Finding the Cartesian Equation of a Perpendicular Line

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SUMMARY

The discussion focuses on finding the vector equation of a line that passes through the point Q(2,0,-5) and is perpendicular to the vectors m=(0,1,4) and n=(-2,-1,3). The solution involves using the cross product of the two vectors to determine the direction vector (a,b,c) for the line. The equations derived from the dot product of the direction vector with the given vectors are b + 4c = 0 and -2a - b + 3c = 0. The correct approach confirms that the cross product yields the necessary vector for both vector and Cartesian forms of the line.

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Homework Statement


Find the vector equation of the line that passes through the point Q(2,0,-5) and
is perpendicular to both the vectors m=(0,1,4) and n=(-2,-1,3).

Homework Equations


vector equation of a line: (x, y, z)=(x0,y0,z0) + t(a,b,c)
cartesian equation of a line: (x-x0)/a=(y-y0)/b=(z-z0)/c

The Attempt at a Solution


(x0,y0,z0)=(2,0,5)
To find (a,b,c)I know I can get two equations because the dot product of (a,b,c) with the two perpendicular lines equals zero:
b+4c=0
-2a-b+3c=0
But two equations isn't enough to solve for three variables. Also, shouldn't the point (2,0,-5) also dot product with u or v to equal zero, since it's on the same line?
Is it correct to assume that the points (0,1,4) and (-2,-1,3) are also points on the line? In which case I can easily find the direction of the line by subtracting one from the other.
 
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The cross product of <0, 1, 4> and <-2, -1, 3> will give you a vector that is perpendicular to both. That's the vector you need to write your line in either its vector form or in Cartesian form.
 
Man, I can't believe I didn't think of that myself. thanks for your help!
 

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