Vectors. Determine the eqn. of plane M?

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SUMMARY

The discussion focuses on determining the equation of plane M that contains the vectors a = i - j + 2k and b = 2i + 3j - k, along with the point P(2,1,-1). The solution involves calculating the direction vector v = ab = i + 4j - 3k and applying the normal vector formula for planes. The final equation of the plane is established as -x + y + z = -2, derived from the cross product of the vectors and the point's coordinates.

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Homework Statement


Question is:
Given the vectors a= i - j + 2k and b= 2i + 3j - k, determine the equation of the plane M that contains the vectors a and b and the point P(2,1,-1).

Homework Equations


x = xo + at
y = yo + bt
z = zo + ct
PoP = tv ---> (x-xo)i + (y-yo)j + (z-zo)k = t(ai + bj + ck)

where Po (xo, yo, zo)
P(x, y, z)

The Attempt at a Solution


pt= (2,1,-1)
direction: v = ab= i + 4j - 3k

equation of line:
x=xo + at --> x= 2 + t
y=yo + bt --> y= 1 + 4t
z=zo + ct --> z= -1 - 3t

and that's where I get stuck, and I don't know how to continue.

The answer is -x + y + z = -2.
I am unsure how to get to that answer.

Got the answer nevermind.
 
Last edited:
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You have a lot about vectors and lines there, but nothing about planes. You need to know two fundamental facts:
1) The plane with normal vector [itex]a\vec{i}+ b\vec{j}+ c\vec{k}[/itex] and containing point [itex](x_0, y_0, z_0)[/itex] has equation [itex]a(x- x_0)+ b(y- y_0)+ c(z- z_0)= 0[/itex].

2) A normal vector to the plane containing vectors [itex]a\vec{i}+ b\vec{j}+ c\vec{k}[/itex] and [itex]d\vec{i}+ e\vec{j}+ f\vec{k}[/itex] is the cross product of the two vectors [itex](bf- ce)\vec{i}- (cd- af)\vec{j}+ (ae- bd)\vec{j}[/itex] which can be remembered by writing it (as a mnemonic) as if it were a determinant:
[tex]\left|\begin{array} {ccc}\vec{i} & \vec{j} & \vec{k} \\ a & b & c \\ d & e & f\end{array}\right|[/tex]
 

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