Find All Possible Values of x When 3 Vectors are Linearly Dependent

[fiksx]

Homework Statement

>There are three vector $$ \vec a ,\vec b, \vec c$$ in three-dimensional real vector space, and the inner product between them $$\vec a . \vec a=\vec b.\vec b=\vec a.\vec c=1, \vec a.\vec b=0, \vec c.\vec c=4 $$ When setting $$x = \vec b.\vec c$$ ,
(dot here means dot product)
answer the following question: when $$ \vec a ,\vec b, \vec c$$ are linearly dependent, find all possible values of $$ x$$

Homework Equations

3. The Attempt at a Solution

For dependent condition

$$\begin{align}
(a×b)·c &= 0\\
a·(b×c) &= 0\\
a(bc \sin θ)&=0
\end{align}$$

So $\theta= 0$ and $\pi$

Then
$$\begin{align}
x&=|b||c| \cos \theta \\
x&=2 \cos \theta \\
\implies x &= 2 \cos 0, x = 2 \cos \pi \\
x &= \mp 2
\end{align}$$

Am I right?

 

[Gene Naden]

You haven't said what you mean by $\theta$ and your equation $a(bc (sin(\theta))=0$ looks suspect to me. Where do you take into account the angle between $\vec b \times \vec c$ and $\vec a$?

 

[fiksx]

You haven't said what you mean by $\theta$ and your equation $a(bc (sin(\theta))=0$ looks suspect to me. Where do you take into account the angle between $\vec b \times \vec c$ and $\vec a$?

$$\theta $$ is angle?
\begin{align}
(a×b)·c &= 0\\
a·(b×c) &= 0\\
a(bc \sin θ)&=0
\end{align}

i think this is cross product? i tried to apply scalar triple product like the one when you find area of parallelogram ? but i wasnt sure

 

[Ray Vickson]

Homework Statement

>There are three vector $$ \vec a ,\vec b, \vec c$$ in three-dimensional real vector space, and the inner product between them $$\vec a . \vec a=\vec b.\vec b=\vec a.\vec c=1, \vec a.\vec b=0, \vec c.\vec c=4 $$ When setting $$\vec x = \vec b.\vec c$$ ,
(dot here means dot product)
answer the following question: when $$ \vec a ,\vec b, \vec c$$ are linearly dependent, find all possible values of $$\vec x$$

Homework Equations

The Attempt at a Solution

For dependent condition

$$\begin{align}
(a×b)·c &= 0\\
a·(b×c) &= 0\\
a(bc \sin θ)&=0
\end{align}$$

So $\theta= 0$ and $\pi$

Then
$$\begin{align}
x&=|b||c| \cos \theta \\
x&=2 \cos \theta \\
\implies x &= 2 \cos 0, x = 2 \cos \pi \\
x &= \mp 2
\end{align}$$

Am I right?[/B]

(1) Remove the bold font; it looks like you are yelling.
(2) Remove the arrow above $x$, because $x$ is not a vector---it is a scalar.

 

[fiksx]

(1) Remove the bold font; it looks like you are yelling.
(2) Remove the arrow above $x$, because $x$ is not a vector---it is a scalar.

ok, done

 

[Dick]

ok, done

There's no cross product in your given relations. You know $|a|=1$ and rotating all of your vectors by the same amount doesn't change anything. So why not set $\vec a=(1,0,0)$ and see what the relations are telling you in terms of coordinates?

 

[fiksx]

There's no cross product in your given relations. You know $|a|=1$ and rotating all of your vectors by the same amount doesn't change anything. So why not set $\vec a=(1,0,0)$ and see what the relations are telling you in terms of coordinates?

thanks! but why set $\vec a=(1,0,0)$ and set to what? i though i can use cross product to test dependency?

 

[Dick]

thanks! but why set $\vec a=(1,0,0)$ and set to what? i though i can use cross product to test dependency?

The question isn't asking you about dependency, it's asking for a range of values. And like I said, the point to setting $\vec a=(1,0,0)$ is just for definiteness. You could set it to any vector with norm 1. Then put $\vec b=(b_x, b_y, b_z)$ and $\vec c=(c_x, c_y, c_z)$ and see what the relations say in terms of those coordinates.

 

[fiksx]

The question isn't asking you about dependency, it's asking for a range of values. And like I said, the point to setting $\vec a=(1,0,0)$ is just for definiteness. You could set it to any vector with norm 1. Then put $\vec b=(b_x, b_y, b_z)$ and $\vec c=(c_x, c_y, c_z)$ and see what the relations say in terms of those coordinates.

do you mean i can pick any vector that match the inner product?
$\vec a=(1, 0, 0)$ , $\vec b=(0, 1, 0)$ (?)
is it possible when $a.c cos \theta= 1$
$1.2 cos \theta=1$ and find the $\theta$?

 

[Dick]

do you mean i can pick any vector that match the inner product?
$\vec a=(1, 0, 0)$ , $\vec b=(0, 1, 0)$ (?)
is it possible when $a.c cos \theta= 1$
$1.2 cos \theta=1$ and find the $\theta$?

The picture you should have in your head is taking the three vectors and rotating them all together. And yes, since $\vec a$ and $\vec b$ are unit vectors and orthogonal you could choose them that way. But now you can't fix $\vec c$ since it can have a variety of positions with respect to the other two. Just see what the dot product relations tell you about the coordinates of $\vec c$.

 

[fiksx]

The picture you should have in your head is taking the three vectors and rotating them all together. And yes, since $\vec a$ and $\vec b$ are unit vectors and orthogonal you could choose them that way. But now you can't fix $\vec c$ since it can have a variety of positions with respect to the other two. Just see what the dot product relations tell you about the coordinates of $\vec c$.

does this mean that a and b and c is in the same plane? and since a and b are independent to each other, it span the plane? and c is inside it?
so i can write c as linear combination of a and b so c= ka+sb?

 

[Dick]

does this mean that a and b and c is in the same plane? and since a and b are independent to each other, it span the plane? and c is inside it?
so i can write c as linear combination of a and b so c= ka+sb?

I don't see anything in the problem that would indicate the vectors are coplanar. Why do you think that?

 

[fiksx]

I don't see anything in the problem that would indicate the vectors are coplanar. Why do you think that?

it is said that a.b=0 ?
if $\vec a=(1, 0, 0)$ and $\vec b=(0, 1, 0)$ ,it is independent to each other? so c is in the span of two of them since three of them are dependent? and they are on the same plane?

 

[Dick]

it is said that a.b=0 ?
if $\vec a=(1, 0, 0)$ and $\vec b=(0, 1, 0)$ ,it is independent to each other? so c is in the span of two of them since three of them are dependent? and they are on the same plane?

You're told they are three dimensional vectors. Two independent vectors don't span the space. Nothing in the problem indicates the three are dependent.

 

[fiksx]

You're told they are three dimensional vectors. Two independent vectors don't span the space. Nothing in the problem indicates the three are dependent.

it is said that when a, b, c are linearly dependent find possible value of x ?

You're told they are three dimensional vectors. Two independent vectors don't span the space. Nothing in the problem indicates the three are dependent.

it is said in the problem a,b,c are linear dependence
$1.2 cos \theta=1$ ,from here the angle between a and c is 60 degree? since a and b is orthogonal to each other b and c will be 30 degree? am i right?

 

[Dick]

it is said that when a, b, c are linearly dependent find possible value of x ?

it is said in the problem a,b,c are linear dependence
$1.2 cos \theta=1$ ,from here the angle between a and c is 60 degree? since a and b is orthogonal to each other b and c will be 30 degree? am i right?

All right. Sorry, now I see that it does say they are linearly dependent. So almost true. There's another choice of angle that works as well.

 

[fiksx]

All right. Sorry, now I see that it does say they are linearly dependent. So almost true. There's another choice of angle that works as well.

can i use the fact that a and b span the plane and c is inside the plane? so i can express c as linear combination of them altough it is in 3 dimentional?
x=b.c cos theta since theta is 30 , $x=b.c cos 30 = \sqrt 3$ ?

 

[Dick]

can i use the fact that a and b span the plane and c is inside the plane? so i can express c as linear combination of them altough it is in 3 dimentional?
x=b.c cos theta since theta is 30 , $x=b.c cos 30 = \sqrt 3$ ?

That's one possibility. There's another. There are two different positions in the plane where there's a 60 degree angle between $\vec a$ and $\vec c$. Draw a picture.

 

[fiksx]

That's one possibility. There's another. There are two different positions in the plane where there's a 60 degree angle between $\vec a$ and $\vec c$. Draw a picture.

$\vec a=(1,0,0)$ $\vec b=(0,1,0)$ $\vec c=(1,1,0)$ , i choose this point . and clearly c is in span of a and b, can you give me more hint?

 

[ehild]

What is the magnitude of $\vec c$? What can be the components then?

 

[fiksx]

What is the magnitude of $\vec c$? What can be the components then?

i was wrong , $\vec c=(1,1,\sqrt2)$

 

[fiksx]

That's one possibility. There's another. There are two different positions in the plane where there's a 60 degree angle between $\vec a$ and $\vec c$. Draw a picture.

sorry i was wrong let $\vec c=(1,1,\sqrt2)$

 

[fiksx]

What is the magnitude of c##? What can be the components then?

##\vec

can you point how can i see the angle here?

 

[ehild]

sorry i was wrong let $\vec c=(1,1,\sqrt2)$

Are the vectors $\vec a (1,0,0)$, $\vec b (0,1,0)$, $\vec c (1,1,√2)$ dependent?

 

[fiksx]

Are the vectors $\vec a (1,0,0)$, $\vec b (0,1,0)$, $\vec c (1,1,√2)$ dependent?

a and b independent but c is dependent to a and b ?

 

[ehild]

a and b independent but c is dependent to a and b ?

Two vectors of the three are independent, and the third can be combined from the two others. You can write vector c as linear combination of a and b vectors, and find the coefficients from the given data.

 

[fiksx]

Two vectors of the three are independent, and the third can be combined from the two others. You can write vector c as linear combination of a and b vectors, and find the coefficients from the given data.

yes but is the picture that i drew right? how can i find the angle between them?

##\vecView attachment 226262

 

[ehild]

You define angles between two vectors, a and c, b and c. But you draw the vector c out of the plane spread by a and b, as if it had a component perpendicular to both a and b, so independent from them. Remember, they are dependent in 3 dimension. That means they are in one plane.

 

[ehild]

##\vecView attachment 226262
can you point how can i see the angle here?

There are two angles here, between a and c and between b and c. Are the vectors in the drawing dependent?

 

[fiksx]

You define angles between two vectors, a and c, b and c. But you draw the vector c out of the plane spread by a and b, as if it had a component perpendicular to both a and b, so independent from them. Remember, they are dependent in 3 dimension. That means they are in one plane.

View attachment 226279

yeah in my picture i chose $c=(1,1, \sqrt2)$ , is it wrong?

There are two angles here, between a and c and between b and c. Are the vectors in the drawing dependent?
View attachment 226280

i think here i was wrong, the three vector are independent to each other in my drawing because they are not in same plane right?
so what coordinate should i choose for c? is $c=(1,1, \sqrt2)$ wrong?

 

[fiksx]

You define angles between two vectors, a and c, b and c. But you draw the vector c out of the plane spread by a and b, as if it had a component perpendicular to both a and b, so independent from them. Remember, they are dependent in 3 dimension. That means they are in one plane.

View attachment 226279

how can i draw like you drew c is in the span of a and b? what coordinate should i choose?

 

[fiksx]

There are two angles here, between a and c and between b and c. Are the vectors in the drawing dependent?
View attachment 226280

i think here i was wrong, the three vector are independent to each other in my drawing because they are not in same plane right?
so what coordinate should i choose for c? is $c=(1,1, \sqrt2)$ wrong?

 

[ehild]

yeah in my picture i chose $c=(1,1, \sqrt2)$ , is it wrong?

Is $\vec c=(1,1, \sqrt2)$ dependent of (1,0,0) and (0,0,1)? How can you decide if three vectors are independent or dependent?

i think here i was wrong, the three vector are independent to each other in my drawing because they are not in same plane right?
so what coordinate should i choose for c? is $c=(1,1, \sqrt2)$ wrong?

If the plane is spread by a and b, what is the third coordinate of all vectors in that plane?

 

[ehild]

i think here i was wrong, the three vector are independent to each other in my drawing because they are not in same plane right?
so what coordinate should i choose for c? is $c=(1,1, \sqrt2)$ wrong?

It is wrong, of course.
If a=(1,0,0) and b=(0,1,0), what should be the third coordinate of c so it is a linear combination of a and b?
The third coordinate is a linear combination of the third coordinates of both a and b, two zeros. What do you get if you combine two zeros?

 

[fiksx]

Is $\vec c=(1,1, \sqrt2)$ dependent of (1,0,0) and (0,0,1)? How can you decide if three vectors are independent or dependent?

the determinant is -1 so it is independent to each other?

If the plane is spread by a and b, what is the third coordinate of all vectors in that plane?

if the plane is spread by a and b, the third coordinate is linear combination of a and b

 

[fiksx]

It is wrong, of course.
If a=(1,0,0) and b=(0,1,0), what should be the third coordinate of c so it is a linear combination of a and b?
The third coordinate is a linear combination of the third coordinates of both a and b, two zeros. What do you get if you combine two zeros?

$c=k(1,0,0)+b(0,1,0)$

 

[ehild]

if the plane is spread by a and b, the third coordinate is linear combination of a and b

The coordinate is not linear combination of vectors.

 

[ehild]

$c=k(1,0,0)+b(0,1,0)$

It is better, but do not use b for coefficient: b was a vector.
So you can write $\vec c = k(1,0,0)+ m(0,1,0).$ The scalar product $\vec a \vec c$ =1 was given. What do you get for k?

 

[fiksx]

It is better, but do not use b for coefficient: b was a vector.
So you can write $\vec c = k(1,0,0)+ m(0,1,0).$ The scalar product $\vec a \vec c$ =1 was given. What do you get for k?

k is 1?
i just want to know what c coordinate should be and how can i draw it.

 

[ehild]

k is 1?
i just want to know what c coordinate should be and how can i draw it.

What is the magnitude of c?

 

[fiksx]

What is the magnitude of c?

2

 

[ehild]

2

yes. how do you get the magnitude of a vector from its components?

 

[fiksx]

yes. how do you get the magnitude of a vector from its components?

by squaring the coordinate and take square root of it
if i choose $c(1,1,\sqrt2)$, i got the magnitude 2 and it is also satisfy $a.c=1$ , is it right or wrong?

 

[ehild]

by squaring the coordinate and take square root of it
if i choose $c(1,1,\sqrt2)$, i got the magnitude 2 and it is also satisfy $a.c=1$ , is it right or wrong?

you know that $c(1,1,\sqrt2)$ is wrong, as it is independent of a and b.

 

[fiksx]

you know that $c(1,1,\sqrt2)$ is wrong, as it is independent of a and b.

so you mean i have to find c coordinate that is dependent to a and b and satisfy $a.c=1$ ?

 

[ehild]

so you mean i have to find c coordinate that is dependent to a and b and satisfy $a.c=1$ ?

yes, and |c|=2.

 

[fiksx]

so you mean i have to find c coordinate that is dependent to a and b and satisfy $a.c=1$ ?

first i chose $a (1,0,0) , b (0,1,0)$ because $a.b=0;b.b=0;a.a=0$
and suppose
$c(a,b,c)$
where $c.c = (a^2+b^2+c^2)$
and $a.c =1 , (1,0,0).(a,b,c)=1$
$a=1$,
and $c.c=4$
so $a^2+b^2+c^2=4$
$1+b^2+c^2=4$
$b^2+c^2=3$
since $b.b=1$
$c^2=2$
so $c= \sqrt 2$
is there another way to find c, beside this way?

 

[ehild]

first i chose $a (1,0,0) , b (0,1,0)$ because $a.b=0;b.b=0;a.a=0$
and suppose
$c(a,b,c)$
where $c.c = (a^2+b^2+c^2)$

You know already, that the third component of vector c must be zero.
And it is misleading if you denote the components and vectors with the same letter.

and $a.c =1 , (1,0,0).(a,b,c)=1$
$a=1$,
and $c.c=4$
so $a^2+b^2+c^2=4$
$1+b^2+c^2=4$
$b^2+c^2=3$
since $b.b=1$

You see, "b" as component of vector c is not the same as the magnitude of the vector b.

$c^2=2$
so $c= \sqrt 2$
is there another way to find c, beside this way?

This is all wrong, as was said to you a thousand times, You show a vector again, that is independent with a and b. Start with $\vec c$ a vector in the plane of $\vec a$ and $\vec b$ , but denote the components with other letters, not with a, b, c,

 

[fiksx]

You know already, that the third component of vector c must be zero.
And it is misleading if you denote the components and vectors with the same letter.You see, "b" as component of vector c is not the same as the magnitude of the vector b.This is all wrong, as was said to you a thousand times, You show a vector again, that is independent with a and b. Start with $\vec c$ a vector in the plane of $\vec a$ and $\vec b$ , but denote the components with other letters, not with a, b, c,

i know and i understood,
$c=\alpha a+ \beta b$
$a.c=1$
$a(\alpha a+ \beta b)=1$
$\alpha=1$
$c.c=(\alpha a+ \beta b).(\alpha a+ \beta b)=\alpha^2+\beta^2=4$
$\beta^2=+/- \sqrt 3$
$x=b.c$
$x=b(\alpha a+ \beta b)$
$x=\beta$

$\alpha^2+\beta^2=1+x^2=4$
do you mean this?

------since $\alpha=1 , \beta=+/-\sqrt 3$ ,$c= \alpha .a+\beta.b$
$c=1.(1,0,0)++/- \sqrt 3 (0,1,0)$
so $c= (1, +/-\sqrt 3, 0)$

it is clearly angle between b and c is 60 degree,
ah i got it. i see from the picture, angle b and c is 150 degree
another question is there another wy to find c without using this way?

 

[ehild]

i know and i understood,
$c=\alpha a+ \beta b$
$a.c=1$
$a(\alpha a+ \beta b)=1$
$\alpha=1$
$c.c=(\alpha a+ \beta b).(\alpha a+ \beta b)=\alpha^2+\beta^2=4$
$\beta^2=+/- \sqrt 3$

That is wrong. can be β² negative?
$\alpha^2+\beta^2=4$ and α=1, what is β²?

$x=b.c$
$x=b(\alpha a+ \beta b)$
$x=\beta$

But what values can β have?

$\alpha^2+\beta^2=1+x^2=\sqrt 3$[
do you mean this?

No, how come that $1+x^2=\sqrt 3$ ? This must be $\vec c \cdot \vec c$ which is 4.

------since $\alpha=1 , \beta=\sqrt 3$ ,

Can not have β some other value?

$c= \alpha .a+\beta.b$
$c=1.(1,0,0)+\sqrt 3 (0,1,0)$
so $c= (1, +/-\sqrt 3, 0)$

If β²=3 β can be √3 or -√3, but √3 is positive. At the end you wrote the correct solution, but take more care to the derivations.