MHB Scalar Triple Product and Coplanarity

AI Thread Summary
The discussion focuses on the scalar triple product and its relation to coplanarity. It emphasizes the importance of understanding cross and dot products, providing a specific example with vectors v and w. The cross product is calculated using a determinant, resulting in the vector 5i - j - 7k. The next step involves taking the dot product of this result with another vector u. Understanding these operations is crucial for solving problems related to vector coplanarity.
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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.
 
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 $u= \vec{i}+ 2\vec{j}+ 3\vec{k}$.
 
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