Supriya is decorating her bedroom. She plans to hang strings of twinkle lights all the way around her walls. She measures the wa
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Answer:
(1) No matter what's the value of ,
is never in the span of
and
.
(2) The three vectors ,
, and
are always linearly dependent for all real
.
Step-by-step explanation:
<h3>(a)</h3>If is in the span of
and
, there need to exist real
and
such that
.
Assume that such and
do exist.
In other words,
.
.
.
Rewrite as an augmented matrix and row-reduce:
.
(Add four times row one to row two and times row one to row three.)
.
Note that in row two,
- Left-hand side:
;
- Right-hand side:
.
In other words, this system is inconsistent. There's no real and
that would satisfy the condition
.
Hence .
There's no real that allows
,
to be part of the span of
and
.
If the three vectors are linearly dependent, at least one of them can be expressed as the linear combination of the other two.
Note that
. In other words,
can be written as the linear combination of the other two vectors. Additionally, since the coefficient in front of
is zero, neither the exact value of
nor the value of
will make a difference. Therefore, for all
, the three vectors
,
, and
are linearly dependent.
- Slope-Intercept Form: y = mx + b, with m = slope and b = y-intercept.
If two lines are perpendicular, then they will have slopes that are <u>negative reciprocals</u> to each other. An example of negative reciprocals are 2 and -1/2
<h2>6.</h2>Now with line 2, I have to convert it to slope intercept form. Firstly, subtract 2x on both sides of the equation:
Next, divide both sides by -5 and your slope-intercept form is
Now since 2/5 is <em>not</em> the negative reciprocal of -2/5, <u>these lines are not perpendicular.</u>
<h2>7.</h2>It's pretty much the same process; convert to slope-intercept and determine if negative reciprocal. This time I'll brush through them:
Now since 2 <em>is</em> the negative reciprocal of -1/2, <u>these lines are perpendicular.</u>