Find coordinates of vertex in a trapezoid.

Question

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 $=(x, y, z, w)$.
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.

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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 $=\cos 60^{o} = 0.5$

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

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Edit 3: Excellent post made for the same problem.

Answer 1

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}

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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 $P$.

Answer 2

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 $p$

Answer 3

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 $120°$.

Answer 4

Alternative solution. Continue $CB$ and $PA$ to cross at $S$. The triangle $CPS$ is equilateral. $$|\vec{CB}|=6;|\vec{CP}|=2\cdot|\vec{CB}|\cdot \cos 60^\circ+|\vec{AB}|=6+12=18;\\ \vec{CS}=\vec{CB}+|\vec{BS}|\cdot \frac{\vec{CB}}{|\vec{CB}|}=\\(-4,2,0,4)+12\cdot (-\frac23,\frac13,0,\frac23)=(-12,6,0,12)\Rightarrow S(-4,8,7,11).\\ \vec{SP}=\vec{SA}+|\vec{AP}|\cdot \frac{\vec{SA}}{|\vec{SA}|}=\\(6,-2-10,-2)+6\cdot (\frac12,-\frac16,-\frac56,-\frac16)=(9,-3,-15,-3)\Rightarrow P(5,5,-8,8).$$