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Please show all your work! Total points: 40

- What
**constraints**on b_{1}, b_{2}, and b_{3}are necessary in order for the following system to have any solutions?__[5]__2x - y + 3z = b

_{1}

3x - (3/2)y + (9/2)z = b_{2}

-4x + 2y - 6z = b_{3}*b*_{2}= (3/2) b_{1}and

b_{3}= -2 b_{1}(or equivalent) - Calculate the
**determinant**of the following matrix:__[6]__2 0 1 0 4 1 0 -2 1 2 2 0 0 -1 0 1 *det = 5.*

One way is to do: 2R_{4}+R_{2}, then -(1/2)R_{3}+R_{1}, then -R_{2}+R_{1}, then you get a lower-triangular matrix. Multiply down the diagonal: (-5/2)(-1)(2)(1) = 5. - A surface is being rendered in a 3D graphics system.
If the surface faces away from the camera, "
**backface culling**" means the surface does not need to be rendered, saving processing time. The vector from the surface to the camera is u = (3, -2, 8). The normal vector to the surface is (-1/2, 0, 0.2). Is the surface facing toward or away from the camera?__[4]__*Dotproduct = -1.5 + 0 + 1.6 = 0.1 > 0,*

so it faces toward the camera. - Let A be a
**skew-symmetric**n x n matrix; i.e., A^{T}= -A.- Find the
**trace**tr(A).__[3]__*For the diagonal entries a*_{ii}, we must have a_{ii}= -a_{ii}, which is only possible if they are zero. Hence all the diagonal entries are zero, so the trace is zero. - If n is odd, prove that det(A) = 0.
__[5]__*det(A) = det(A*^{T}) = det(-A) = (-1)^{n}det(A).

If n is odd, then det(A) = -det(A), so it must be zero.

- Find the
- Let u and v be any non-collinear (non-parallel) vectors in 3-space.
Let w = v × (u × v).
- Draw a sketch illustrating how u, v, and w are
**oriented**relative to each other.__[4]__*All three of u, v, w are in the same plane, and w is perpendicular to v.* - Find the
**dot product**(v ⋅ w).__[2]__*Since w is perpendicular to v, v ⋅ w is zero.* - What can you say about the
**dot product**(u ⋅ w)? (Simplify)__[5]__*||u × v|| = ||u|| ||v|| sinθ.*

Since the angle between v and (u × v) is 90deg,

||w|| = ||v|| (||u × v||) sin(90deg)

= ||v|| (||u × v||)

= ||u|| ||v||^{2}sinθ.

Now, because u, v, w are all in the same plane and v and w are perpendicular, the angle φ between u and w and the angle θ between u and v add up to 90deg. Hence, cosφ = sinθ.

So u ⋅ w = ||u|| ||w|| cosφ

= ||u|| ( ||u|| ||v||^{2}sinθ ) cosφ

= ||u||^{2}||v||^{2}sin^{2}θ

= ||u × v||^{2}.

- Draw a sketch illustrating how u, v, and w are
- Let y = f(x) be a cubic polynomial in x, i.e.,
y = c
_{1}x^{3}+ c_{2}x^{2}+ c_{3}x + c_{4}. The graph of y has a horizontal tangent when it goes through the point (0,1), and goes through the point (2,3) with a tangent slope of -2. Find y = f(x) (this is called a**cubic Hermite spline**).__[6]__*f(0) = 1 ⇒ c*_{4}= 1

f'(0) = 0 ⇒ c_{3}= 0

f(2) = 3 ⇒ 8c_{1}+ 4c_{2}+ 2c_{3}+ c_{4}= 3

f'(2) = -2 ⇒ 12c_{1}+ 4c_{2}+ c_{3}= -2

⇒ 8c_{1}+ 4c_{2}+ 1 = 3 and 12c_{1}+ 4c_{2}= -2

⇒ c_{1}= -1, c_{2}= 5/2

⇒ y = - x^{3}+ (5/2) x^{2}+ 1