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Unformatted text preview: + 2t, 2 + t, −4 − 2t
4. r(t) = 4 − 2t, 2 − t, 4 − 2t
5. r(t) = 2 + 4t, 1 + 2t, 2 − 4t
6. r(t) = 4 + 2t, −2 + t, 4 − 2t correct
Explanation:
A line passing through a point P (a, b, c)
and having direction vector v is given in vector form by
r(t) = a + tv , a= 3. 2x2 + y 2 + 3z 2 = 1 Explanation:
The graphs of each of the given quadratic
relations is a quadric surface in standard position. We have to decide which quadric surface
goes with which equation  a good way of doing that is by taking plane slices parallel to
the coordinate planes, i.e., by setting respectively x = a, y = a and z = a in the equations
once we’ve decided what the graphs of those
plane slices should be.
Now as slicing of a, b, c . Now for the given line, its direction vector will
be parallel to the normal to the plane
2x + y − 2z = 6 ,
so
v = 2, 1, 2 ; shows, slices of this hyperbolic paraboloid by
z = a are hyperbolas, while slices by x = a
and y = a are parabolas, opening in diﬀerent
directions. Only the graph of on the other hand,
a = 4, 2, 4 ,
since the line passes through P (4, −2, 4).
Consequently, in vector form the line is given
by
r(t) = 4 + 2t, −2 + t, 4 − 2t . z = y 2 − x2
has these properties. geyer (dag2798) – HW09 – rusin – (55735) 7
z 013 10.0 points x Which one of the following equations has
graph But this circle has radius 1 because the cylinder has radius 1, and so its equation is
x2 + (z + 1)2 = 1
when the circular cylinder...
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 Spring '07
 Sadler
 Determinant, Multivariable Calculus

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