Final answer:
The equation of the line through point u and parallel to vector v is comprised of parametric equations. The equation of the line through points v and w involves determining a direction vector. The equations of the respective planes and sphere are derived using the normal vector, cross product, and the given center and radius.
Step-by-step explanation:
Vector Equations and Geometry
To solve these vector and geometry problems, we can apply knowledge of vector operations and equations of lines and planes in three dimensions.
(a) Equation of the Line Through u Parallel to v
A line parallel to vector v passing through point u can be represented as r = u + tv, where t is a scalar parameter. Substituting the given values, we have the parametric equations of the line:
x = 2 + t,
y = 1,
z = -3 + t.
(b) Equation of the Line Through v and w
For a line through points v and w, we find the direction vector by subtracting their coordinates: d = w - v. The parametric equations are:
x = 1 + s(0-1),
y = s(-1),
z = 1 + s(3-1), yielding
x = 1 - s,
y = -s,
z = 1 + 2s, where s is the parameter.
(c) Equation of the Plane Through u Perpendicular to v
A plane perpendicular to vector v can be given by the dot product (n · (r - u) = 0), where n is a normal vector to the plane. Here, n is parallel to v and thus n = v. Substituting v and u into the plane equation gives:
1(x - 2) + 0(y - 1) + 1(z + 3) = 0, simplifying to
x + z + 1 = 0.
(d) Equation of the Plane Through w Parallel to u and v
To find a plane parallel to vectors u and v, we calculate their cross product u × v to get a normal vector to the plane. The plane's equation is found similarly to part (c): n · (r - w) = 0. After computing u × v and substituting w, we obtain an equation of the form ax + by + cz = d.
(e) Equation of the Sphere With Center u and Radius 2
The equation of a sphere with center (h, k, l) and radius R is (x - h)² + (y - k)² + (z - l)² = R². Substituting u as the center and 2 as the radius, we get:
(x - 2)² + (y - 1)² + (z + 3)² = 4.