Finding equation of a plane with 3 points

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To find the equation of a plane given three points, the user struggles to arrive at the correct equation, which is 2x + y + 7z = -3. The main issue identified is an incorrect calculation of the cross-product, which is essential for determining the normal vector of the plane. Specifically, the user miscalculated components related to the i and k unit vectors. Clarification on these calculations is requested to identify the exact error. Understanding the correct method for calculating the cross-product is crucial for solving the problem accurately.
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Homework Statement


attachment.php?attachmentid=68806&stc=1&d=1397897233.png

Homework Equations


N/A

The Attempt at a Solution


attachment.php?attachmentid=68807&stc=1&d=1397897233.jpg


The problem is that I don't get the right answer which is:
2x + y + 7z = -3.

Can you please help me find where I went wrong?
 

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Can you explain what you calculated there?
 
uzman1243 said:

Homework Statement


[IMG ]https://www.physicsforums.com/attachment.php?attachmentid=68806&stc=1&d=1397897233[/PLAIN]

Homework Equations


N/A

The Attempt at a Solution


attachment.php?attachmentid=68807&stc=1&d=1397897233.jpg


The problem is that I don't get the right answer which is:
2x + y + 7z = -3.

Can you please help me find where I went wrong?
You calculated the cross-product incorrectly.

##\left( -2 + 0 \right)\hat i \ne 0\hat i ##

##\hat k \left( \ (1) (-3) - (-2)(-2)\ \right)\ne \hat k \left(-3 + 4 \right)##
 

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