Vector calc area of triangle and equation of a plane

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January 27th 2009, 08:41 AM
Frostking

Vector calc area of triangle and equation of a plane

I have three points on a plane A: (-1, 3, 0), B: (3, 2, 4) and C: (1, -1,5) and I determined that the equation for this plane is 11x - 12y - 14z = -47 To get this I figured out AB vector and AC vector and then by cross product got n vector. With this I used point (-1, 3, 0) to get my equation for the plane.

I am now suppose to obtain the area of the triangle determined by these three points. ????

I also need to find the equation for the plane that is through the origin and perpendicular to
the planes x - y + z = 5 and 2x + y - 2z = 7

By using these two planes I got two normal vectors and obtained the cross product of them to get a vector to use with the point (0,0,0) . I got that the plane perpendicular to the ones above containing the origin would have the equation x + 4y + 3z = 0 Could someone confirm this or tell me where I am in error? Thanks to all of you who help on this site. Frostking
January 27th 2009, 08:57 AM
Plato

Quote:

Originally Posted by Frostking

I have three points on a plane A: (-1, 3, 0), B: (3, 2, 4) and C: (1, -1,5) and I determined that the equation for this plane is 11x - 12y - 14z = -47 CORRECT
To get this I figured out AB vector and AC vector and then by cross product got n vector. With this I used point (-1, 3, 0) to get my equation for the plane.

I am now suppose to obtain the area of the triangle determined by these three points. ????

If vectors $p\;\&\;q$ are adjacent sides of a triangle then the area of the triangle is $\frac{{\left\| {p \times q} \right\|}}<br /> {2}$
January 27th 2009, 12:23 PM
Soroban

Hello, Frostking!

Quote:

I have three points on a plane A: (-1, 3, 0), B: (3, 2, 4) and C: (1, -1,5)
and I determined that the equation for this plane is: . $11x - 12y - 14z \:=\: -47$ . . . . yes!

To get this, I figured out $\overrightarrow{AB}\text{ and }\overrightarrow{AC}$ , and then by cross product got $\vec n.$
With this, I used point (-1, 3, 0) to get my equation for the plane. . . . . good!

I am now suppose to obtain the area of the triangle determined by these three points.

We're expected to know that: the absolute value of the cross product of two vectors
. . is the area of the parallelogram determined by the two vectors.

You already found that: . $\overrightarrow{AB} \times \overrightarrow{AC} \:=\:\langle 11,\text{-}12,\text{-}14\rangle$

. . Hence, the area of the parallelgram: . $\sqrt{11^2 + (\text{-}12)^2 + (\text{-}14)^2} \:=\:\sqrt{461}$

Therefore, the area of the triangle is: . $\tfrac{1}{2}\sqrt{461}$

Ahh, Plato is too fast for me . . .
.
January 27th 2009, 12:53 PM
Bilbo Baggins

Quote:

Originally Posted by Frostking

I also need to find the equation for the plane that is through the origin and perpendicular to
the planes x - y + z = 5 and 2x + y - 2z = 7

By using these two planes I got two normal vectors and obtained the cross product of them to get a vector to use with the point (0,0,0) . I got that the plane perpendicular to the ones above containing the origin would have the equation x + 4y + 3z = 0 Could someone confirm this or tell me where I am in error? Thanks to all of you who help on this site. Frostking

Your approach is correct. If two planes are perpendicular, then the normal vector of the one will lie within the second. Therefore, the two normal vectors to the two given planes both lie within the plane of interest, and their cross product yields the normal vector of interest.