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

[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.

 

[fiksx]

That is wrong. can be β² negative?
$\alpha^2+\beta^2=4$ and α=1, what is β²?But what values can β have?No, how come that $1+x^2=\sqrt 3$ ? This must be $\vec c \cdot \vec c$ which is 4.

Can not have β some other value?

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.

i just wanted to show that $x=\beta$ ,
anyway what i learn from this problem is from a.b=0, dot product=0 show that a and b is orthogonal to each other. so it imply a and b is independent to each other and it span the plane(?) while c is dependent to both a and b, it is in the span of a and b?
but without finding c coordinate, how can you know the angle between b and c can be 150 degree?

 

[ehild]

i just wanted to show that $x=\beta$ ,
anyway what i learn from this problem is from a.b=0, dot product=0 show that a and b is orthogonal to each other. so it imply a and b is independent to each other and it span the plane(?) while c is dependent to both a and b, it is in the span of a and b?
but without finding c coordinate, how can you know the angle between b and c can be 150 degree?

Yes, vector c is in the span of a and b. What is the angle between a and c (both of them)? From that, what can be the angle between b and c?
But I do not see that the angle was asked in the problem.

 

[fiksx]

Yes, vector c is in the span of a and b. What is the angle between a and c (both of them)? From that, what can be the angle between b and c?
But I do not see that the angle was asked in the problem. View attachment 226295

yes from a and c 60 degree, and b and c is 30 degree, i mean there is another way to solve this using, $x=b.c \cos \theta$ ,
$x=$
from $a.c \cos \theta=1$, $1.2 cos \theta=1$, angle between a and c is 60 degree.
$x=b.c \cos \theta$ $x=2.1 cos 30 = \sqrt3$
without finding coordinate, how do you know another angle is 150 degree?

 

[fiksx]

Yes, vector c is in the span of a and b. What is the angle between a and c (both of them)? From that, what can be the angle between b and c?
But I do not see that the angle was asked in the problem. View attachment 226295

ah i think i got it, there is possibility that 60 degree is obtuse or acute angle with b from a

 

[ehild]

yes from a and c 60 degree, and b and c is 30 degree, i mean there is another way to solve this using, $x=b.c \cos \theta$ ,
$x=$
from $a.c \cos \theta=1$, $1.2 cos \theta=1$, angle between a and c is 60 degree.

$1\cdot 2 cos \theta=1$ means that cosθ = 1/2. θ can be 60° or -60°. The angle between b and c can be 30°or 90+60= 150 °. See figure in the previous post.

 

[fiksx]

$1\cdot 2 cos \theta=1$ means that cosθ = 1/2. θ can be 60° or -60°. The angle between b and c can be 30°or 90+60= 150 °. See figure in the previous post.

ok thankyou so much for the help :D learn so much from your answer! sorry I am asking too much !

 

[ehild]

ok thankyou so much for the help :D learn so much from your answer! sorry I am asking too much !

You are welcome