- #1
brinlin
- 12
- 0
Scalar Triple Product and Coplanarity
- Context: MHB
- Thread starter brinlin
- Start date
Click For Summary
Discussion Overview
The discussion centers around the scalar triple product and its relation to coplanarity, including calculations involving cross and dot products. Participants explore the mathematical operations involved and seek clarification on specific aspects of the topic.
Discussion Character
- Technical explanation
- Mathematical reasoning
Main Points Raised
- One participant requests clarification on the original question, suggesting that the poster should explain what is unclear.
- Another participant provides a link to a resource on the scalar triple product, potentially as a reference for further understanding.
- A participant demonstrates how to calculate the cross product of two vectors, v and w, using a determinant, and provides the resulting vector.
- The same participant suggests taking the dot product of the resulting vector with another vector, u, indicating a step in the process of exploring the scalar triple product.
Areas of Agreement / Disagreement
The discussion does not appear to have reached a consensus, as participants are at different stages of understanding and calculation, with some seeking clarification while others provide technical details.
Contextual Notes
There are limitations in the clarity of the original question, and the discussion may depend on the definitions of the terms used in the context of the scalar triple product.
- #2
Evgeny.Makarov
Gold Member
MHB
- 2,434
- 4
Welcome to the forum. Before we go into the discussion of your particular question, please read the https://mathhelpboards.com/help/forum_rules/, especially "Show the nature of your question in your thread title" and "Show some effort". In this case, you should probably explain what exactly is not clear to you.
- #3
skeeter
- 1,103
- 1
- #4
HOI
- 921
- 2
Do you not know how to do a cross product and a dot product?
With v= <2, 3, 1> and w= <3, 1, 2> the cross product, v x w, can be calculated as the determinant
$\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{array}\right|= \vec{i}\left|\begin{array}{cc}3 & 1 \\ 1 & 2\end{array}\right|- \vec{j}\left|\begin{array}{cc}2 & 1 \\ 3 & 2\end{array}\right|+ \vec{k}\left|\begin{array}{cc} 2 & 3 \\ 3 & 1 \end{array}\right|$
$= (6- 1)\vec{i}- (4- 3)\vec{j}+ (2- 9)\vec{k}= 5\vec{i}- \vec{j}- 7\vec{k}$
Now take the dot product of that with $u= \vec{i}+ 2\vec{j}+ 3\vec{k}$.
With v= <2, 3, 1> and w= <3, 1, 2> the cross product, v x w, can be calculated as the determinant
$\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{array}\right|= \vec{i}\left|\begin{array}{cc}3 & 1 \\ 1 & 2\end{array}\right|- \vec{j}\left|\begin{array}{cc}2 & 1 \\ 3 & 2\end{array}\right|+ \vec{k}\left|\begin{array}{cc} 2 & 3 \\ 3 & 1 \end{array}\right|$
$= (6- 1)\vec{i}- (4- 3)\vec{j}+ (2- 9)\vec{k}= 5\vec{i}- \vec{j}- 7\vec{k}$
Now take the dot product of that with $u= \vec{i}+ 2\vec{j}+ 3\vec{k}$.
Similar threads
- · Replies 12 ·
- · Replies 19 ·
- · Replies 2 ·
- · Replies 4 ·
- · Replies 4 ·
- · Replies 1 ·
- · Replies 5 ·
- · Replies 1 ·
- · Replies 3 ·
- · Replies 2 ·