside the region bounded by OB, except the point A looks like a possibility for P.

no idea how to prove tho haha.
Take a point such that . It follows that must be located on a line such that the perpendicular distance from to produced must be 200. There is no point on the circle that satisfies this requirement and so cannot be located at any position where . In fact, there must be a circle of exclusion around , though I don't believe that is the entire prohibited region.
 

CM_Tutor

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Modified question

Suppose was located at the origin, is at , and is restricted to . The lines are parallel to and each meets the circle at two points if , at exactly one point (either or ) if , and does not meet the circle for . In the first two cases, with being any of the points of intersection of the lines and the circle, has a base of and a perpendicular height of , and its area can be set to 10 u2:



So, for (Case 1), we get and so if is located at , then can be located at or

If (Case 2), then the possible positions of are at and so



and since



So, for , we get and so if is located at , then can be located at or . Note that nothing here requires and so can lie on or outside the circle and valid positions for will be produced.

The remaining possibility (Case 3) occuurs if , in which case the possible positions of are the solutions of :



In other words, there is no possible position for and thus the required triangle cannot exist under these circumstances.

Now, these calculations have considered lying on the radius from to and has shown that must be located units or more from . Since any radius is identical to every other radius under rotation, the same result occurs for any other location of .

So, if can be located anywhere on our outside the circle and a corresponding position for can be found.

---

This simplification establishes that can be anywhere except for a circular region around for the case where the centre of the circle on which must lie is fixed at . The radius of this region, , is determined by the radius of the circle onto which is constrained, , and the area of the required triangle, , by the equation .

I am convinced that, for the original question with located within the circle on which must lie, will also be permitted anywhere except in an exclusion zone around . However, with not at , the exclusion zone will not be circular, though it will have reflection symmetry across produced.

The situations where is a fixed point on the same circle as , or if is located outside that circle, have not been considered.
 
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