DLA Discovery guide

Discovery guide 14.1 Discovery guide
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Discovery 14.1.

Begin with a set of

x y

-axes. Draw the vector

x_{0} = (3, 0)

with its tail at the origin, and then draw the vector

p = (2, 1)

with its tail at the head of

x_{0} .

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(a)

Consider the expression

x = x_{0} + t p

in the parameter t. Think of

x

as a variable vector: using different values of

t,

x

evaluates to different vectors. Draw the vector

x

for

t = 1

on your diagram with its tail at the origin and using a dashed line for the shaft of the arrow. Then do the same for

t = 2,

t = - 1,

t = 1 / 2,

t = - 3 .

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(b)

Suppose you continued sketching in the different possible

x

vectors forever, using every possible value for the parameter

t .

What shape would be traced out by all of the points at the heads of the different versions of

x ?

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Discovery 14.2.

The equation

x - 2 y = 3

defines a line

ℓ

R^{2} .

We can also view this equation as a system of linear equations. Its solution requires one parameter.

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(a)

Set

y = t

and then compute the parametric equation for

x .

Set

x

to be the variable vector

x = [\begin{matrix} x \\ y \end{matrix}] .

Fill in the vectors at below. Then compare with Discovery 14.1.

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x = [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} t \end{matrix}] = [\begin{matrix}  \end{matrix}] + t [\begin{matrix}  \end{matrix}]

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(b)

Use the line equation

x - 2 y = 3

to verify that the point

(4, 1 / 2)

lies on

ℓ .

Then determine the value of the parameter

t

so that

x = (4, 1 / 2) .

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Discovery 14.3.

Consider the two planes

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\begin{aligned} Π_{1} & : 2 x - y + 5 z = - 5, & Π_{2} & : x + 2 y - 5 z = 10 \end{aligned}

R^{3} .

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(a)

Verify that

Π_{1}

and

Π_{2}

are not parallel.

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Hint.

Compare their normal vectors.

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(b)

Two nonparallel planes must intersect in a line. Describe the line of intersection of

Π_{1}

and

Π_{2}

in the form

x = x_{0} + t p .

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Hint.

Any point in the intersection must lie on both planes at once. That is, any point in the intersection must be a solution to the system of equations formed by the two plane equations.

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Discovery 14.4.

The equation

x - y + 5 z = - 5

defines a plane in

R^{3} .

We can also view this equation as a system of linear equations.

🔗

(a)

Similarly to Discovery 14.2, determine vectors

x_{0},

p_{1},

and

p_{2}

so that

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x = [\begin{matrix} x \\ y \\ z \end{matrix}] = x_{0} + s p_{1} + t p_{2}

describes all solutions to the equation (and hence all points on the plane).

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(b)

Use the plane’s equation

x - y + 5 z = - 5

to verify that the point

(1, 1, - 1)

lies on the plane.

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Then determine the values of the parameters

s

and

t

so that the formula

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x = x_{0} + s p_{1} + t p_{2}

results in this point

x = (1, 1, - 1) .

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Discovery 14.5.

Draw a grid over the

x y

-plane, with a vertical line at each integer value of

x

and a horizontal line at each integer value of

y .

Then draw

e_{1}

and

e_{2}

on your diagram.

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What does the decomposition

(3, 2) = 3 e_{1} + 2 e_{2}

look like on your grid?

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How about

(- 1, 2) = (- 1) e_{1} + 2 e_{2} ?

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How about

(3 / 2, - 2) = (3 / 2) e_{1} + (- 2) e_{2} ?

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Discovery 14.6.

Draw a “grid” over the

x y

-plane as follows: at each integer value along the

x

-axis, draw both a vertical line and a slant line parallel to the line

y = x .

Then draw

u = (1, 1)

and

e_{2}

on your diagram.

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What does the decomposition

(3, 2) = 3 u + (- 1) e_{2}

look like on your grid?

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How about

(- 1, 2) = (- 1) u + 3 e_{2} ?

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How about

(3 / 2, - 2) = (3 / 2) u + (- 7 / 2) e_{2} ?

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Discovery 14.7.

The set of all solutions to the homogeneous equation

x - 2 y + 3 z = 0

forms a plane in

R^{3} .

We can solve this equation by assigning parameters

y = s

and

z = t,

so that all solutions can be described parametrically by

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(x, y, z) = s (2, 1, 0) + t (- 3, 0, 1) .

Discuss how the vectors

p_{1} = (2, 1, 0)

and

p_{2} = (- 3, 0, 1)

create a “grid” on the plane defined by

x - 2 y + 3 z = 0,

similarly to the grids you worked with in Discovery 14.5 and Discovery 14.6.

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Discovery 14.8.

Determine the point of intersection of the line

ℓ,

described parametrically below left, and the plane

Π,

described algebraically below right.

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\begin{aligned} ℓ & : x = (2, 0, 3) + t (- 1, 1, 1) & Π & : 2 x + y - 3 z = 7 \end{aligned}

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Hint.

The point of intersection is simultaneously on the line and on the plane.

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Discovery 14.9.

Set up a system of equations whose solution is the point of intersection of the line

ℓ

and the plane

Π,

described parametrically below.

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\begin{aligned} ℓ & : x = (2, 0, 3) + t (- 1, 1, 1) & Π & : x = (3, 1, 0) + r (1, 1, 1) + s (3, 0, 2) \end{aligned}

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Hint.

The point of intersection is simultaneously on the line and on the plane.

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Discovery guide 14.1 Discovery guide A4US

Discovery 14.1.

(a)

(b)

Discovery 14.2.

(a)

(b)

Discovery 14.3.

(a)

(b)

Discovery 14.4.

(a)

(b)

Discovery 14.5.

Discovery 14.6.

Discovery 14.7.

Discovery 14.8.

Discovery 14.9.

Discovery guide 14.1 Discovery guide
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