Using the Cross Product to Find the Equation of a Plane
We know from Property (f) that the cross product ๐ฎ ร ๐ฏ is orthogonal to both of the vectors ๐ฎ and ๐ฏ. Therefore, if ๐ฎ and ๐ฏ are a pair of direction vectors for a plane, then ๐ฎ ร ๐ฏ is a normal vector for the plane.
An example similar to, yet different from, the one given below can be found in the video Determinants: Properties and Applications 2 starting at 7:40.
Example: Consider the plane ๐ in โยณ that contains the three points ๐ฉโ=[1,0,1], ๐ฉโ=[1,0,0] and ๐ฉโ=[0,1,0].
Let ๐ฎ=๐ฉโโ๐ฉโ= and let ๐ฏ=๐ฉโโ๐ฉโ= .
Correct answer, well done.
Marks for this submission: 1.00/1.00.
Correct answer, well done.
Marks for this submission: 1.00/1.00.
Since ๐ฎ and ๐ฏ are scalar multiples of each other, they are parallel
and hence ๐ฎ and ๐ฏ serve as direction vectors for ๐.
Correct answer, well done.
Marks for this submission: 1.00/1.00.
Correct answer, well done.
Marks for this submission: 1.00/1.00.
Using the cross product of the vectors ๐ฎ and ๐ฏ, we obtain that a normal vector for this plane is
๐ง=๐ฎ ร ๐ฏ=
Incorrect answer.
The entries underlined in red below are those that are incorrect.
[0 0 1]
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Now, using this and the fact that the point ๐ฉโ is in the plane we obtain that the general equation for this plane is
(Use ๐ฅ, ๐ฆ, ๐ง as the variables in your computation for this equation.)
There is yet another way to use the determinant to obtain the general equation of a plane in โยณ. It essentially amounts to the same calculation as with the cross product above, but the argument is different: see the first half of the video Determinants: Properties and Applications 2.