Unformatted text preview: 1.7. APPLICATIONS OF CROSS PRODUCTS 101 and hence parallel to their crossproduct
→→
− =− ×− .
v
n 1 →2
n
Thus, given any one point P0 (x0 , y0 , z0 ) on ℓ (i.e., one solution of the pair
of equations) the line ℓ can be parametrized using P0 as a basepoint and
→−
− = → × − as a direction vector.
v
n 1 →2
n
For example, consider the two planes
3x − 2y + z = 1 2x + y − z = 0. The ﬁrst has normal vector
→
− = 3− − 2− + −
→
→→
n1
ı k
while the second has →
− = 2− + − − − .
→→→
n2
ı k Thus, a direction vector for the intersection line is
→→
− =− ×−
v
n 1 →2
n
→→→
−−−
ı k
= 3 −2 1
2
1 −1 −2 1
→3 1
−− 2 −1
1 −1
→
−
→
→
= − + 5− + 7 k .
ı →
=−
ı →
− 3 −2
+k
21 One point of intersection can be found by adding the equations to
eliminate z
5x − y = 1
and, for example, picking
x=1
which forces
y = 4. ...
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 Fall '08
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 Calculus, Equations, Dot Product, Force, Bivector, direction vector

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