Solutions to Problem Set 2

Math 311-01/02

1-26-2018

[Vectors]

1. Vectors $\vec{a}$ and $\vec{b}$ are shown below.

$$\hbox{\epsfysize=1in \epsffile{vectors-problems-1.eps}}$$

Sketch the vectors $2 \vec{b}$ , $2 \vec{a} + 3 \vec{b}$ , and $\vec{a} - 2 \vec{b}$ .

$$\hbox{\epsfysize=2.5in \epsffile{vectors-problems-2.eps}}\quad\halmos$$


[Dot product]

2. Compute the dot product $(2,
   -1, 5) \cdot (3, 1, -4)$ .

$$(2, -1, 5) \cdot (3, 1, -4) = (2)(3) + (-1)(1) + (5)(-4) = -15.\quad\halmos$$


3. Compute the dot product $(1,
   8, -3, 0, 4) \cdot (3, 1, 1, -2, 7)$ .

$$(1, 8, -3, 0, 4) \cdot (3, 1, 1, -2, 7) = (1)(3) + (8)(1) + (-3)(1) + (0)(-2) + (4)(7) = 36.\quad\halmos$$

We won't be using dot products in dimensions other than 2 or 3 in this course. The point of this problem is to show that it doesn't work any differently with higher-dimensional vectors.


4. Compute the dot product $(-4
   \ihat + 3 \jhat + \khat) \cdot (7 \ihat + 2 \jhat + 5 \khat)$ .

$$(-4 \ihat + 3 \jhat + \khat) \cdot (7 \ihat + 2 \jhat + 5 \khat) = (-4)(7) + (3)(2) + (1)(5) = -17.\quad\halmos$$


5. Find the exact value of the cosine of the angle between $(3, -5, 1)$ and $(2, 1, 1)$ .

Tell whether the vectors are orthogonal; if not, tell whether the angle between them is acute or obtuse.

$$\cos \theta = \dfrac{(3, -5, 1) \cdot (2, 1, 1)}{\|(3, -5, 1)\| \|(2, 1, 1)\|} = \dfrac{6 - 5 + 1}{\sqrt{35} \sqrt{6}} = \dfrac{2}{\sqrt{210}}.$$

The angle is acute.


6. Find the exact value of the cosine of the angle between $(3, -2, 1)$ and $(2, 1, -4)$ .

Tell whether the vectors are orthogonal; if not, tell whether the angle between them is acute or obtuse.

$$\cos \theta = \dfrac{(3, -2, 1) \cdot (2, 1, -4)}{\|(3, -2, 1)\| \|(2, 1, -4)\|} = \dfrac{6 - 2 - 4}{\sqrt{14} \sqrt{21}} = 0.$$

The vectors are orthogonal.


7. Find the exact value of the cosine of the angle between $\ihat + 4 \jhat + 3 \khat$ and $2
   \ihat + \jhat - 4 \khat$ .

Tell whether the vectors are orthogonal; if not, tell whether the angle between them is acute or obtuse.

$$\cos \theta = \dfrac{(\ihat + 4 \jhat + 3 \khat) \cdot (2 \ihat + \jhat - 4 \khat)}{\|\ihat + 4 \jhat + 3 \khat\| \|2 \ihat + \jhat - 4 \khat\|} = \dfrac{2 + 4 - 12}{\sqrt{26} \sqrt{21}} = \dfrac{-6}{\sqrt{546}}.$$

The angle is obtuse.


8. Find the scalar component of $(-4, 5, 1)$ in the direction of $(3, 1, 1)$ .

$$\comp_{(3, 1, 1)} (-4, 5, 1) = \dfrac{(-4, 5, 1) \cdot (3, 1, 1)}{\|(3, 1, 1)\|\|} = \dfrac{-12 + 5 + 1}{\sqrt{11}} = -\dfrac{6}{\sqrt{11}}.\quad\halmos$$


9. Find the scalar component of $\vec{u} = 2 \ihat - \jhat + 4 \khat$ in the direction of $\vec{v} = \ihat - \jhat + 4 \khat$ .

$$\comp_{\vec{v}} \vec{u} = \dfrac{(2 \ihat - \jhat + 4 \khat) \cdot (\ihat - \jhat + 4 \khat)}{\|\ihat - \jhat + 4 \khat\|} = \dfrac{2 + 1 + 16}{\sqrt{18}} = \dfrac{19}{\sqrt{18}}.\quad\halmos$$


10. Find the vector component of $(4, -1)$ in the direction of $(3, -4)$ .

$$\proj_{(3, -4)} (4, -1) = \dfrac{(4, -1) \cdot (3, -4)}{(3, -4) \cdot (3, -4)} (3, -4) = \dfrac{12 + 4}{25} (3, -4) = \dfrac{16}{25} (3, -4).\quad\halmos$$


11. Find the vector component of $(3, 0, -1)$ in the direction of $(5, 1, 2)$ .

$$\proj_{(5, 1, 2)} (3, 0, -1) = \dfrac{(3, 0, -1) \cdot (5, 1, 2)}{(5, 1, 2) \cdot (5, 1, 2)} (5, 1, 2) = \dfrac{15 + 0 - 2}{30} (5, 1, 2) = \dfrac{13}{30} (5, 1, 2).\quad\halmos$$


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