Solutions to Problem Set 1

Math 311-01/02

1-25-2018

[Euclidean space]

1. Plot the points $(1, 2, 4)$ , $(-1, 2, -1)$ , and $(-3, -1, 3)$ .

$$\hbox{\epsfysize=1.5in \epsffile{euclidean-space-problems-1.eps}}$$

$$\hbox{\epsfysize=1.5in \epsffile{euclidean-space-problems-2.eps}}$$

$$\hbox{\epsfysize=1.5in \epsffile{euclidean-space-problems-3.eps}}\quad\halmos$$


2. Find the exact distance between $(2, -4, 1)$ and $(3, 1, -2)$ .

$$\sqrt{(2 - 3)^2 + (-4 - 1)^2 + (1 - (-2))^2} = \sqrt{1 + 25 + 9} = \sqrt{35}.\quad\halmos$$


3. Find the exact distance between $(6, 1, 0)$ and $(-1, 1, -3)$ .

$$\sqrt{(6 - (-1))^2 + (1 - 1)^2 + (0 - (-3))^2} = \sqrt{49 + 0 + 9} = \sqrt{58}.\quad\halmos$$


[Vectors]

4. Suppose

$$\vec{a} = (-3, -5, 1), \quad \vec{b} = (-6, -4, 5).$$

Find:

(a) $3 \cdot \vec{a}$ .

(b) $5 \cdot \vec{a} + 4 \cdot
   \vec{b}$

(c) $3 \cdot \vec{a} - 2 \cdot
   \vec{b}$

(a) $3 \cdot \vec{a} = (-9, -15,
   3)$ .

(b)

$$5 \cdot \vec{a} + 4 \cdot \vec{b} = (-15, -25, 5) + (-24, -16, 20) = (-39, -41, 25).\quad\halmos$$

(c)

$$3 \cdot \vec{a} - 2 \cdot \vec{b} = (-9, -15, 3) - (-12, -8, 10) = (3, -7, -7).\quad\halmos$$


5. Suppose

$$\vec{a} = 2 \ihat - 4 \jhat + 5 \khat, \quad \vec{b} = -3 \ihat + 2 \jhat - 7 \khat.$$

Find:

(a) $6 \cdot \vec{a}$ .

(b) $2 \cdot \vec{a} + 3 \cdot
   \vec{b}$

(c) $4 \cdot \vec{a} - \vec{b}$

(a) $6 \cdot \vec{a} = 12 \ihat -
   24 \jhat + 30 \khat$ .

(b)

$$2 \cdot \vec{a} + 3 \cdot \vec{b} = (4 \ihat - 8 \jhat + 10 \khat) + (-9 \ihat + 6 \jhat - 21 \khat) = -5 \ihat - 2 \jhat - 11 \khat.\quad\halmos$$

(c)

$$4 \cdot \vec{a} - \vec{b} = (8 \ihat - 16 \jhat + 20 \khat) - (-3 \ihat + 2 \jhat - 7 \khat) = 11 \ihat - 18 \jhat + 27 \khat.\quad\halmos$$


6. Compute

$$2 \cdot (1, 3, 4, 0) + 5 \cdot (-1, 2, 7, 1).$$

$$2 \cdot (1, 3, 4, 0) + 5 \cdot (-1, 2, 7, 1) = (-3, 16, 43, 5).\quad\halmos$$


7. Suppose $\vec{a} = (2, -4,
   1)$ .

(a) Find $\|\vec{a}\|$ .

(b) Find a unit vector in the same direction as $\vec{a}$ .

(c) Find a unit vector in the opposite direction to $\vec{a}$ .

(a) $\|\vec{a}\| = \sqrt{4 + 16 +
   1} = \sqrt{21}$ .

(b) $\dfrac{1}{\sqrt{21}} (2, -4,
   1)$ .

(c) $-\dfrac{1}{\sqrt{21}} (2,
   -4, 1)$ .


8. Find a vector with length 6 and the same direction as $(-4, -1, 8)$ .

$$\|(-4, -1, 8)\| = \sqrt{16 + 1 + 64} = 9.$$

Therefore, $\dfrac{1}{9} (-4, -1,
   8)$ is a unit vector with the same direction as $(-4, -1, 8)$ . Hence, $\dfrac{6}{9} (-4, -1, 8) =
   \dfrac{2}{3} (-4, -1, 8)$ is a vector with length 6 and the same direction as $(-4, -1, 8)$ .


9. (a) For the points $A(2, -1,
   5)$ and $B(3, 2, 7)$ , find $\bvec{A
   B}$ .

(b) Find a unit vector in the opposite direction to $\bvec{A B}$ .

(a) $\bvec{A B} = (1, 3, 2)$ .

(b) $\|\bvec{A B}\| = \sqrt{1 + 9
   + 4} = \sqrt{14}$ , so $-\dfrac{1}{\sqrt{14}} (1, 3, 2)$ is a unit vector in the opposite direction to $\bvec{A B}$ .


10. Prove that the vectors $(-4,
   -6, 8)$ and $(10, 15, -20)$ are parallel.

$$-\dfrac{5}{2} \cdot (-4, -6, 8) = (10, 15, -20).$$

Hence, the vectors are parallel.


11. Prove that the vectors $(2,
   4, 6)$ and $(3, -6, 9)$ are NOT parallel.

Suppose

$$k \cdot (2, 4, 6) = (3, -6, 9).$$

Then

$$(2 k, 4 k, 6 k) = (3, -6, 9).$$

This gives

$$2 k = 3, \quad 4 k = -6, \quad 6 k = 9.$$

The first equation gives $k =
   \dfrac{3}{2}$ , but the second gives $k = -\dfrac{3}{2}$ . This contradiction shows there is no such k. Hence, the vectors aren't parallel.


Life shrinks or expands in proportion to one's courage. - Anaïs Nin


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