DLA Discovery guide

In this activity, make sure you can answer the questions for all dimensions, and make sure you can justify your answer using the formula for norm from Discovery 12.2, not just geometrically.

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(a)

Can

‖ v ‖

ever be negative?

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(b)

What is

‖ 0 ‖ ?

0

the only vector that has this value for its norm?

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(c)

Complete the formulas:

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$‖ 2 v ‖ = ‖ v ‖$
$‖ - 2 v ‖ = ‖ v ‖$
$‖ k v ‖ = ‖ v ‖$

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Discovery 12.4.

A unit vector is one whose norm is equal to

1 .

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(a)

Verify that the standard basis vectors are all unit vectors, in all dimensions.

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(b)

Fill in the blanks with an appropriate scalar multiple.

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If $‖ u ‖ = 1 / 2,$ then $u$ is a unit vector.
If $‖ w ‖ = 2,$ then $w$ is a unit vector.
For every nonzero $v,$ $k v$ is a unit vector for both $k =$ and $k = .$

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Discovery 12.5.

Plot points

P (1, 3)

and

Q (4, - 1)

in the

x y

-plane. Now draw in the vectors

u

and

v

that correspond to

\vec{O P}

and

\vec{O Q} .

Complete the triangle by drawing a vector between

P

and

Q .

Do you remember how to express this vector as a combination of

u

and

v ?

Now compute the distance between

P

and

Q

by computing the norm of this third vector.

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Recall that in math we measure angles in radians. Here are some common conversions:

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\begin{aligned} 30^{\circ} & = π / 6 rad, & 90^{\circ} & = π / 2 rad, \\ 45^{\circ} & = π / 4 rad, & 180^{\circ} & = π rad . \\ 60^{\circ} & = π / 3 rad, \end{aligned}

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Discovery 12.6.

(a)

In the

x y

-plane, what is the angle between

e_{1}

and

e_{2} ?

… between

e_{1}

and

u = (1, 1) ?

… between

e_{1}

and

2 e_{1} ?

… between

e_{1}

and

- e_{2} ?

… between

e_{1}

and

v = (1, - 1) ?

… between

e_{1}

and

- e_{1} ?

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(b)

Fill in the blanks: an angle

θ

between a pair of two-dimensional vectors should satisfy

\leq θ \leq .

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Discovery 12.7.

In the diagram below, consider

u

and

v

to be two-dimensional vectors. Label the third vector with the appropriate combination of

u

and

v,

just as you did in Discovery 12.5.

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Vector diagram of the law of cosines.

There is a version of Pythagoras that applies here even though

θ \neq 90^{\circ},

called the law of cosines:

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a^{2} + b^{2} - c^{2} = 2 a b \cos θ,

where

a

is the length of

u,

b

is the length of

v,

and

c

is the length of the “hypotenuse” across from

θ .

(If

θ

were

90^{\circ},

the right-hand side of this equality would be zero and this law would “collapse” to the same equality as Pythagoras.)

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Use the formulas from Discovery 12.2 to rewrite the left-hand side of the law of cosines in terms of the components of

u = (u_{1}, u_{2})

and

v = (v_{1}, v_{2}),

then simplify until you get

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2 \times (simple formula) .

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Using the new expression

2 \times (simple formula)

from Discovery 12.7 as the left-hand side in the law of cosines, and dividing both sides by

2 a b,

we get

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\cos θ = \frac{(simple formula)}{‖ u ‖ ‖ v ‖} .

(Remember that

a

and

b

are the lengths of

u

and

v,

respectively.)

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The “simple formula” part of this angle formula turns out to be an important one — it is called the Euclidean inner product or standard inner product (or just simply the dot product) of

u

and

v,

and written

u \cdot v .

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Discovery 12.8.

Let’s extend the computational pattern from Discovery 12.7. In the two-dimensional case in Task a below, you should just enter the “simple formula” you discovered above. In the subsequent tasks in higher dimensions, use the pattern from the two-dimensional case to create a similar higher-dimensional formula.

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