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State where possible vector, parametric, and symmetric equations for each of the following lines.
a. The line passing through the point P (-1,2,1) with direction vector (3,-2,1)
b. The line passing through the point B (-2,3,0) and parallel to the line passing through the
points M(-2,-2,1) and N(-2,4,7)
c. The line passing through the points Q (1,2,4) and parallel to the z-axis
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(a) Parametric equation is
(x,y,z) = (-1,2,1) + t*(3,-2,1), where t is any real number.
From the formula, it is immediately seen that the point P lies in this line (at t= 0),
and that the direction vector is (3,-2,1).
In vector form, this parametric equation is the set of these three scalar (component) equations
x = -1 + 3t, y = 2 - 2t, z = 1 + t.
(b) The direction vector for this line is vector MN connecting the points
MN = N_bar - M_bar = (-2,4,7) - (-2,-2,1) = (0,6,6).
Hence, similar to (a), the parametric equation of the desired line is
(x,y,z) = (-2,3,0) + t*(0,6,6), where t is any real number.
In vector form, this parametric equation is the set of these three scalar (component) equations
x = -2, y = 3 + 6t, z = 6t.
(c) The line passing through the points Q(1,2,4) and parallel to z-axis in parametric form is
(x,y,z) = (1,2,t), where t is any real number.
In vector form, this parametric equation is the set of these three scalar (component) equations
x = 1, y = 2, z = t.
Solved.
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A student SHOULD be able to complete this assignment on his or her own as soon as he (or she)
get familiar with definitions of all basic conceptions, participating in the problem's description.
Well, may be, having minimal additional training.
The solution does not require a flight of thought in higher spheres - only good understanding
of the learned conceptions.