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There is a problem here in Q. $4$ on the last page.

It states to find coordinates of vertex $p$ in a trapezoid with four vertices: A$=(2,6,-3,9)$, B$=(4,4,7,3)$, C $=(8,2,7,-1)$, p $=(x, y, z, w)$.
Image Also, let the angles be named correspondingly; so $\angle pAB = \angle A, \angle ABC = \angle B, \angle BCA = \angle C, \angle CpA = \angle p$.

As the sum of all internal angles in a convex polygon is $360^o$, so on taking angle $\angle A = \theta$, $\angle B = 240^o - \theta$. But seems that no use can be made of the same, as some information regarding tilt of polygon is missing. Please suggest.

Use can be made of two parallel lines (here, BA, Cp) in trapezoid.

Direction vector of line BA $=<-2, 2, -10, 6>$, & the parametric form for any point on the line is $r(t) = (4,4,7,3) + t(-2, 2, -10, 6) = (4-2t, 4+2t, 7-10t, 3 +6t), t\in \mathbb{R}$; as line is assumed to be extend unlimited in both directions; with $t=1$ yielding vertex A.

So, parametric form for any point on the line Cp is $r_1(t) = (8,2,7,-1) + s(-2, 2, -10, 6) = (8-2s, 2+2s, 7-10s, 3 +6s), s\in \mathbb{R}$; with $s=0$ yielding vertex C.
Similarly, $s=1$ should give the vertex p; i.e. $(8-2s, 2+2s, 7-10s, 3 +6s)\implies (6, 4, -3, 9)$.

Need to verify the above, by taking additional equations.

Say, to find intersection of the lines Ap, Cp; with the direction vector of the line Cp $= <x-8, y-2, z-7, w +1>$, & of line Ap $=<x-2, y-6, z+3, w-9>$.

But, it is not working. Need help.


Edit : comment to the answer by @Dhanvi Sreenivasan :

coordinates of vertex $C=(8,2,7,-1)$;
dv of $\vec{BA} = (-2, 2, -10, 6)$;
dv of $\vec{Cp} =$ dv of $\vec{BA}$;
parametric coordinates of vertex $p = (8-2s, 2+2s, 7-10s, -1+6s)$;
coordinates of vertex $A = (2, 6 , -3, 9)$;
parametric coordinates of $\vec{pA} =(-6+2s, 4-2s, -10+10s, 10-6s)$;
parametric coordinates of $\vec{Cp} = \vec{BA}$;

$\vec{pA}.\vec{Cp}= (-6+2s, 4-2s, -10+10s, 10-6s).(-2s, 2s, -10s, 6s)$$= (12s -4s^2)+(8s-4s^2)+(100s-100s^2)+(60s - 36s^2)$
$=180s -144s^2=(12s)(3)(5 -4s)=4s(9)(5 -4s)= 4s(45-36s)$

Similarly, $ |\vec{pA}|= \sqrt{(-6+2s, 4-2s, -10+10s, 10-6s).(-6+2s, 4-2s, -10+10s, 10-6s)}$
$= \sqrt{((36-24s+4s^2)+(16-16s+4s^2)+(100-200s+100s^2)+(100+36s^2-120s))}$
$= \sqrt{144s^2-360s+252}$$= 3.2\sqrt{4s^2 -10s +7}$.

$|\vec{Cp}|= \sqrt{(-2s, 2s, -10s, 6s).(-2s, 2s, -10s, 6s)} $$= \sqrt{(4s^2+4s^2+100s^2+36s^2)}$$= \sqrt{144s^2}$$=12s=2s.6$

$|\vec{pA}|.|\vec{Cp}|=(3.2\sqrt{4s^2 -10s +7})(2s.6)= 4s.18\sqrt{4s^2 -10s +7}= 4s.6\sqrt{36s^2 -90s +63}$

The division $\frac{\vec{pA}.\vec{Cp}}{|\vec{pA}|.|\vec{Cp}|}= \frac{5 - 4s}{ 2\sqrt{4s^2 -10s +7}}$ which is not $=\cos 60^{o} = 0.5$


Edit 2: The selected answer has given how to get $p=(5, 5, -8, 8) $ from here.


Edit 3: Excellent post made for the same problem.

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Drop perpendiculars of $B$ onto $Cp$, call it $B'$, similarly drop perpendicular of $A$ onto $Cp$, call it $A'$.

$\triangle CBB'$ is congruent to $\triangle PAA'$ and we have $|CB'|=|A'P|$.

$$|CP| = |AB|+2 |CB'|=|AB|+2 |BC|\cos 60^\circ=|AB|+|BC|$$

\begin{align}\vec{OP}&=\vec{OC}+|\vec{CP}|\frac{\vec{BA}}{|\vec{BA}|}\\&=(8,2,7,-1) +\left(1+\frac{|\vec{BC}|}{|\vec{BA}|} \right)\vec{BA}\\ &=(8,2,7,-1) + \left( 1+\frac{6}{12}\right)(-2,2,-10,6) \\ &=(8,2,7,-1) + 3(-1,1,-5,3)\\ &=(8,2,7,-1) +(-3,3,-15, 9)\\ &=(5,5,-8,8)\end{align}


Edit:

For your another approach:

You can compute $$\frac{\vec{PA}\cdot \vec{BA}}{|\vec{PA}||\vec{BA}|} = \cos 120^\circ=-\frac12$$

$$\frac{45-36s}{6\sqrt{63-90s+36s^2}}=-\frac12$$

$$5-4s=-\sqrt{7-10s+4s^2}$$

Hence, we need $5-4s\le 0$

$$25-40s+16s^2=7-10s+4s^2$$

$$2s^2-5s+3=0$$ $$(2s-3)(s-1)=0$$

Hence $s=\frac32$, now you can obtain the coordiante for $P$.

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  • $\begingroup$ So the property matters that $\triangle CBB'$ is congruent to $\triangle PAA'$. The two lines $\vec{BA}$, $\vec{CP}$ are parallel, leading to same lengths of $\vec{BB'}$, $\vec{AC'}$ (I think you better had named $C'$ as $A'$ to maintain similarity with $B'$). Also, the two equal angles $\angle{C} = \angle{P}$ provide the rest needed to make the projections of $\vec{CB}$, $\vec{PA}$ the same. However, even if the problem is changed to have unequal angles say, $\angle{C} = 60^o, \angle{P}= 45^o$ still would work with modifications to compute $|CB'|, |A'P|$. $\endgroup$ – jiten Jul 10 at 5:46
  • $\begingroup$ Request be in chatroom (chat.stackexchange.com/rooms/110379/new-room-108248) for related issue. $\endgroup$ – jiten Jul 10 at 5:53
  • $\begingroup$ You will admire the solution at: math.stackexchange.com/q/3751853/424260. Request an upvote there. Really deserves multiples. Will try the mathematica code given there at tio in python. $\endgroup$ – jiten Jul 10 at 11:11
  • $\begingroup$ Please see a simpler post of mine at: math.stackexchange.com/q/3754364/424260. I feel it is simple, but not solvable. It concerns historical proof of irrational quantities being there by Hippasus. It has also not attracted any response so far. $\endgroup$ – jiten Jul 12 at 13:18
  • $\begingroup$ Kindly see my post for an induction based proof at : math.stackexchange.com/q/3760948/424260. $\endgroup$ – jiten Jul 18 at 5:46
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Alternative solution. Continue $CB$ and $PA$ to cross at $S$. The triangle $CPS$ is equilateral. $$|\vec{CB}|=6;|\vec{CP}|=2\cdot|\vec{CB}|\cdot \cos 60^\circ+|\vec{AB}|=6+12=18;\\ \vec{CS}=\vec{CB}+|\vec{BS}|\cdot \frac{\vec{CB}}{|\vec{CB}|}=\\(-4,2,0,4)+12\cdot (-\frac23,\frac13,0,\frac23)=(-12,6,0,12)\Rightarrow S(-4,8,7,11).\\ \vec{SP}=\vec{SA}+|\vec{AP}|\cdot \frac{\vec{SA}}{|\vec{SA}|}=\\(6,-2-10,-2)+6\cdot (\frac12,-\frac16,-\frac56,-\frac16)=(9,-3,-15,-3)\Rightarrow P(5,5,-8,8).$$

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You can use the fact that the trapezium is isosceles since the base angles are equal.

Using your notation, let $\vec D = \vec C + \vec{BA}$.

Then $\triangle DPA$ is equilateral since the angles are all equal to $60°$.

Hence,

$$\vec P = \vec C + \vec{BA} + |\vec{CB}|\frac{\vec{BA}}{|\vec{BA}|}$$

Plugging in the values gives $P=(5,5,-8,8)$.

Note concerning your solution:

The point $(6, 4, -3, 5)$ you found is exactly my above mentioned point $D$. However, at this point the angle between the direction vector $\vec{BA}$ and $\vec{DA}$ is $60°$ and not $120°$.

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Why would $s=1$ give you $p$? The lengths of the parallel sides are not equal. What you should do instead is get use the parametric form of point $p$ in terms of $s$, and also use the information of the angle being $60$ degrees with $pA$

Hence if $p = (8-2s,2+2s,7-10s,-1+6s)$ then we have

$$\frac{\vec{pA}.\vec{Cp}}{|\vec{pA}|.|\vec{Cp}|} = \cos 60$$

Now solve this equation to get $s$ and hence point $p$

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  • $\begingroup$ You'd get an s term in Cp $\endgroup$ – Dhanvi Sreenivasan Jul 10 at 6:35
  • $\begingroup$ I think the parametric coordinates for $p= (8-2s, 2+2s, 7-10s, -1+6s)$. $\endgroup$ – jiten Jul 10 at 10:49

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