- #1
brinlin
- 12
- 0
Scalar Triple Product and Coplanarity
- Context: MHB
- Thread starter brinlin
- Start date
Click For Summary
SUMMARY
The discussion focuses on the Scalar Triple Product and its application in determining coplanarity of vectors. The cross product of vectors v = <2, 3, 1> and w = <3, 1, 2> is calculated using the determinant method, resulting in the vector 5i - j - 7k. Subsequently, the dot product of this cross product with the vector u = i + 2j + 3k is introduced as a next step in the analysis. Understanding these operations is crucial for applications in vector mathematics and physics.
PREREQUISITES- Understanding of vector operations, specifically cross product and dot product.
- Familiarity with determinants and their calculation.
- Basic knowledge of vector notation and components.
- Concept of coplanarity in vector spaces.
- Learn how to compute the Scalar Triple Product and its geometric interpretation.
- Study the implications of coplanarity in three-dimensional space.
- Explore applications of vector products in physics, particularly in mechanics.
- Investigate the properties of determinants in relation to vector operations.
Students of mathematics, physics enthusiasts, and anyone studying vector calculus or linear algebra will benefit from this discussion.
- #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 ·