Tough problem... Vectors, points, and a cube

**Ares_D1** · August 3rd 2009, 02:32 AM

Problem:
A strange building has been built to look like part of a cube sticking out of the ground. You are required to find the 3d coordinates for points B and D, given the points A(13,30,9) and C(2,4,17). Points B and D are at equal height above the ground. The ground coincides with the x,y-plane

This picture provides two views of the building.

Relevant Equations:
Vector properties (dot product, cross product, length of vector v = |v|.

Anyone know how to approach this?

**Harry1W** · August 3rd 2009, 07:15 AM

I haven't worked this through fully... but here are my thoughts:

- If B and D are at equal height above the ground, they will have the same z-coordinate (why?). So, $z = z_3$

- The three vectors that define each corner of the cube are mutually perpendicular. So the vectors $\vec{AD}$ and $\vec{AB}$ are perpendicular, as are $\vec{CD}$ and $\vec{BC}$ . Can you think about how you might use the vector dot or cross product to process this information?

Hope that helps a little.

**Opalg** · August 3rd 2009, 07:50 AM

If the points B and D have the same z-coordinate then so does any point on the line BD. In particular, the midpoint of BD, which is also the midpoint of AC, has that same z-coordinate. But the coordinates of the midpoint of AC are given by $\tfrac12(13,30,9) + \tfrac12(2,4,17)$ . All you have to do is to compute that third coordinate.

**Media_Man** · August 3rd 2009, 07:59 AM

The distance AC=29.343. Since this is the hypotenuse of a 45-45-90 triangle, the sides of the cube are 20.748.

Since AB and AD are both this length, B and D must satisfy the equation $(x-13)^2+(y-30)^2+(z-9)^2=20.748^2$ . Similarly, Since CB and CD are both this length, B and D must satisfy the equation $(x-2)^2+(y-4)^2+(z-17)^2=20.748^2$

Setting these two equations equal to each other, we get $22x+52y-16z=841$ . Since B and D are the same height (same value for z), they must be the same height as AC's midpoint (try to visualize this). The z-value of AC's, and therefore B's and D's coordinates are therefore $\frac12(9+17)=13$ . Letting $z=13$ in all further equations, we arrive at $22x+52y=1049$ , or $y=20.173-.423x$ .

Now we pick the simpler of the two equations above and substitute y and z in terms of x: $(x-2)^2+(16.173-.423x)^2+(13-17)^2=20.748^2$ , or $1.178x^2-16.503x-148.934=0$ . The discriminant of this quadratic is 31.22, therefore the two possible values of x are 20.24 and -6.24. Plugging these back in for y, we get y=11.61 or 22.81.

So our final answers are $B=(20.24,11.61,13)$ and $D=(-6.24,22.81,13)$

Aside from a few possible algebraic and/or arithmetic and/or rounding errors, this is (in my opinion) the most elementary approach, if not the fastest. Make sure to run through these equations to make sure you understand them and see where their solutions come from.

**Ares_D1** · August 3rd 2009, 09:01 AM

Originally Posted by Media_Man

The distance AC=29.343. Since this is the hypotenuse of a 45-45-90 triangle, the sides of the cube are 20.748.

Could you tell me how you got the length of AC? By finding the AC vector and then |AC| to find it's length, I'm getting 29.698. Trivial difference, I know, but I have to be spot on. Are you rounding before your final result?

**Media_Man** · August 3rd 2009, 09:18 AM

Originally Posted by Ares_D1

Could you tell me how you got the length of AC? By finding the AC vector and then |AC| to find it's length, I'm getting 29.698. Trivial difference, I know, but I have to be spot on. Are you rounding before your final result?

$\sqrt{(13-2)^2+(30-4)^2+(9-17)^2}=$ $\sqrt{(11)^2+(26)^2+(-8)^2}=\sqrt{121+676+64}=\sqrt{861}=20.3428015$

You are getting $29.69848481=\sqrt{882}$ . Check your math.

**Ares_D1** · August 3rd 2009, 09:28 AM

Originally Posted by Media_Man

$\sqrt{(13-2)^2+(30-4)^2+(9-17)^2}=$ $\sqrt{(11)^2+(26)^2+(-8)^2}=\sqrt{121+676+64}=\sqrt{861}=20.3428015$

You are getting $29.69848481=\sqrt{882}$ . Check your math.

Yep, exactly what I did. You've just rounded before your final result, hence the discrepancy.

Not sure if I understand the second sentence in your first post, but it's probably because it's 3:30am here. :O I'll have a look over it tomorrow and ask questions if need be.

Thanks so much for the help everyone.

**Media_Man** · August 12th 2009, 07:51 AM

hellohello123 -

A = (13.0492, 30.9498, 9.01115),
B = (21.1958196, 12.05024361, 13.187175),
C = (1.97687, 4.68868, 17.3632),
D = (-6.169749596, 23.58823639, 13.187175),
O = (7.513035, 17.81924, 13.187175)

At the moment I'm having trouble finding the 4 points A', B', C' and D' where the cube's corners intersect the ground. Another two properties I need to find are the area that this A'B'C'D' plane on the ground encloses and also what volume of the cube is above or below the ground. I'm so sorry to be bothering you but any advice at all in approaching this would be of huge benefit to me.

Finding Coordinates of A',B',C',D'

Find the normal vector N of the plane ABCD using cross product: $N=ABxAD=(-46,-108,-421)$

Each line $AA',BB',CC',DD'$ has the same direction as N, so the parametric equations are $A'(t)=A+Nt, B'(t)=B+Nt, C'(t)=C+Nt, D'(t)=D+Nt$ . These four lines intersect the plane $A'B'C'D'$ when their z-value is zero (by construction). Therefore set $A_z'(t)=A_z+N_zt=0$ to solve for $t$ , and plugging $t$ back into $A'(t)$ gives you the other two coordinates of $A'$ . Repeat to find $B',C',D'$ .

Finding the Area of A'B'C'D'

The area of a parallelogram $Area=|A'B'xA'D'|$ , the magnitude of the cross product of two adjacent sides.

Finding the Volume of the contained area

The volume of the figure is simply the area of $ABCD$ times the "height" $OO'$ . So, using the parametric equation technique described above, $O'(t)=O+Nt$ , find the coordinates of $O'$ , use the length of $OO'$ as your height, and multiply by $S^2$ , where $S$ is the length of the side of the top face of the cube.

**hellohello123** · August 12th 2009, 08:47 PM

Thank you so very much, I will get working on this ASAP.

**Ares_D1** · August 14th 2009, 11:37 PM

Originally Posted by Media_Man

hellohello123 -
Finding the Volume of the contained area

The volume of the figure is simply the area of $ABCD$ times the "height" $OO'$ . So, using the parametric equation technique described above, $O'(t)=O+Nt$ , find the coordinates of $O'$ , use the length of $OO'$ as your height, and multiply by $S^2$ , where $S$ is the length of the side of the top face of the cube.

Media_Man,

I managed to get the rest of the problem, but not sure I'm following this bit. Specifically, where the points O and O' come from. How do I get their values so I can solve the parametric for t?

**Media_Man** · August 15th 2009, 05:40 AM

Sorry, I thought we used that in a previous thread. I'm using the point O to represent the center of the cube face, easily gotten by taking the midpoint of AC, so O=(7.5,17,13). Now O' is the center of the "shadow" shape A'B'C'D'. Therefore, find O' the same way you found the other "shadow" points, using O'(t)=O+Nt, setting the z coordinate equal to zero to find t, and plugging t back into O'(t) to find all three points. If you imagine turning this object upside down, you have a perfect right cube sitting on its face with its top sliced off. The height at which this slice occurs (average height) determines the remaining volume. Thus the height can be gotten as the distance between the centers of the top and bottom.

**jacquie** · August 16th 2009, 01:43 AM

Originally Posted by Media_Man

hellohello123 -

Finding Coordinates of A',B',C',D'

Find the normal vector N of the plane ABCD using cross product: $N=ABxAD=(-46,-108,-421)$

Each line $AA',BB',CC',DD'$ has the same direction as N, so the parametric equations are $A'(t)=A+Nt, B'(t)=B+Nt, C'(t)=C+Nt, D'(t)=D+Nt$ . These four lines intersect the plane $A'B'C'D'$ when their z-value is zero (by construction). Therefore set $A_z'(t)=A_z+N_zt=0$ to solve for $t$ , and plugging $t$ back into $A'(t)$ gives you the other two coordinates of $A'$ . Repeat to find $B',C',D'$ .

Finding the Area of A'B'C'D'

The area of a parallelogram $Area=|A'B'xA'D'|$ , the magnitude of the cross product of two adjacent sides.

Finding the Volume of the contained area

The volume of the figure is simply the area of $ABCD$ times the "height" $OO'$ . So, using the parametric equation technique described above, $O'(t)=O+Nt$ , find the coordinates of $O'$ , use the length of $OO'$ as your height, and multiply by $S^2$ , where $S$ is the length of the side of the top face of the cube.

would you be able to explain to me how you got (-46,-108,-421) for the normal vector? i know how to use the cross product. but i think my AB and AD values might be different.

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