MATH-4600
Logan Carter
Due: Monday January 13, 2025
Problem Set 1
Submissions are due in the LMS by the end of the day.
1. Given two vectors v and w.
(a) Determine v∥ (component of v parallel to w) and v⊥ (component of v perpendicular to
w).
(b) What is the area of the triangle with sides v and w?
Solution part a
The component for v along w is given as follows:
v·w
v∥ = ∥w∥
Using the Pythagorean theorem, we have:
∥v∥ ∥2 + ∥v⊥ ∥2 = ∥v∥2
Solving for ∥v⊥ ∥, we get
∥v⊥ ∥ =
q
∥v∥2 − ∥v∥ ∥2
s
∥v⊥ ∥ =
∥v∥2 −
v·w
∥w∥
2
Solution (Part b)
The area of the parallelogram defined by sides v and w is the length of their normal vector,
.
∥v × w∥, so the area of the triangle with sides v and w is half that, ∥v×w∥
2
2. Suppose the 3 points A = (a1 , a2 , a3 ), B = (b1 , b2 , b3 ), and C = (c1 , c2 , c3 ) in R3 are given.
(a) Determine an equation for the plane defined by A, B, and C, as well as an equation for
its normal.
(b) Determine an equation for the area of the triangle with vertices A, B, and C assuming
a3 = b3 = c3 = 0.
Solution part a Solving for the Plane Equation:
The target plane equation is given as:
n1 x + n2 y + n 3 z = D
To compute n1 , n2 , n3 , we use the cross product of vectors v = ⟨b1 − a1 , b2 − a2 , b3 − a3 ⟩ and
w = ⟨c1 − a1 , c2 − a2 , c3 − a3 ⟩:
i
j
k
v × w = b1 − a1 b2 − a2 b3 − a3
c1 − a1 c2 − a2 c3 − a3
1
Expanding the determinant:

  
(b2 − a2 )(c3 − a3 ) − (b3 − a3 )(c2 − a2 )
n1
v × w = (b3 − a3 )(c1 − a1 ) − (b1 − a1 )(c3 − a3 ) = n2 
(b1 − a1 )(c2 − a2 ) − (b2 − a2 )(c1 − a1 )
n3
To find D, we can substitute any point into the plane equation n1 x + n2 y + n3 z = D, lets
pick A = (a1 , a2 , a3 ):
D = n1 · a1 + n2 · a2 + n3 · a3
Finding D:
We start with the equation for D:
D = (b2 −a2 )(c3 −a3 )−(b3 −a3 )(c2 −a2 ) a1 + (b3 −a3 )(c1 −a1 )−(b1 −a1 )(c3 −a3 ) a2 + (b1 −a1 )(c2 −a2 )−(b2 −a2 )
Expanding each term:
1. First term:
(b2 − a2 )(c3 − a3 ) − (b3 − a3 )(c2 − a2 ) a1
(b3 − a3 )(c1 − a1 ) − (b1 − a1 )(c3 − a3 ) a2
(b1 − a1 )(c2 − a2 ) − (b2 − a2 )(c1 − a1 ) a3
2. Second term:
3. Third term:
Combine all terms:
D = a1 (b2 −a2 )(c3 −a3 )−(b3 −a3 )(c2 −a2 ) +a2 (b3 −a3 )(c1 −a1 )−(b1 −a1 )(c3 −a3 ) +a3 (b1 −a1 )(c2 −a2 )−(b2 −a
Simplifying further, we expand and rearrange terms:
D = a1 b2 c3 − a1 b3 c2 + a2 b3 c1 − a2 b1 c3 + a3 b1 c2 − a3 b2 c1
Which is the determinant of the following matrix:
a1 b1 c1
D = a2 b2 c2
a3 b3 c3
Thus, the scalar D is calculated as the determinant of the matrix formed by the coordinates
of the three points A = (a1 , a2 , a3 ), B = (b1 , b2 , b3 ), and C = (c1 , c2 , c3 ).
The equation for the plane defined by points A, B, and C is as follows:
a1 b1 c1
n1 x + n2 y + n3 z = a2 b2 c2
a3 b3 c3
where:
n1 = (b2 − a2 )(c3 − a3 ) − (b3 − a3 )(c2 − a2 ),
2
n2 = (b3 − a3 )(c1 − a1 ) − (b1 − a1 )(c3 − a3 ),
n3 = (b1 − a1 )(c2 − a2 ) − (b2 − a2 )(c1 − a1 ).
Solution part b
We are tasked to find the area of the triangle with vertices A = (a1 , a2 , 0), B = (b1 , b2 , 0),
and C = (c1 , c2 , 0) in R2 .
Let’s define the vectors:
v = ⟨c1 − a1 , c2 − a2 , 0⟩,
w = ⟨b1 − a1 , b2 − a2 , 0⟩
The area of the triangle is given by:
Area =
∥v × w∥
2
The cross product is as follows:
i
j
k
v × w = c1 − a1 c2 − a2 0
b1 − a1 b2 − a2 0
Expanding the determinant:
v × w = i(0) − j(0) + k [(b2 − a2 )(c1 − a1 ) − (b1 − a1 )(c2 − a2 )]
v × w = ⟨0, 0, (b2 − a2 )(c1 − a1 ) − (b1 − a1 )(c2 − a2 )⟩
The magnitude of the cross product is:
∥v × w∥ = |(b2 − a2 )(c1 − a1 ) − (b1 − a1 )(c2 − a2 )|
Substituting in the formula for the area:
Area of triangle =
|(b2 − a2 )(c1 − a1 ) − (b1 − a1 )(c2 − a2 )|
∥v × w∥
=
2
2
3. Consider surfaces S1 defined by sin2 (x)ecos(y) + z 3 = 2, and S2 defined by x2 + y 2 = 1.
(a) Determine an equation for the tangent plane to S1 at the point ( π2 , π2 , 1).
(b) Determine an equation for the intersection of S2 and the tangent plane determined in
part (a) above.
3
Solution part a
We are tasked with finding the tangent plane to the surface Sa defined by:
sin2 (x)ecos(y) + z 3 = 2
at the point
π π
2, 2,1 .
2
Let f (x, y, z) = sin (x)ecos(y) + z 3 − 2. Then, the gradient of f is:
∂f ∂f ∂f
∇f =
,
,
∂x ∂y ∂z
Computing each partial derivative:
∂f
= 2 sin(x) cos(x)ecos(y)
∂x
∂f
= − sin2 (x) sin(y)ecos(y)
∂y
∂f
= 3z 2
∂z
Evaluating the Gradient at (x0 , y0 , z0 ) = π2 , π2 , 1 :
π π π
∂f
= 2 sin
cos
ecos( 2 )
∂x
2
2
0
=2·1·0·e =0
π π π
∂f
= − sin2
sin
ecos( 2 )
∂y
2
2
= −12 · 1 · e0 = −1
∂f
= 3z 2 = 3 · 12 = 3
∂z
Thus:
∇f
π π , , 1 = ⟨0, −1, 3⟩
2 2
The equation of the tangent plane is:
∂f
∂f
∂f
(x − x0 ) +
(y − y0 ) +
(z − z0 ) = 0
∂x
∂y
∂z
Substituting in the values:
0(x −
π
π
) + (−1)(y − ) + 3(z − 1) = 0
2
2
−y +
π
+ 3z − 3 = 0
2
y − 3z + 3 −
4
π
=0
2
The equation for the tangent plane is as follows:
y − 3z + 3 −
π
=0
2
Solution part b
Finding intersections of: x2 + y 2 = 1 and y − 3z + 3 − π2 = 0
Rearranging the plane equation,
y = 3z − 3 +
π
.
2
Substituting this expression for y into the cylinder equation x2 + y 2 = 1 gives:
2
x2 + 3z − 3 + π2 = 1.
Solving for x, we get
x=±
q
1 − 3z − 3 + π2
2
.
Thus, the intersection of the two surfaces is given by the set of all points (x, y, z) satisfying:

y = 3z − 3 + π2 ,
q
2

x = ± 1 − 3z − 3 + π2 .
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