Solve the problem that involves vector displacement

[chwala]

Homework Statement

See attached

Relevant Equations

Vectors

My interest is on part $a$ only. Is the markscheme correct?

I have $BA= -b+a= a-b.$ It therefore follows that $|a-b|$ is the Length of $BA$...or are we saying it does not matter even if we have ... $AB.$ Cheers

 

[chwala]

and for part (b), going with ms
$AB=b-a.$ Let the midpoint of $AB=M$, then $⇒AM=\dfrac{1}{2}(b-a)$ unless i am missing something, there seems to be something wrong with the guide response...

 

[PeroK]

Homework Statement:: See attached
Relevant Equations:: Vectors

View attachment 302155

My interest is on part $a$ only. Is the markscheme correct?

View attachment 302156

I have $BA= -b+a= a-b.$ It therefore follows that $|a-b|$ is the Length of $BA$...or are we saying it does not matter even if we have ... $AB.$ Cheers

$|AB| = |BA|$.

 

[PeroK]

and for part (b), going with ms
$AB=b-a.$ Let the midpoint of $AB=M$, then $⇒AM=\dfrac{1}{2}(b-a)$ unless i am missing something, there seems to be something wrong with the guide response...

The textbook is correct. Try some example vectors to see where are $\frac 12 (\mathbf b + \mathbf a)$ and $\frac 12 (\mathbf b - \mathbf a)$

 

[Delta2]

It is $AM=\frac{1}{2}|\mathbf{b}-\mathbf{a}|$ (notice we take the magnitude of b-a)

however (b) asks for the vector $\frac{1}{2}(\mathbf{a}+\mathbf{b})$. The full detailed response should be that this vector is equal to the position vector of M.

 

[chwala]

Thanks, let me look at this again...

 

[chwala]

$|AB| = |BA|$.

I am aware of this...i was just concerned on the order of operation...noted with regards.

 

[Delta2]

When is (vector equality follows) $$\frac{1}{2}(\mathbf{b}-\mathbf{a})=\frac{1}{2}(\mathbf{a}+\mathbf{b})$$?

 

[chwala]

When is (vector equality follows) $$\frac{1}{2}(\mathbf{b}-\mathbf{a})=\frac{1}{2}(\mathbf{a}+\mathbf{b})$$?

If they have the same magnitude...

 

[Delta2]

If they have the same magnitude...

Nope. Hint: Solve the equation for $\mathbf{a}$.

 

[chwala]

Nope. Hint: Solve the equation for $\mathbf{a}$.

Ok...I'll look at this later...thanks...

 

[chwala]

Nope. Hint: Solve the equation for $\mathbf{a}$.

I do not seem to understand your question, a quick look at those vectors informs me that we have to use Modulus! For them to be equal. You may inform me on this...

 

[Delta2]

I do not seem to understand your question, a quick look at those vectors informs me that we have to use Modulus! For them to be equal. You may inform me on this...

well you are kind of right, for all $\mathbf{a}$ those two vectors don't seem to be equal, except for one unique trivial $\mathbf{a}=?$

 

[chwala]

well you are kind of right, for all $\mathbf{a}$ those two vectors don't seem to be equal, except for one unique trivial $\mathbf{a}=?$

except when $a=0$

 

[Delta2]

Would also be interesting to find out what condition should hold between a,b so that $$|\mathbf{b}-\mathbf{a}|=|\mathbf{b}+\mathbf{a}|$$

There are 3 conditions, the two are trivial cases, and to find the third take the square of the above equality

 

[chwala]

Would also be interesting to find out what condition should hold between a,b so that $$|\mathbf{b}-\mathbf{a}|=|\mathbf{b}+\mathbf{a}|$$

There are 3 conditions, the two are trivial cases, and to find the third take the square of the above equality

yes, the two trivial cases would be given by;

Let

$b-a=b+a$
$b-b=2a$
$2a=0 ⇒a=0$

also

Let
$b-a=-(b+a)$
$2b=a-a$
$2b=0 ⇒ b=0$

For the third one we have;
$b^2-2ab+a^2=b^2+2ab+a^2$
$2ab+2ab=0, ⇒4ab=0 , ⇒ab=0$

 

[Delta2]

what does it mean for the dot product $\mathbf{a}\cdot\mathbf{b}$ to be zero?

 

[chwala]

what does it mean for the dot product $\mathbf{a}\cdot\mathbf{b}$ to be zero?

$i. j=0$ implies vectors are perpendicular to each other, we also have $i.i=1$ This is basic to me...