given vector a=3i+4j, vector b=8i+6j and vector c=2i+kj find the value of k, here vector c makes equal angle with vectors a and b.

Question

solomonzelman · Answer

$\color{black}{\displaystyle  {\bf a}=\langle 3,4\rangle}$
$\color{black}{\displaystyle   {\bf b}=\langle 3m,4m\rangle~~~{\rm such~that}\,~k>0}$
$\color{black}{\displaystyle  {\bf c}=\langle 2,k\rangle}$

$\color{blue}{\displaystyle   {\bf a}\cdot  {\bf c}=\left|{\bf a}\right|\left|{\bf c}\right|\cos \theta_1 }$
$\color{blue}{\displaystyle  6+4k=5\sqrt{k^2+4}\cos\theta_1 }$

$\color{green}{\displaystyle   {\bf b}\cdot  {\bf c}=\left|{\bf b}\right|\left|{\bf c}\right|\cos \theta_2 }$
$\color{green}{\displaystyle   6m+4mk=5m\sqrt{k^2+4}\cos\theta_2}$



$\color{blue}{\displaystyle \cos\theta_1= \frac{6+4k}{5\sqrt{k^2+4}} }$

$\color{green}{\displaystyle \cos\theta_2= \frac{6m+4mk}{5m\sqrt{k^2+4}}=\frac{6+4k}{5\sqrt{k^2+4}} }$

solomonzelman · Answer

So, you can see that in the above case (which is the same as yours, but a bit more abstract), regardless of the $m$, the angles are same.

solomonzelman · Answer

This is because $\bf a$ and $\bf b$  are in the same direction,
so, regardless of their magnitudes (as long as the corresponding
components of $\bf a$ and $\bf b$ are proportional), they will always make
the same angle with any other vector $\bf c$, no matter what this vector
$\bf c$ is (well, asides from $\bf 0$).

solomonzelman · Answer

Oh, I am sorry, they are not proportional .... it is 8,6 not 6,8 :(

solomonzelman · Answer

Well, you can follow the same process with your vectors:

$\color{black}{\displaystyle  {\bf a}=\langle 3,4\rangle}$
$\color{black}{\displaystyle   {\bf b}=\langle 8,6\rangle~~~{\rm such~that}\,~k>0}$
$\color{black}{\displaystyle  {\bf c}=\langle 2,k\rangle}$

$\color{blue}{\displaystyle   {\bf a}\cdot  {\bf c}=\left|{\bf a}\right|\left|{\bf c}\right|\cos \theta_1 }$
$\color{blue}{\displaystyle  6+4k=5\sqrt{k^2+4}\cos\theta_1 }$

$\color{green}{\displaystyle   {\bf b}\cdot  {\bf c}=\left|{\bf b}\right|\left|{\bf c}\right|\cos \theta_2 }$
$\color{green}{\displaystyle   16+6k=10\sqrt{k^2+4}\cos\theta_2}$



$\color{blue}{\displaystyle \cos\theta_1= \frac{6+4k}{5\sqrt{k^2+4}} }$

$\color{green}{\displaystyle \cos\theta_2= \frac{16+6k}{10\sqrt{k^2+4}} }$

solomonzelman · Answer

Then, set $\color{black}{\displaystyle \cos\theta_1= \cos\theta_2 }$.

$\color{black}{\displaystyle \frac{6+4k}{5\sqrt{k^2+4}} =\frac{16+6k}{10\sqrt{k^2+4}}}$

samhang · Answer

let me try

samhang · Answer

I got k = 2...

samhang · Answer

Thanks a lot

solomonzelman · Answer

$\color{black}{\displaystyle \frac{6+4k}{5\sqrt{k^2+4}} =\frac{16+6k}{10\sqrt{k^2+4}}}$

$\color{black}{\displaystyle 6+4k=8+3k}$

$\color{black}{\displaystyle k=2}$

solomonzelman · Answer

Correct ... You welcome!