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Calculus 3 intersection of planes?
I'm having trouble with the following question:
Find the parametric equations and symmetric equations for the line:
The line of intersection of the planes x+y+z=1 and x+z=0.
I can visualize what the problem is asking. How do I translate the planes into vectors so that I can use cross-multiplication to find the line (assuming this is the right approach to solving the problem)?
2 Answers
- NorthstarLv 71 decade agoFavorite Answer
The directional vector v, of the line of intersection of the two planes is orthogonal to the normal vectors n1 and n2 of the two given planes. Take the cross product.
v = n1 X n2 = <1, 1, 1> X <1, 0, 1> = <1, 0, -1>
Now find a point P, on the line. It will lie in both planes. By inspection we see that one such point is P(0, 1, 0). The vector equation of the line is:
r(t) = <0, 1, 0> + t<1, 0, -1>
where parameter t is any real number
The parametric form of the equation of the line is:
r(t):
x = t
y = 1
z = -t
The symmetric form of the equation of the line is:
t = x/1 = z/(-1), and y = 1
- Anonymous4 years ago
The directional vector v, of a line perpendicular to the aircraft is an comparable by using fact the classic vector of the aircraft. v = <3, -2, 7> With the directionla vector of the line and a point on the line O(0,0,0) we are waiting to jot down the equation of the aircraft. r(t) = O + television r(t) = <0, 0, 0> + t<3, -2, 7> by using fact the element on the line is the commencing place we are waiting to simplify the equation of the line further. r(t) = t<3, -2, 7> the placement parameter t levels over the genuine numbers