MHB Scalar Triple Product and Coplanarity

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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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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 $u= \vec{i}+ 2\vec{j}+ 3\vec{k}$.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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