Question

Questions about vector
i will attach the questions

Answer 1

.

Answer 2

Hint:
\[\Large \begin{bmatrix}4 & -6 & -8\end{bmatrix}\begin{bmatrix}x & y & z\end{bmatrix}^{T} = 0\]
\[\Large \begin{bmatrix}4 & -6 & -8\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix} = 0\]
\[\Large 4x-6y-8z = 0\]

Answer 3

Another hint: isolate x in the first equation
\[\Large x+2y-3z = 0\]
\[\Large x+2y-3z-2y = 0-2y\]
\[\Large x-3z = -2y\]
\[\Large x-3z+3z = -2y+3z\]
\[\Large x = -2y+3z\]

Answer 4

Still have no idea howcto do this

Answer 5

So we have these two equations:
x = -2y+3z
4x-6y-8z = 0

Answer 6

Focus on the second equation. Replace x with -2y+3z

4x-6y-8z = 0

4(-2y+3z)-6y-8z = 0

-8y+12z-6y-8z = 0

-14y+4z = 0

Agreed so far?

Answer 7

Ok, just want to ask are the two vectors u just mentioned orthogonal to the plane?

Answer 8

I haven't mentioned any vectors yet. Those are equations.

Answer 9

What do you get when you isolate z in -14y+4z = 0 ?

Answer 10

No x direction?

Answer 11

no x will go away after you plug in x = -2y+3z

Answer 12

I don't know

Answer 13

solve for z
\[\Large -14y+4z = 0\]
\[\Large -14y+4z+14y = 0+14y\]
\[\Large 4z = 14y\]
\[\Large \frac{4z}{4} = \frac{14y}{4}\]
\[\Large z = \frac{7y}{2}\]
Does this make sense?

Answer 14

Yes

Answer 15

so we will plug that into x = -2y+3z

Answer 16

\[\Large x = -2y+3z\]
\[\Large x = -2y+3\left(\frac{7y}{2}\right)\]
\[\Large x = -2y+\frac{3*7y}{2}\]
\[\Large x = -2y+\frac{21y}{2}\]
\[\Large x = \frac{-4y}{2}+\frac{21y}{2}\]
\[\Large x = \frac{-4y+21y}{2}\]
\[\Large x = \frac{17y}{2}\]

Answer 17

making sense?

Answer 18

Yes

Answer 19

To summarize, we have
x = 17y/2
and
z = 7y/2


if y = k, then we'd have this vector \[\Large \begin{bmatrix}x & y & z\end{bmatrix}^{T} = \begin{bmatrix}\frac{17k}{2} & k & \frac{7k}{2}\end{bmatrix}^{T}\]

Answer 20

Another way to write it using different notation would be \[\Large \left<x,y,z\right> = \left<\frac{17k}{2},k,\frac{7k}{2}\right> \]

Answer 21

k is any real number

Answer 22

But that's the answer? I dont understand why is it orthogonal to(4,-6,-8)

Answer 23

Pick any number you want for k. Let's say you pick k = 2

That would mean
\[\Large \left<\frac{17k}{2},k,\frac{7k}{2}\right>=\left<\frac{17*2}{2},2,\frac{7*2}{2}\right>=\left<17,2,7\right> \]
Do you agree?

Answer 24

Ok

Answer 25

now just dot product that vector with the given vector. Tell me what you get

Answer 26

Zero

Answer 27

so that tells us <17,2,7> and <4,-6,-8> are orthogonal vectors

Answer 28

pick another value of k. Plug it into \[\Large \left<\frac{17k}{2},k,\frac{7k}{2}\right>\] and then dot product that with <4,-6,-8>. Whatever you pick for k, the dot product result should be 0.

Answer 29

But when i substitute (4,-6,-8) into x+2y-3z, it is also equal to zero. Is that means these two vectors are orthogonal too?

Answer 30

`when i substitute (4,-6,-8) into x+2y-3z, it is also equal to zero`
you made an error somewhere

Answer 31

Sorry typo, i mean substitute(17,2,7)

Answer 32

yes any vector that is in the plane will dot product with <4,-6,-8> to get 0

Answer 33

that's essentially what \[\Large \begin{bmatrix}4 & -6 & -8\end{bmatrix}\begin{bmatrix}x & y & z\end{bmatrix}^{T} = 0\] means

Answer 34

But (17,2,7) should be on the plane of x+2y-3z=0 , so if the dot product of (17,2,7)and(1,2,-3) is zero, that means they are orthogonal?

Answer 35

<1,2,-3> is the normal vector of the plane x+2y-3z=0

if <x,y,z> is some vector and <x,y,z> dot product with <1,2,-3> = 0, then the vector <x,y,z> will be in the plane

Answer 36

https://services.math.duke.edu/education/ccp/materials/mvcalc/linesplanes/3dPlane1.gif

in that image, the blue vector is the normal vector
the red vectors are all in the plane. Pick any red vector and dot product it with the blue vector. You will get 0 as the result

Answer 37

Thxxxxxxxxx i got it!

Answer 38

btw I want to know can I express the answer using x as a free variable?

Answer 39

and please help me deal with the second question!!!