oliver (smo998) – HW 02 – rusin – (52850) 1

This print-out should have 20 questions.

Multiple-choice questions may continue on

the next column or page – ﬁnd all choices

before answering.

001 10.0 points

If ais a vector parallel to the xy-plane and

bis a vector parallel to k, determine ka×bk

when kak= 3 and kbk= 2.

1. ka×bk= 0

2. ka×bk=−3

3. ka×bk= 3

4. ka×bk= 6 correct

5. ka×bk=−6

6. ka×bk=−3√2

7. ka×bk= 3√2

Explanation:

For vectors aand b,

ka×bk=kakkbksin θ

when the angle between them is θ, 0 ≤θ < π.

But θ=π/2 in the case when ais parallel

to the xy-plane and bis parallel to kbe-

cause kis then perpendicular to the xy-plane.

Consequently, for the given vectors,

ka×bk= 6 .

keywords: cross product, length, angle,

002 10.0 points

A line ℓpasses through the point P(2,3,3)

and is perpendicular to the plane

x+ 4y+ 3z= 6 .

At what point Qdoes ℓintersect the xy-

plane?

1. Q(3,7,0)

2. Q(3,0,−1)

3. Q(1,−1,0) correct

4. Q(7,0,3)

5. Q(1,0,7)

6. Q(−1,1,0)

Explanation:

Since the xy-plane is given by z= 0, we

have to ﬁnd an equation for ℓand then set

z= 0.

Now a line passing through a point

P(a, b, c) and having direction vector vis

given parametrically by

r(t) = a+tv,a=ha, b, c i.

But for ℓ, its direction vector will be parallel

to the normal to the plane

x+ 4y+ 3z= 6 .

Thus

a=h2,3,3i,v=h1,4,3i,

and so

r(t) = h2 + t, 3 + 4t, 3 + 3ti.

In this case, z= 0 when t=−1. Conse-

quently, ℓintersects the xy-plane at

Q(1,−1,0) .

keywords: line, parametric equations, direc-

tion vector, point on line, intercept, coordi-

nate plane

003 10.0 points

Which one of the following shaded-regions

in the plane consists of all points whose polar

coordinates satisfy the inequalities

0≤r < 3,1

6π≤θ≤7

12π?

oliver (smo998) – HW 02 – rusin – (52850) 2

1.

2 4

2.

2 4

3.

2 4

4.

2 4

correct

5.

2 4

6.

2 4

Explanation:

Using the deﬁnition of polar coordinates

(r, θ), we see that the region deﬁned by the

inequalities

0≤r < 3,1

6π≤θ≤7

12π

is

2 4

keywords: polar coordinates, inequalities, po-

lar graph,

004 10.0 points

Find a polar equation for the curve given

by the Cartesian equation

4y2=x .

1. r= 4 csc θtan θ

2. 4r= sec θtan θ

3. 4r= sec θcot θ

4. 4r= csc θcot θcorrect

5. r= 4 csc θcot θ

6. r= 4 sec θtan θ

Explanation:

We have to substitute for x, y in

4y2=x

using the relations

x=rcos θ , y =rsin θ .

In this case the Cartesian equation becomes

4r2sin2θ=rcos θ .

oliver (smo998) – HW 02 – rusin – (52850) 3

Consequently, the polar form of the equation

is

4r= csc θcot θ .

005 10.0 points

Use the graph in Cartesian coordinates

r

θ

π2π

of ras a function of θto determine which

one of the following is the graph of the corre-

sponding polar function?

1.

2.

3.

4.

5. correct

6.

Explanation:

As the Cartesian graph of rnever crosses

the θ-axis, there is no value of θin [0,2π]

at which r= 0; in particular, therefore, the

polar graph never passes through the origin.

This eliminates 3 of the graphs, leaving only