Examples On Vector Equations Of Lines Set-3

Go back to  'Vectors and 3-D Geometry'

Example – 48

Using vector methods, prove this result from plane geometry : in a triangle, the angle bisector of any angle divides the opposite side in the ratio of the sides containing the angle.

Solution: Assume a triangle ABC with position vectors of the vertices as indicated :

Using the result of the previous example, the equation of AD can be written as

\[\begin{align}&\qquad\;\; \vec{r}=\vec{0}+\lambda \left( \hat{b}+\hat{c} \right)\;\;,\qquad\qquad\lambda \in \mathbb{R} \\ & \Rightarrow \quad \vec{r}=\lambda \left( \frac{{\vec{b}}}{\left| {\vec{b}} \right|}+\frac{{\vec{c}}}{\left| {\vec{c}} \right|} \right)\;\;,\qquad\qquad\lambda \in \mathbb{R} \\ \end{align}\]

Assume that D divides BC in the ratio \(\mu :1.\) We wish to determine \(\mu .\) The position vector of D is

\[D\equiv \frac{\mu \vec{c}+\vec{b}}{\mu +1}\]

Thus, for some \(\lambda ,\,\,\mu \in \mathbb{R},\) we must have

\[\begin{align}&\qquad\quad \lambda \left( \frac{{\vec{b}}}{\left| {\vec{b}} \right|}+\frac{{\vec{c}}}{\left| {\vec{c}} \right|} \right)=\frac{\mu \vec{c}+\vec{b}}{\mu +1} \\\\ & \Rightarrow \quad\left( \frac{\lambda }{\left| {\vec{b}} \right|}-\frac{1}{\mu +1} \right)\vec{b}+\left( \frac{\lambda }{\left| {\vec{c}} \right|}-\frac{\mu }{\mu +1} \right)\vec{c}=\vec{0} \\ \end{align}\]

Since \(\vec{b}\,\,and\,\,\vec{c}\) are non collinear, we have

\[\begin{align} & \frac{\lambda }{\left| {\vec{b}} \right|}=\frac{1}{\mu +1}\qquad\qquad\qquad...\left( 1 \right) \\ & \frac{\lambda }{\left| {\vec{c}} \right|}=\frac{\mu }{\mu +1}\qquad\qquad\qquad...\left( 2 \right) \\ \end{align}\]

Dividing (2) by (1), we have

\[\mu =\frac{\left| {\vec{b}} \right|}{\left| {\vec{c}} \right|}=\frac{AB}{AC}\]

Thus, D divides BC in the ratio AB : AC ; this proves the theorem.

Example – 49

Find the perpendicular distance of the point \(A(\vec{a})\) from the line \(\vec{r}=\vec{b}+\lambda \vec{c}.\)


\[\begin{align}& \;AP\;=\sqrt{A{{B}^{2}}-B{{P}^{2}}} \\ \\ & \qquad=\sqrt{A{{B}^{2}}-{{(\overrightarrow{AB}\cdot \hat{C})}^{2}}} \\\\ &\qquad =\sqrt{{{\left| \vec{a}-\vec{b} \right|}^{2}}-(\vec{a}-\vec{b})\cdot \hat{c}} \\\\  &\qquad =\sqrt{{{\left| \vec{a}-\vec{b} \right|}^{2}}-\frac{(\vec{a}-\vec{b})\cdot \vec{c}}{\left| {\vec{c}} \right|}} \\ \end{align}\]

This is one way to specify AP. Another way could be as follows:

\[\begin{align}&\qquad\quad Area(\Delta ABP) = \frac{1}{2} \times BP \times AP \hfill \\\\& \Rightarrow \quad \frac{1}{2}\left| {\left( {\vec a - \vec b} \right) \times \left( {{{\vec r}_P} - \vec b} \right)} \right| = \frac{1}{2}\left| {{{\vec r}_P} - \vec b} \right|(AP) \hfill \\\\&\Rightarrow \quad \frac{1}{2}\left| {\left( {\vec a - \vec b} \right) \times \lambda \vec c} \right| = \frac{1}{2}\left| {\lambda \vec c} \right|\left( {AP} \right)&\left\{ \begin{gathered} \because {\mkern 1mu} {\mkern 1mu} {{\vec r}_P} - \vec b = \lambda \vec c \hfill \\\\ {\text{for}}{\mkern 1mu} {\text{some}}\lambda  \in \mathbb{R} \hfill \\\\ \end{gathered}  \right\} \hfill\\ \\& \Rightarrow \quad AP = \frac{{\left| {\left( {\vec a - \vec b} \right) \times \vec c} \right|}}{{\left| {\vec c} \right|}} \hfill \\\\&\qquad\qquad= \left| {\left( {\vec a - \vec b} \right) \times \hat c} \right| \hfill \\ \end{align} \]

We could have arrived at this last result even more easily:

\[\begin{align}&AP = AB\sin {\theta } \hfill \\\\&\quad\; = \left| {\vec a - \vec b} \right|\sin \theta  \hfill \\\\&\quad\;= \left| {\,\left( {\vec a - \vec b} \right) \times \hat c\,} \right| \hfill \\ \end{align} \]

Example – 50

Two straight lines in space are called skew lines if they are neither parallel nor intersecting. Find the shortest distance between the two skew lines

\[\begin{align}& {L_1}:\vec r = \vec a + \lambda \vec b \hfill \\& {L_2}:\vec r = \vec c + \mu \vec d \hfill \\ \end{align} \]

Solution: First of all, convince yourself that there will be only one line along which the distance between L1 and L2 is minimum. Such a line will be perpendicular to both L1 and L2.

It should be obvious that since AB is perpendicular to both L1 and L2, i.e, since AB is perpendicular to both \(\vec b\) and \(\vec d\)it must be collinear with the vector \(\vec b \times \vec d.\)Thus, a unit vector along the direction AB is given by

\[\hat n = \frac{{\vec b \times \vec d}}{{\left| {\vec b \times \vec d} \right|}}\]

The distance AB is now simply the projection of the line segment joining the points \(\vec a\) and \(\vec c\) along the (extended) line AB. Again, visualise this in your mind; you must be very clear why this is so. We thus have,

\[\begin{align}& \,d = AB = \left| {\left( {\vec a - \vec c} \right) \cdot \hat n} \right| \hfill \\\\&\;\;  = \left| {\frac{{\left( {\vec a - \vec c} \right) \cdot \left( {\vec b \times \vec d} \right)}}{{\left| {\vec b \times \vec d} \right|}}} \right| \hfill \\ \end{align} \]

We can deduce a very useful corollary from this result. The two straight lines L1 and L2 intersect (in other words, they are coplanar) if

\[\begin{align}&\quad\qquad  d = 0 \hfill \\  & \Rightarrow \quad  \left( {\vec a - \vec c} \right) \cdot \left( {\vec b \times \vec d} \right) = 0 \hfill \\&   \Rightarrow \quad  \vec a \cdot \left( {\vec b \times \vec d} \right) = \vec c \cdot \left( {\vec b \times \vec d} \right) \hfill \\ &\Rightarrow \quad  \left[ {\vec a\,\,\,\;\;\vec b\;\;\,\,\,\vec d} \right] = \left[ {\vec c\,\,\,\;\;\vec b\;\;\,\,\,\vec d} \right] \hfill \\ \end{align} \]

Note that if L1 and L2 are parallel, then the distance between them can be evaluated simply as the perpendicular distance of \(\vec a\) from L2 (or  \(\vec c\) from L1).

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