1. Consider the points A = (2, 2, 0), B = (1, 2, 1), C = (1, 0, h), where h is a constant
(which may appear in your answers below).
(a) Find the coordinates of the point D for which ABCD is a parallelogram.
We draw the segments AB and AC and note that the point D that completes the
parallelogram will arise by starting at A and first following the vector AB and then
the vector AC. AB = h−1, 0, 1i, AC = h−1, −2, hi, so AB + AC = h−2, −2, hi,
and thus starting at A and following this vector, we get to the point D = (0, 0, h).
(b) Find the area of the triangle ABC.
The area of the parallelogram
ABCD is|AB × AC|, and so the area of the triangle
i
j k p
1
will be half this. Area = 2 −1 0 1 = 12 | h2, h − 1, 2i | = 12 8 + (h − 1)2
−1 −2 h
(c) For which value of h is the triangle ABC a right triangle (with B being
the right angle).
We want the angle between BA and BC to be π/2, so we compute: BA = h1, 0, 1i,
BC = h0, −2, h − 1i, and if we let θ be the angle between them, then cos θ =
BA·BC
√ h−1 2 . If θ is to be equal to π/2, then cos(θ) must be 0, so we
|BA||BC| =
need to
2(4+(h−1) )
solve √ h−1 2
2(4+(h−1) )
= 0, which is clearly the case when h = 1.
(d) Find an equation for the plane through A, B, and C.
We need to find a point on the plane and a normal vector to the plane. The point
is easy–A will do. For the normal, we use AB × AC, which we computed above to
be n = h2, h − 1, 2i. Then the equation for the plane will be n · hx, y, zi = n · A,
i.e. 2x + (h − 1)y + 2z = 4 + 2(h − 1).
(e) For which calue of h does the plane through ABC contain the origin?
The plane described by the above equation contains the origin if and only if x =
y = z = 0 is a solution to the equation, i.e. if 0 = 4 + 2(h − 1), which we simplify
to 0 = 2 + 2h, so h = −1.
2. Given points A = (−1, 1, 4), B = (0, 2, 2), C = (2, 1, 1), find the following:
AB·AC
(a) Angle BAC. AB = h1, 1, −2i , AC = h3, 0, −3i. θ = cos−1 ( |AB||AC|
) = cos−1 ( √69√18 ) =
9
cos−1 ( 6√
) = cos−1 (
3
√
3
2 ).
(b) The equation of the plane containing the points A, B, C.
Need a point and a normal. Point is A = (−1, −1, 4), vector is n = AB × AC =
h−3, −3, −3i, so the equation is n · X = n · A, i.e. −3x − 3y − 3z = 3 + 3 − 12 = −6
(which, if we like, we can simplify to x + y + z = 2
(c) The point where the line through B perpendicular to the line AC meets
the line AC.
We could go through the rigamarole of finding the equations of the line and on and
on, but if we draw a picture, we realise that the point we want is really just the
endpont of the projection of AB onto AC, when we think of this vector starting
from A. And the projection is given by AB·AC
AC = 21 h3, 0, −3i, so the point is
|AC|2
(−1 + 23 , 1 + 0, 4 − 32 ) = ( 12 , 1, 52 ).
1
3. Consider the parametric curve r(t) = t, t2 , t3 defining a curve C in 3-space.
(a) Show that the point (1, 1, 1) lies on the curve.
r(1) = h1, 1, 1i.
(b) Find an equation for the tangent line to the curve C at the point (1, 1, 1).
We have a point on this line, namely (1, 1, 1), so we just need a vector in the right
direction, which will be r0 (1) = h1, 2, 3i, so the line is x = 1+t, y = 1+2t, z = 1+3t.
(c) Find the intersection of the tangent line you found above with the xyplane. (If you didn’t find an answer for the tangent line, you may work
with the (wrong) equation x = 2 + 6t, y = 3 + 7t, z = 4 + 8t for full credit.)
The xy plane is the set of points where z = 0, so setting z = 0 in the equations of
the line, we get t = −1
3 , and so the intersection point is the point on the line where
t = − 13 , i.e. ( 32 , 31 , 0).
4. Given vectors u = h1, 2, 3i and v = h1, 1, 0i, verify that u×(u×v) = (u·v)u−(u·u)v.
We just compute the left and right sides separately and note that they are equal.
Left side: u × h−3, 3, −1i = h−11, −8, 9i
Right side: 3 h1, 2, 3i − 14 h1, 1, 0i = h−11, −8, 9i.
Behold the equality!
5. Consider the parametric curve given by r(t) = hcos t + t sin t, sin t − t cos ti.
(a) Let A be the point on the curve with t = π/4. Find a parametric representation for the tangent line to the curve at the point A.
√
We need a point on the line, namely r(π/4) = 22 (1 + π4 , 1 − π4 ), and a tangent vector, namely r0 (pi/4). First, r√0 (t) = h− sin t + sin t + t cos t, cos t − cos t + t sin ti =
t hcos t, sin ti, so r0 (π/4) = π4 22 h1, 1i, and thus the line is:
x=
√
2
2 (1
+ π4 ) +
√
π 2
8 t, y
√
=
2
2 (1
− π4 ) +
√
π 2
8 t.
(b) For which value of t with 0 ≤ t ≤ π is the tangent vector to our parametric
curve perpendicular to the vector v = h1, −1i?
The tangent vector is r0 (t) = ht cos t, t sin ti as computed above, and so to find
when this is perpendicular to h1, −1i, we dot the two and see when it is zero. That
is, we need a t such that t cos t − t sin t = 0, i.e. t cos t = t sin t. This happens when
t = 0 or when cos t = sin t, which happens when t = π4 .
(c) Find the length of the curve segment corresponding to the parameter
values 0 ≤ t ≤ π/2
R π/2
R π/2
2
L = 0 |r0 (t)|dt = 0 tdt = π8 .
2