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

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

Now lets compute the magnitude

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

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

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

Just solve from here

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

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

Now lets compute the magnitude

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

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

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

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 $\bold{a}=$ and $\bold{b}=<7,6>$, then $\bold{a}-\bold{b}==$. Therefore:

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

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

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

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

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

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

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

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

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

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

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