Solve the given problem involving vectors

[chwala]

Homework Statement

see attached

Relevant Equations

vectors

This is a textbook problem, the text book solution is $λ=7$ I think this may not be correct, unless the thinking of gradient was applicable, i.e since they lie on same straight line then the gradients should be the same

i.e $\dfrac{0.4}{0.3} = \dfrac{4}{λ -4}$

in this case then

$\dfrac{4}{3} = \dfrac{4}{λ -4}$


$λ=7$.

Otherwise, using algebra, unit vector of $\hat {AB} = \dfrac {\vec {AB}}{ |\vec{AB}|}$

i got the equations,

for ith component,

$\dfrac{λ-4}{\sqrt{(λ-4)^2+16} }= 0.3$

and

for jth component,

$\dfrac{4}{\sqrt{(λ-4)^2+16}} = 0.4$

...
$\sqrt{(λ-4)^2+16}=\dfrac{4}{0.4}$

$\sqrt{(λ-4)^2+16}=10$

$λ-4 = ± \sqrt 84$

not an integer value...

 

[anuttarasammyak]

A(4,-2) B(l,2)
AB vector (l-4,4)=(3,4)=10(0.3,0.4)
l=7

 

[chwala]

On the other hand, using scale factor, we have the unit vector as $0.3i + 0.4 j$, its magnitude is $0.5$

The vector $\vec AB$ is 10 times the unit vector. Therefore the magnitude of $\vec AB = 10 × 0.5 = 5$

Therefore,

${\sqrt{(λ-4)^2+16} } = 5$

$(λ-4)^2+16 = 25$

$(λ-4)^2 = 9$

$λ_1= 4+3=7$ and $λ_2=4-3=1$

 

[anuttarasammyak]

The root 1 gives AB vector (-3,4) which is not proportional to given (0.3,0.4). This happens because you see only magnitude of the vector.

 

[Mark44]

On the other hand, using scale factor, we have the unit vector as $0.3i + 0.4 j$, its magnitude is $0.5$

$0.3i + 0.4 j$ is not a unit vector for precisely the reason you show. The magnitude of a unit vector is 1.

 

[chwala]

$0.3i + 0.4 j$ is not a unit vector for precisely the reason you show. The magnitude of a unit vector is 1.

That is exactly what was causing my confusion, it truly is not a unit vector...unless i look at it from this perspective $\dfrac{1}{2} \left[ \dfrac{3}{5} i + \dfrac{4}{5}j\right]$ with the constant $0.5$ outside bracket being a scalar value.

 

[Mark44]

That is exactly what was causing my confusion, it truly is not a unit vector...unless i look at it from this perspective $\dfrac{1}{2} \left[ \dfrac{3}{5} i + \dfrac{4}{5}j\right]$ with the constant $0.5$ outside bracket being a scalar value.

The vector above is not a unit vector, but the vector in brackets is. The bracketed vector can also be written as $0.6\hat i + 0.8 \hat j$.

 

[chwala]

The vector above is not a unit vector, but the vector in brackets is. The bracketed vector can also be written as $0.6\hat i + 0.8 \hat j$.

Then it seems that the problem was to determine the unit vector from what was originally given ( from the definition of unit vector) and moreso to identify that $0.5$ is just but a scalar value. I think the problem required that understanding...with that said the problem is solvable unless you think otherwise. Cheers man!

 

[anuttarasammyak]

The problem might be using oblique coordinates :
\mathbf{i}=a\mathbf{e_i}
\mathbf{j}=b\mathbf{e_j}
|\mathbf{e_i}|=|\mathbf{e_j}|=1
\mathbf{e_i}\cdot\mathbf{e_j}=\cos\theta
which satisfy
0.09a^2+0.16b^2-0.12*2ab\cos\theta=1though it does not affect the answer.

 

[Mark44]

The problem might be using oblique coordinates

I doubt that the problem is using oblique coordinates.

The main problem I see is in the problem itself, where it says, "The unit vector in the direction of $\overrightarrow{AB}$ is $0.3\hat i + 0.4\hat j$. This is not a unit vector. The correct unit vector is $0.6\hat i + 0.8\hat j$. The value given for $\lambda$ seems otherwise correct.

 

[anuttarasammyak]

The main problem I see is in the problem itself, where it says, "The unit vector in the direction of AB→ is 0.3i^+0.4j^. This is not a unit vector. The correct unit vector is 0.6i^+0.8j^. The value given for λ seems otherwise correct.

I see the same main problem with you. In order that "the unit vector" in the statement is not false but strictly right, we may have to forget the usual sense that vector i and j are unit vectors and/or i and j are perdendicular.
[EDIT]As already said in the last of #9, it is a kind of digression and have nothing to do with the answer where length of the vector does not matter.

 

[Mark44]

we may have to forget the usual sense that vector i and j are unit cectors and/or i and j are perdendicular.

I think it's just a typo in the book the problem came from.

 

[chwala]

I think it's just a typo in the book the problem came from.

I do not think so... I gave a reason on post $8$. The trick was to isolate the scalar...in which case, one was to use understanding of the true definition of a unit vector to identify that.

 

[Mark44]

I do not think so... I gave a reason on post $8$. The trick was to isolate the scalar...in which case, one was to use understanding of the true definition of a unit vector to identify that.

I don't think you understood what I was saying. The textbook said that 0.3i + 0.4j is a unit vector -- this is incorrect because the magnitude of this vector is 1/2. A unit vector is one whose magnitude, or length, is 1. I quoted the exact text from the screenshot you attached. There are no tricks or scaling needed. For the textbook to say that this is a unit vector is a mistake, pure and simple. Again, their answer that $\lambda = 7$ is correct, but the problem statement includes the typo that I have identified.