1. ## Vector Problem

a = vector a
b = vector b
[ ] = numbers in squared brackets are vector components
|a - b| = the magnitude of vector a minus vector b

Problem:

If a=[k, 2] and b=[7,6], where k is a real number, determine all values of k such that |a - b| = 5.
The answer in the book says, "10 or 4."

2. Originally Posted by Morphayne
a = vector a
b = vector b
[ ] = numbers in squared brackets are vector components
|a - b| = the magnitude of vector a minus vector b

Problem:

If a=[k, 2] and b=[7,6], where k is a real number, determine all values of k such that |a - b| = 5.

Lets start by doing the computations

$\displaystyle |(k,2)-(7,6)|=|(k-7,-4)|$

Now lets compute the magnitude

$\displaystyle |(k-7),-4|=\sqrt{(k-7)^2+(-4)^2}$

Now we need to set the above equal to 5 and solve.

$\displaystyle \sqrt{(k-7)^2+(-4)^2}=5$

Just solve from here

3. Originally Posted by TheEmptySet
Lets start by doing the computations

$\displaystyle |(k,2)-(7,6)|=|(k-7,-4)|$

Now lets compute the magnitude

$\displaystyle |(k-7),-4|=\sqrt{(k-7)^2+(-4)^2}$

Now we need to set the above equal to 5 and solve.

$\displaystyle \sqrt{(k-7)^2+(-4)^2}=5$

Just solve from here
Do you mind solving it? I'm a struggling student. I gave you +Thanks.

4. Originally Posted by Morphayne
a = vector a
b = vector b
[ ] = numbers in squared brackets are vector components
|a - b| = the magnitude of vector a minus vector b

Problem:

If a=[k, 2] and b=[7,6], where k is a real number, determine all values of k such that |a - b| = 5.

If $\displaystyle \bold{a}=<k,2>$ and $\displaystyle \bold{b}=<7,6>$, then $\displaystyle \bold{a}-\bold{b}=<k-7,2-6>=<k-7,-4>$. Therefore:

$\displaystyle {\left|\bold{a}-\bold{b}\right|}=\sqrt{(k-7)^2+(-4)^2}=\sqrt{k^2-14k+49+16}=\sqrt{k^2-14k+65}$

Now find where
$\displaystyle {\left|\bold{a}-\bold{b}\right|}=5$.

$\displaystyle 5=\sqrt{k^2-14k+65}$
$\displaystyle \rightarrow 25=k^2-14k+65$
$\displaystyle \rightarrow k^2-14k+40=0$
$\displaystyle (k-4)(k-10)=0$
$\displaystyle \rightarrow k=4$ or $\displaystyle k=10$.

You can plug these back in to verify the k values.

$\displaystyle 5=\sqrt{4^2-14(4)+65}=\sqrt{16-56+65}=\sqrt{25}={\color{red}5}$
$\displaystyle 5=\sqrt{10^2-14(10)+65}=\sqrt{100-140+65}=\sqrt{25}={\color{red}5}$

5. Originally Posted by Chris L T521
If $\displaystyle \bold{a}=<k,2>$ and $\displaystyle \bold{b}=<7,6>$, then $\displaystyle \bold{a}-\bold{b}=<k-7,2-6>=<k-7,-4>$. Therefore:

$\displaystyle {\left|\bold{a}-\bold{b}\right|}=\sqrt{(k-7)^2+(-4)^2}=\sqrt{k^2-14k+49+16}=\sqrt{k^2-14k+65}$

Now find where
$\displaystyle {\left|\bold{a}-\bold{b}\right|}=5$.

$\displaystyle 5=\sqrt{k^2-14k+65}$
$\displaystyle \rightarrow 25=k^2-14k+65$
$\displaystyle \rightarrow k^2-14k+40=0$
$\displaystyle (k-4)(k-10)=0$
$\displaystyle \rightarrow k=4$ or $\displaystyle k=10$.

You can plug these back in to verify the k values.

$\displaystyle 5=\sqrt{4^2-14(4)+65}=\sqrt{16-56+65}=\sqrt{25}={\color{red}5}$
$\displaystyle 5=\sqrt{10^2-14(10)+65}=\sqrt{100-140+65}=\sqrt{25}={\color{red}5}$