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brinlin
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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.
- #2
Evgeny.Makarov
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MHB
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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.
- #3
skeeter
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- #4
HOI
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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 $u= \vec{i}+ 2\vec{j}+ 3\vec{k}$.
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 $u= \vec{i}+ 2\vec{j}+ 3\vec{k}$.