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C4 Vectors question

Q:

Points A, B and C have coordinates (5, -1, 0), (2, 4, 10) and (6, -1, 4) respectively.

(a) Find the vectors CA⃗\vec{CA} and CB⃗\vec{CB}

(b) Find the area of the triangle ABC

(c) Point D is such that the point A, B, C and D form the vertices of a parallelogram. Find the coordinates of three possible positions of D.

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(a) CA⃗\vec{CA} = (−10−4)\begin{pmatrix} -1 \\ 0 \\ -4 \end{pmatrix}

CB⃗\vec{CB} = (−456)\begin{pmatrix} -4 \\ 5 \\ 6 \end{pmatrix}

(b) Area of triangle = 15.07 units^2

(c) Need help in this part.

@Zacken any help with part (c) please?
Reply 1
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Zacken
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What's the definition of parallelogram?
Reply 2
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SaadKaleem
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Original post by Zacken
What's the definition of parallelogram?


A four sided shape, with opposite sides parallel I guess?
Reply 3
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Zacken
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Original post by SaadKaleem
A four sided shape, with opposite sides parallel I guess?


Yep, so set up your D such that CD is parallel to AB or DA is parallel to another side or etc...
Reply 4
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SaadKaleem
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Original post by Zacken
Yep, so set up your D such that CD is parallel to AB or DA is parallel to another side or etc...


Yes, I managed to sketch three possible shapes, however failed to deduce regardless, the coordinates of D.

Probably missing out on something silly here.

Vectors :argh:

Reply 5
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Zacken
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Original post by SaadKaleem
Yes, I managed to sketch three possible shapes, however failed to deduce regardless, the coordinates of D.

Probably missing out on something silly here.

Vectors :argh:


Set D=(x,y,z)TD = (x, y, z)^{T} and then look at AD,CD,BDAD, CD, BD, or along those lines and equate AD with the opposite side of AD - since it's a parallelogram, then compare components, etc...
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SaadKaleem
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Original post by Zacken
Set D=(x,y,z)TD = (x, y, z)^{T} and then look at AD,CD,BDAD, CD, BD, or along those lines and equate AD with the opposite side of AD - since it's a parallelogram, then compare components, etc...


Alright, got my answers.. This was my first shape-like vector question, so i struggled.

Answers I got:

OD1⃗\vec{OD 1} = (9−6−6)\begin{pmatrix} 9 \\ -6 \\ -6 \end{pmatrix}

OD2⃗\vec{OD 2} = (146)\begin{pmatrix} 1 \\ 4 \\ 6 \end{pmatrix}

OD3⃗\vec{OD 3} = (3414)\begin{pmatrix} 3 \\ 4 \\ 14 \end{pmatrix}

So in coordinates (9, -6, -6), (1, 4, 6) and (3, 4, 14)

Thank you Zacken :wink:
(edited 8 years ago)
Reply 7
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Zacken
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Original post by SaadKaleem

Thank you Zacken :wink:


First class work!

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