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Find point of intersection and the bisectors of the angles made by line $k$ passing through point $(3, -1)$ with direction vector $-\vec i + 2\vec j$ and line $l$ passing through point $(-2, 5)$ with direction vector $3\vec i - 2j$.

Assuming the lines intersect I went about finding the point of intersection like so,

Direction vector of line $k$ and $l$,

$$m_k = (-1, 2), m_l = (3, -2)$$

Vector equation of line $k$ is,

$$r_k = (3, -1) + \lambda (-1, 2) = (-\lambda + 3, 2 \lambda - 1)$$

And Vector equation of line $l$ is,

$$r_l = (-2, 5) + \mu (3, -2) = (-2 + 3\mu, 5 - 2\mu)$$

At intersection,

$$(-\lambda + 3, 2 \lambda - 1) = (-2 + 3\mu, 5 - 2\mu)$$

Which gives the 2 equations,

$$\lambda + 3\mu = 5$$ $$\lambda + \mu = 3$$

Solving for $\lambda$ and $\mu$ gives,

$\lambda = 2, \mu = 1$

And thus the point of intersection is $(1, 1)$

The next part of the problem is eluding me. I figured that if the quadrilateral formed by the direction vectors was a rhombus/kite, the angle bisectors would be the diagonals ie:- $m_k + m_l$ and $m_k - m_l$. But the magnitude of $m_k$ and $m_l$ are $\sqrt{5}$ and $\sqrt{13}$ respectively. So that idea doesn't work.

How do i find the equation of the bisectors?

As always, thanks for all your help, Much appreciated!

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What are the vector equations of your lines $k$ and $l$ when you normalize the given direction vectors such that they have length $1$? –  Christian Blatter Jul 24 '11 at 14:50
    
@Christian I am not sure how to do normalize to 1. I normalized the vectors to $m_k = \dfrac{-\vec i + 2\vec j}{\sqrt 5}$ and $m_l = \dfrac{3\vec i - 2\vec j}{\sqrt{13}}$ –  mathguy80 Jul 24 '11 at 15:26
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Your calculation is correct except for the very end; substituting the values you got for $\lambda$ and $\mu$ into vector equations of the lines yields $(1,3)$. (Perhaps you substituted $1$ for $\lambda$ and $2$ for $\mu$ in calculating the $y$ component?) Regarding the bisectors, you should be able to find them like you wrote, using the normalized direction vectors instead of the unnormalized ones. –  joriki Jul 24 '11 at 15:47
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Your new vectors $m_k=({-1\over\sqrt{5}},{2\over\sqrt{5}})$ and $m_l=\ldots$ have length 1, so they form a kite, as you say. –  Christian Blatter Jul 24 '11 at 15:49
    
@joriki, Yes I messed that up at the very end. I understand now what Christian was saying about normalize to 1. I was thinking of the magnitude of the direction vector not the unit direction vector. –  mathguy80 Jul 24 '11 at 15:56

1 Answer 1

up vote 0 down vote accepted

Adding the answer from the comments to close this question.

The point of intersection is $(1, 3)$. The unit direction vectors then are,

$$ \begin{align} m_k &= \dfrac{\vec m_k}{|m_k|} &= \dfrac{-\vec i + 2\vec j}{\sqrt 5} \\ m_l &= \dfrac{\vec m_l}{|m_l|} &= \dfrac{3\vec i - 2\vec j}{\sqrt{13}} \end{align} $$

Then the quadrilateral formed with the unit direction vectors is a rhombus of unit length, whose diagonals are also its angle bisectors.

Thus, the angle bisectors are,

$$ m_k + m_l = \left(\dfrac{-1}{\sqrt 5} + \dfrac{3}{\sqrt{13}}\right)\vec i + \left(\dfrac{2}{\sqrt 5} - \dfrac{2}{\sqrt{13}}\right) \vec j $$

$$ m_k - m_l = \left(\dfrac{-1}{\sqrt 5} - \dfrac{3}{\sqrt{13}}\right)\vec i + \left(\dfrac{2}{\sqrt 5} + \dfrac{2}{\sqrt{13}}\right) \vec j $$

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