Math 53 Quiz 2 Name:

Problem 1 [3 pts]. Let a and b be vectors. Show that |a b|2 + (a b)2 = |a|2|b|2

Problem 2 [4 pts]. Find a non-zero vector a that is perpendicular to the vectors b = 3, 1, 2 and c = 3, 2, 4.Verify the perpendicularity of a by explicitly calculating a b and a c.

Problem 3 [3 pts]. Consider the parallelogram determined by vectors a and b. Show that the two diagonals ofthe parallelogram are perpendicular only when |a| = |b|, i.e. the parallelogram is a rhombus. Hint: Find vectorrepresentations for the two diagonals.

1

Problem 4. You are walking along the Berkeley Marina one day after Math 53 lecture, when you come across aspiral seashell on the beach. You take it home and discover that its cross-section can be described by the polarequation r = aeb for some positive constants a and b.(a) [4 pts.] Sketch a graph for the polar equation from 0 2pi and find the total cross-sectional area of theshell.(b) [1 pt.] You measure the seashell and find r(0) = 10cm and r(2pi) = 1.25cm. Determine the constants a (incm) and b.

Note: Many spirals found in nature can be modelled using this logarithmic spiral. For instance, the arms of theMilky Way galaxy have a b value of around 0.212.

2

Solutions

Problem 1

|a b|2 + (a b)2 = |a|2|b|2 sin2 + |a|2|b|2 cos2 = |a|2|b|2(sin2 + cos2 ) = |a|2|b|2

Problem 2

To find a vector a perpendicular to both b and c we compute the cross product.

a = b c =x y z3 1 23 2 4

= 0, 18,9 . Any scalar multiple also works.To test perpendicularity we compute dot products.a b = (0)(3) + (18)(1) + (9)(2) = 0 + 18 18 = 0a c = (0)(3) + (18)(2) + (9)(4) = 0 + 36 36 = 0Thus a is perpendicular to both b and c.

Problem 3

Draw a parallelogram with sides a and b. The diagonals are then a+ b and a b. To check perpendicularity wecompute the dot product.(a+ b) (a b) = a a a b+ b a b b = |a|2 |b|2 = 0 which is true iff |a| = |b|

Problem 4

(a) We are computing the curve enclosed by the polar equation r = aeb from 0 to 2pi.

A =

2pi0

1

2r2d =

2pi0

1

2(aeb)2d =

a2

2

2pi0

e2bd = a2

4be2b

2pi0

=a2

4b

(1 e4pib

)

(b) r(0) = aeb0

= a = 10cm

r(2pi) = aeb2pi

= 10cm e2bpi = 1.25cm e2bpi = 0.125 = frac18 2bpi = ln 1/8 = 3 ln 2 b = 3 ln 22pi

3