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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!

  • 0
    Thanks guys that checks out. :)2011-07-24

1 Answers 1

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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 $