Vector Projections Qs Help Plz UGRENT (1 Viewer)

csi · Feb 24, 2021

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!

CM_Tutor · Feb 25, 2021

csi said:
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 $\overrightarrow{a}$ onto $\overrightarrow{b}$ is different from that resulting from the projection of $\overrightarrow{b}$ onto $\overrightarrow{a}$ .

So, for

$\overrightarrow{a} = \begin{pmatrix}3\\4\end{pmatrix} \qquad \text{and} \overrightarrow{b} = \begin{pmatrix}2\\1\end{pmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b} = \begin{pmatrix}3\\4\end{pmatrix}\cdot\begin{pmatrix}2\\1\end{pmatrix} = 3 \times 2 + 4 \times 1 = 10$

$\begin{align*} \text{proj}_{\overrightarrow{b}}\overrightarrow{a} &= \frac{\overrightarrow{a}\cdot\overrightarrow{b}}{\left|\overrightarrow{b}\right|^2}\overrightarrow{b} \\ &= \frac{10}{2^2 + 1^2}\begin{pmatrix}2\\1\end{pmatrix} \\ &= 2\begin{pmatrix}2\\1\end{pmatrix} \\ &= \begin{pmatrix}4\\2\end{pmatrix} \end{align*}$

$\begin{align*} \text{proj}_{\overrightarrow{a}}\overrightarrow{b} &= \frac{\overrightarrow{b}\cdot\overrightarrow{a}}{\left|\overrightarrow{a}\right|^2}\overrightarrow{a} \\ &= \frac{10}{3^2 + 4^2}\begin{pmatrix}3\\4\end{pmatrix} \qquad \text{since $ \overrightarrow{a}\cdot\overrightarrow{b} = \overrightarrow{b}\cdot\overrightarrow{a}$} \\ &= \frac{2}{5}\begin{pmatrix}3\\4\end{pmatrix} \\ &= \begin{pmatrix}\cfrac{6}{5}\\\cfrac{8}{5}\end{pmatrix} \end{align*}$

Question 1: The area of the triangle formed by $\overrightarrow{a}$ , $\overrightarrow{b}$ , and $\text{proj}_{\overrightarrow{a}}\overrightarrow{b}$ . We need two vectors that are perpendicular to each other, $\text{proj}_{\overrightarrow{a}}\overrightarrow{b}$ and $\overrightarrow{b} - \text{proj}_{\overrightarrow{a}}\overrightarrow{b}$

$\begin{align*} \text{Area } &= \frac{1}{2}\left|\text{proj}_{\overrightarrow{a}}\overrightarrow{b}\right| \left|\overrightarrow{b}-\text{proj}_{\overrightarrow{a}}\overrightarrow{b}\right| \\ &= \frac{1}{2}\left|\begin{pmatrix}\cfrac{6}{5}\\\cfrac{8}{5}\end{pmatrix}\right|\left|\begin{pmatrix}2\\1\end{pmatrix}-\begin{pmatrix}\cfrac{6}{5}\\\cfrac{8}{5}\end{pmatrix}\right| \\ &= \frac{1}{2}\sqrt{\left(\cfrac{6}{5}\right)^2+\left(\cfrac{8}{5}\right)^2}\left|\begin{pmatrix}\cfrac{4}{5}\\-\cfrac{3}{5}\end{pmatrix}\right| \\ &= \frac{1}{2}\sqrt{\cfrac{36+64}{25}}\sqrt{\left(\cfrac{4}{5}\right)^2+\left(-\cfrac{3}{5}\right)^2} \\ &= \frac{1}{2}\times \sqrt{4} \times \sqrt{\cfrac{16+9}{25}} \\ &= \frac{2}{2}\times \sqrt{1} \\&= 1 \end{align*}$

Question 2: The area of the triangle formed by $\overrightarrow{a}$ , $\overrightarrow{b}$ , and $\text{proj}_{\overrightarrow{b}}\overrightarrow{a}$ . We need two vectors that are perpendicular to each other, $\text{proj}_{\overrightarrow{b}}\overrightarrow{a}$ and $\overrightarrow{a} - \text{proj}_{\overrightarrow{b}}\overrightarrow{a}$

$\begin{align*} \text{Area } &= \frac{1}{2}\left|\text{proj}_{\overrightarrow{b}}\overrightarrow{a}\right| \left|\overrightarrow{a}-\text{proj}_{\overrightarrow{b}}\overrightarrow{a}\right| \\ &= \frac{1}{2}\left|\begin{pmatrix}4\\2\end{pmatrix}\right|\left|\begin{pmatrix}3\\4\end{pmatrix}-\begin{pmatrix}4\\2\end{pmatrix}\right| \\ &= \frac{1}{2}\sqrt{4^2+2^2}\left|\begin{pmatrix}-1\\2\end{pmatrix}\right| \\ &= \frac{1}{2}\sqrt{20} \times \sqrt{\left(-1\right)^2+2^2} \\ &= \frac{1}{2}\times 2\sqrt{5} \times \sqrt{5} \\ &= \frac{2\left(\sqrt{5}\right)^2}{2} \\&= 5 \end{align*}$

csi · Feb 25, 2021

CM_Tutor said:
The approach is the same, but the answers are differ as the vector produced by projecting $\overrightarrow{a}$ onto $\overrightarrow{b}$ is different from that resulting from the projection of $\overrightarrow{b}$ onto $\overrightarrow{a}$ .

So, for

$\overrightarrow{a} = \begin{pmatrix}3\\4\end{pmatrix} \qquad \text{and} \overrightarrow{b} = \begin{pmatrix}2\\1\end{pmatrix}$

$\overrightarrow{a}\cdot\overrightarrow{b} = \begin{pmatrix}3\\4\end{pmatrix}\cdot\begin{pmatrix}2\\1\end{pmatrix} = 3 \times 2 + 4 \times 1 = 10$

$\begin{align*} \text{proj}_{\overrightarrow{b}}\overrightarrow{a} &= \frac{\overrightarrow{a}\cdot\overrightarrow{b}}{\left|\overrightarrow{b}\right|^2}\overrightarrow{b} \\ &= \frac{10}{2^2 + 1^2}\begin{pmatrix}2\\1\end{pmatrix} \\ &= 2\begin{pmatrix}2\\1\end{pmatrix} \\ &= \begin{pmatrix}4\\2\end{pmatrix} \end{align*}$

$\begin{align*} \text{proj}_{\overrightarrow{a}}\overrightarrow{b} &= \frac{\overrightarrow{b}\cdot\overrightarrow{a}}{\left|\overrightarrow{a}\right|^2}\overrightarrow{a} \\ &= \frac{10}{3^2 + 4^2}\begin{pmatrix}3\\4\end{pmatrix} \qquad \text{since $ \overrightarrow{a}\cdot\overrightarrow{b} = \overrightarrow{b}\cdot\overrightarrow{a}$} \\ &= \frac{2}{5}\begin{pmatrix}3\\4\end{pmatrix} \\ &= \begin{pmatrix}\cfrac{6}{5}\\\cfrac{8}{5}\end{pmatrix} \end{align*}$

Question 1: The area of the triangle formed by $\overrightarrow{a}$ , $\overrightarrow{b}$ , and $\text{proj}_{\overrightarrow{a}}\overrightarrow{b}$ . We need two vectors that are perpendicular to each other, $\text{proj}_{\overrightarrow{a}}\overrightarrow{b}$ and $\overrightarrow{b} - \text{proj}_{\overrightarrow{a}}\overrightarrow{b}$

$\begin{align*} \text{Area } &= \frac{1}{2}\left|\text{proj}_{\overrightarrow{a}}\overrightarrow{b}\right| \left|\overrightarrow{b}-\text{proj}_{\overrightarrow{a}}\overrightarrow{b}\right| \\ &= \frac{1}{2}\left|\begin{pmatrix}\cfrac{6}{5}\\\cfrac{8}{5}\end{pmatrix}\right|\left|\begin{pmatrix}2\\1\end{pmatrix}-\begin{pmatrix}\cfrac{6}{5}\\\cfrac{8}{5}\end{pmatrix}\right| \\ &= \frac{1}{2}\sqrt{\left(\cfrac{6}{5}\right)^2+\left(\cfrac{8}{5}\right)^2}\left|\begin{pmatrix}\cfrac{4}{5}\\-\cfrac{3}{5}\end{pmatrix}\right| \\ &= \frac{1}{2}\sqrt{\cfrac{36+64}{25}}\sqrt{\left(\cfrac{4}{5}\right)^2+\left(-\cfrac{3}{5}\right)^2} \\ &= \frac{1}{2}\times \sqrt{4} \times \sqrt{\cfrac{16+9}{25}} \\ &= \frac{2}{2}\times \sqrt{1} \\&= 1 \end{align*}$

Question 2: The area of the triangle formed by $\overrightarrow{a}$ , $\overrightarrow{b}$ , and $\text{proj}_{\overrightarrow{b}}\overrightarrow{a}$ . We need two vectors that are perpendicular to each other, $\text{proj}_{\overrightarrow{b}}\overrightarrow{a}$ and $\overrightarrow{a} - \text{proj}_{\overrightarrow{b}}\overrightarrow{a}$

$\begin{align*} \text{Area } &= \frac{1}{2}\left|\text{proj}_{\overrightarrow{b}}\overrightarrow{a}\right| \left|\overrightarrow{a}-\text{proj}_{\overrightarrow{b}}\overrightarrow{a}\right| \\ &= \frac{1}{2}\left|\begin{pmatrix}4\\2\end{pmatrix}\right|\left|\begin{pmatrix}3\\4\end{pmatrix}-\begin{pmatrix}4\\2\end{pmatrix}\right| \\ &= \frac{1}{2}\sqrt{4^2+2^2}\left|\begin{pmatrix}-1\\2\end{pmatrix}\right| \\ &= \frac{1}{2}\sqrt{20} \times \sqrt{\left(-1\right)^2+2^2} \\ &= \frac{1}{2}\times 2\sqrt{5} \times \sqrt{5} \\ &= \frac{2\left(\sqrt{5}\right)^2}{2} \\&= 5 \end{align*}$

thanks!!

CM_Tutor · Feb 25, 2021

csi said:
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 $\overrightarrow{OP}$ .

We know that $A$ is a fixed point represented by $3\overrightarrow{i}+\overrightarrow{j}$ and that, since $\left|\overrightarrow{OB}\right| = 4\sqrt{5}$ then $B$ must lie on the circle with centre $O$ and radius $4\sqrt{5}$

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

$d_{AB} = \left|\overrightarrow{AB}\right| = \left|\overrightarrow{OB}\right| - \left|\overrightarrow{OA}\right| = 4\sqrt{5} - \sqrt{10} = \sqrt{5}\left(4 - \sqrt{2}\right)$

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

$\begin{align*} h &= \frac{2A_{\Delta}}{d_{AB}} \\ &= \frac{20}{\sqrt{5}\left(4 - \sqrt{2}\right)} \times \frac{\sqrt{5}}{\sqrt{5}} \\ &= \frac{4\sqrt{5}}{4 - \sqrt{2}} \times \frac{4 + \sqrt{2}}{4 + \sqrt{2}} \\ &= \frac{4\sqrt{5}\left(4 + \sqrt{2}\right)}{4^2 - \left(\sqrt{2}\right)^2} \\ &= \frac{4\sqrt{5}\left(4 + \sqrt{2}\right)}{14} \\ &= \frac{2\sqrt{5}\left(4 + \sqrt{2}\right)}{7} \end{align*}$

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

Or, consider a rectangle $MXYN$ with $MN || AB || XY$ and $A$ and $B$ being the midpoints of $MX$ and $NY$ , respectively. So long as $MA = AX = NB = BY = h$ , then with $P$ at any point on $MN$ or $XY$ , then $\Delta PAB$ has base $AB$ and height $h$ and thus an area of

$\begin{align*} \text{Area} &= \frac{1}{2} \times d_{AB} \times h \\ &= \frac{1}{2} \times \sqrt{5}\left(4 - \sqrt{2}\right) \times \frac{2\sqrt{5}\left(4 + \sqrt{2}\right)}{7} \\ &= \frac{2\left(\sqrt{5}\right)^2\left(4^2-\left(\sqrt{2}\right)^2\right)}{2 \times 7} \\ &= \frac{2 \times 5 (16-2)}{14} \\ &= 10 \text{ u}^2 \end{align*}$

There will also be two line segments parallel to $AB$ along which $P$ can be located for every other possible position of $B$ on the circle $x^2+y^2=80$ . The greater is the distance from $A$ to $B$ then the longer are the two line segments but the shorter is the perpendicular distance between those line segments and $AB$ .

Unless my reasoning is flawed, there is no reasonable way to determine every possible vector $\overrightarrow{OP}$ to answer the question, which is why I think there is a problem.

CM_Tutor · Feb 25, 2021

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

Another proof of infinite possibilities for $P$ . Take $P$ as any point outside the circle $x^2+y^2=80$ and join $AP$ , letting $AP$ produced meet the circle at $X$ and $Y$ . We know from above that $AP$ must be greater than whichever is the smaller of $AX$ and $AY$ and it follows from above that

$AP \geqslant \sqrt{5}\left(4-\sqrt{2}\right) > 5.7819$

and thus that $h$ , the perpendicular distance from $B$ to $AP$ for $\Delta PAB$ , must satisfy

$h \leqslant \frac{2\sqrt{5}\left(4+\sqrt{2}\right)}{7} < 3.46$

Since $h < 4\sqrt{5} \approx 8.944...$ , the radius of the circle, it follows that there must exist at least one point $M$ on the line segment $XY$ (and so within the circle $x^2+y^2=80$ ) where a line perpendicular to $XY$ at $M$ meets the circle at a distance $h$ from M, and this is a point $B$ to complete the required triangle with the required area.

We thus can show that $P$ can be any point satisfying $x^2 + y^2 > 80$ . 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 $P$ being at $A$ is impossible and (ii) if $P$ is sufficiently close to $A$ (say we set $\left|\overrightarrow{AP}\right|=0.1$ ) then constructing a triangle with the required area mandates that the distance $AB$ is too large for it to be possible for $B$ to lie on the required circle. Determining the complete region within which $P$ can lie is an interesting problem, and one I don't know how to solve...

Qeru · Feb 25, 2021

CM_Tutor said:
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 $\overrightarrow{OP}$ .

We know that $A$ is a fixed point represented by $3\overrightarrow{i}+\overrightarrow{j}$ and that, since $\left|\overrightarrow{OB}\right| = 4\sqrt{5}$ then $B$ must lie on the circle with centre $O$ and radius $4\sqrt{5}$

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

$d_{AB} = \left|\overrightarrow{AB}\right| = \left|\overrightarrow{OB}\right| - \left|\overrightarrow{OA}\right| = 4\sqrt{5} - \sqrt{10} = \sqrt{5}\left(4 - \sqrt{2}\right)$

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

$\begin{align*} h &= \frac{2A_{\Delta}}{d_{AB}} \\ &= \frac{20}{\sqrt{5}\left(4 - \sqrt{2}\right)} \times \frac{\sqrt{5}}{\sqrt{5}} \\ &= \frac{4\sqrt{5}}{4 - \sqrt{2}} \times \frac{4 + \sqrt{2}}{4 + \sqrt{2}} \\ &= \frac{4\sqrt{5}\left(4 + \sqrt{2}\right)}{4^2 - \left(\sqrt{2}\right)^2} \\ &= \frac{4\sqrt{5}\left(4 + \sqrt{2}\right)}{14} \\ &= \frac{2\sqrt{5}\left(4 + \sqrt{2}\right)}{7} \end{align*}$

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

Or, consider a rectangle $MXYN$ with $MN || AB || XY$ and $A$ and $B$ being the midpoints of $MX$ and $NY$ , respectively. So long as $MA = AX = NB = BY = h$ , then with $P$ at any point on $MN$ or $XY$ , then $\Delta PAB$ has base $AB$ and height $h$ and thus an area of

$\begin{align*} \text{Area} &= \frac{1}{2} \times d_{AB} \times h \\ &= \frac{1}{2} \times \sqrt{5}\left(4 - \sqrt{2}\right) \times \frac{2\sqrt{5}\left(4 + \sqrt{2}\right)}{7} \\ &= \frac{2\left(\sqrt{5}\right)^2\left(4^2-\left(\sqrt{2}\right)^2\right)}{2 \times 7} \\ &= \frac{2 \times 5 (16-2)}{14} \\ &= 10 \text{ u}^2 \end{align*}$

There will also be two line segments parallel to $AB$ along which $P$ can be located for every other possible position of $B$ on the circle $x^2+y^2=80$ . The greater is the distance from $A$ to $B$ then the longer are the two line segments but the shorter is the perpendicular distance between those line segments and $AB$ .

Unless my reasoning is flawed, there is no reasonable way to determine every possible vector $\overrightarrow{OP}$ 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 · Feb 25, 2021

Qeru said:
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 $P$ can lie.

Can $P$ be located anywhere on the plane so long as

$\frac{20}{\left|\overrightarrow{AP}\right|} < d_{\text{MAX}}$

where $d_{\text{MAX}}$ is the largest of the perpendicular distances from $A$ to a tangent to the circle that is parallel to $AP$ ?

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

idkkdi · Feb 25, 2021

Yo, is everyone having a vectors test or what?

@CM_Tutor chuck a doc with juicy vector questions.

csi · Feb 25, 2021

Qeru said:
Yeah looks good I dont see any issues with your working. @csi can you send a screenshot of the original question?

csi · Feb 25, 2021

CM_Tutor said:
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 $\overrightarrow{OP}$ .

We know that $A$ is a fixed point represented by $3\overrightarrow{i}+\overrightarrow{j}$ and that, since $\left|\overrightarrow{OB}\right| = 4\sqrt{5}$ then $B$ must lie on the circle with centre $O$ and radius $4\sqrt{5}$

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

$d_{AB} = \left|\overrightarrow{AB}\right| = \left|\overrightarrow{OB}\right| - \left|\overrightarrow{OA}\right| = 4\sqrt{5} - \sqrt{10} = \sqrt{5}\left(4 - \sqrt{2}\right)$

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

$\begin{align*} h &= \frac{2A_{\Delta}}{d_{AB}} \\ &= \frac{20}{\sqrt{5}\left(4 - \sqrt{2}\right)} \times \frac{\sqrt{5}}{\sqrt{5}} \\ &= \frac{4\sqrt{5}}{4 - \sqrt{2}} \times \frac{4 + \sqrt{2}}{4 + \sqrt{2}} \\ &= \frac{4\sqrt{5}\left(4 + \sqrt{2}\right)}{4^2 - \left(\sqrt{2}\right)^2} \\ &= \frac{4\sqrt{5}\left(4 + \sqrt{2}\right)}{14} \\ &= \frac{2\sqrt{5}\left(4 + \sqrt{2}\right)}{7} \end{align*}$

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

Or, consider a rectangle $MXYN$ with $MN || AB || XY$ and $A$ and $B$ being the midpoints of $MX$ and $NY$ , respectively. So long as $MA = AX = NB = BY = h$ , then with $P$ at any point on $MN$ or $XY$ , then $\Delta PAB$ has base $AB$ and height $h$ and thus an area of

$\begin{align*} \text{Area} &= \frac{1}{2} \times d_{AB} \times h \\ &= \frac{1}{2} \times \sqrt{5}\left(4 - \sqrt{2}\right) \times \frac{2\sqrt{5}\left(4 + \sqrt{2}\right)}{7} \\ &= \frac{2\left(\sqrt{5}\right)^2\left(4^2-\left(\sqrt{2}\right)^2\right)}{2 \times 7} \\ &= \frac{2 \times 5 (16-2)}{14} \\ &= 10 \text{ u}^2 \end{align*}$

There will also be two line segments parallel to $AB$ along which $P$ can be located for every other possible position of $B$ on the circle $x^2+y^2=80$ . The greater is the distance from $A$ to $B$ then the longer are the two line segments but the shorter is the perpendicular distance between those line segments and $AB$ .

Unless my reasoning is flawed, there is no reasonable way to determine every possible vector $\overrightarrow{OP}$ 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 · Feb 25, 2021

csi said:
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 · Feb 25, 2021

CM_Tutor said:
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 · Feb 25, 2021

csi said:
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 · Feb 25, 2021

CM_Tutor said:
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 · Feb 25, 2021

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 · Feb 25, 2021

idkkdi said:
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 $P$ such that $\left|\overrightarrow{AP}\right|=0.1$ . It follows that $B$ must be located on a line such that the perpendicular distance from $B$ to $AP$ produced must be 200. There is no point on the circle $x^2+y^2=80$ that satisfies this requirement and so $P$ cannot be located at any position where $\left|\overrightarrow{AP}\right|=0.1$ . In fact, there must be a circle of exclusion around $A$ , though I don't believe that is the entire prohibited region.

CM_Tutor · Feb 26, 2021

Modified question

Suppose $A$ was located at the origin, $P$ is at $(d, 0)$ , and $B$ is restricted to $x^2 + y^2 =80$ . The lines $y=\pm h, h>0$ are parallel to $AP$ and each meets the circle at two points if $h<4\sqrt{5}$ , at exactly one point (either $\left(0, 4\sqrt{5}\right)$ or $\left(0, -4\sqrt{5}\right)$ ) if $h=4\sqrt{5}$ , and does not meet the circle for $h>4\sqrt{5}$ . In the first two cases, with $B$ being any of the points of intersection of the lines and the circle, $\Delta PAB$ has a base of $d$ and a perpendicular height of $h$ , and its area can be set to 10 u²:

$\begin{align*} A_{\Delta PAB} = \frac{1}{2}dh &= 10 \\ d&=\frac{20}{h} \\ \text{If $h=4\sqrt{5}$:} \qquad d&=\frac{20}{4\sqrt{5}} \\ &= \frac{5}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} \\ &= \frac{5\sqrt{5}}{5} \\ &= \sqrt{5} \end{align*}$

So, for $h=4\sqrt{5}$ (Case 1), we get $d=\sqrt{5}$ and so if $P$ is located at $\left(\sqrt{5}, 0\right)$ , then $B$ can be located at $\left(0, 4\sqrt{5}\right)$ or $\left(0, -4\sqrt{5}\right)$

If $h<4\sqrt{5}$ (Case 2), then the possible positions of $B$ are at $y=\pm h$ and so

$\begin{align*} x^2 + \left(\pm h\right)^2 &= 80 \\ x^2 &= 80 - h^2 > 0 \text{ as $h^2<80$} \\ x &= \pm\sqrt{80-h^2} \end{align*}$

and since

$\begin{align*} h &< 4\sqrt{5} \\ \frac{20}{d} &< 4\sqrt{5} \qquad \text{as $dh = 20$} \\ \frac{d}{20} &> \frac{1}{4\sqrt{5}} \qquad \text{as $d>0$} \\ d &>\frac{20}{4\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} \\ &> \frac{5\sqrt{5}}{\left(\sqrt{5}\right)^2} \\ &> \sqrt{5} \end{align*}$

So, for $h<4\sqrt{5}$ , we get $d>\sqrt{5}$ and so if $P$ is located at $\left(d, 0\right)$ , then $B$ can be located at $\left(\pm\sqrt{80-h^2}, h\right)$ or $\left(\pm\sqrt{80-h^2}, -h\right)$ . Note that nothing here requires $d<4\sqrt{5}$ and so $P$ can lie on or outside the circle $x^2+y^2 = 80$ and valid positions for $B$ will be produced.

The remaining possibility (Case 3) occuurs if $h>4\sqrt{5}$ , in which case the possible positions of $B$ are the solutions of $y=\pm h$ :

$\begin{align*} x^2 + \left(\pm h\right)^2 &= 80 \\ x^2 &= 80 - h^2 < 0 \text{ as $h^2>80$} \\ \text{So, } x &\notin \mathbb{R} \end{align*}$

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

Now, these calculations have considered $P$ lying on the radius from $(0, 0)$ to $\left(4\sqrt{5}, 0\right)$ and has shown that $P$ must be located $\sqrt{5}$ units or more from $O$ . Since any radius is identical to every other radius under rotation, the same result occurs for any other location of $P$ .

So, if $P$ can be located anywhere on our outside the circle $x^2+y^2=5$ and a corresponding position for $B$ can be found.

---

This simplification establishes that $P$ can be anywhere except for a circular region around $A$ for the case where the centre of the circle on which $B$ must lie is fixed at $A$ . The radius of this region, $r$ , is determined by the radius of the circle onto which $B$ is constrained, $R$ , and the area of the required triangle, $A_{\Delta}$ , by the equation $rR=2A_{\Delta}$ .

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

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

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