(Precalc) gang help me out here on the second Question

[pure mason]

For the first question I got x is about equal to 2.09489..

But for the second question I'm not sure how to approach this one? I think I have to use dot product right? Otherwise how would I make the vector orthogonal...

[eager rune]

Yeah dot or scalar product however it's called and set it equal to 0

However it's important to know how the dot product is defined for functions, i think it was some integral

But now I am thinking maybe the notation of the vectors are off and those are actual vectors

from R²

[pure mason]

Huh wait how is the notation of the vectors off?

[eager rune]

Usually they look like this

p = \left(
\begin{array}{c}
x\
5\
\end{array}
\right)

[stiff oriole]

adonhs
Compile Error! Click the reaction for more information.
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[eager rune]

and the notation on your paper resembles more the dot product but anyway

[pure mason]

ah

I don't think we've learned that notation yet

I'll stick to dot product then

[eager rune]

yeah whatever the point is to do b you have to make the dot product of p and q and make it equal to 0

$\vec{p} \cdot \vec{q} = 0$

[stiff oriole]

adonhs

[pure mason]

Ohhh

Ok

[eager rune]

If the dot product is 0 then the vectors are orthogonal/perpendicular

So you do this and solve for x

You should get a quadratic equation to solve

[stable imp]

i have no clue what vectors are other than they r arrows that come from the origin

so if they are perpendicular then doesnt that mean the slope of one of the vectors has to be -1/(slope of the other vector) ?

so -1/(-(6/5)/3x) = 5/x

and then solve for x

i got x = +- sqrt2

[pure mason]

Yall give me a min

[pure mason]

I got the quad equation you were talking about tho

[eager rune]

You should get the same result as your friend Jeff

[pure mason]

Yup

Thanks guys!

you too Jeff

[stable imp]

np lil bro

[pure mason]

.close