Vectors Ques 105
- Let $A B C$ and $P Q R$ be any two triangles in the same plane. Assume that the perpendiculars from the points $A, B, C$ to the sides $Q R, R P, P Q$ respectively are concurrent. Using vector methods or otherwise, prove that the perpendiculars from $P, Q, R$ to $B C, C A, A B$ respectively are also concurrent.
$(2000,4\ M)$
Show Answer
Answer:
Correct Answer: 105.$p=r=\frac{1}{\sqrt{1+2 \cos \theta}}, q=\frac{-2 \cos \theta}{\sqrt{1+2 \cos \theta}}$
Solution:
- Let the position vectors of $A, B, C$ be $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ respectively and that of $P, Q, R$ be $\overrightarrow{\mathbf{p}}, \overrightarrow{\mathbf{q}}$ and $\overrightarrow{\mathbf{r}}$, respectively. Let $\overrightarrow{\mathbf{h}}$ be the position vector of the orthocentre $H$ of the $\triangle P Q R$. We have, $H P \perp Q R$. Equation of straight line passing through $A$ and perpendicular to $Q R$ i.e. parallel to $\overrightarrow{\mathbf{H P}}=\overrightarrow{\mathbf{P}}-\overrightarrow{\mathbf{h}}$ is
$$ \overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{a}}+t _1(\overrightarrow{\mathbf{p}}-\overrightarrow{\mathbf{h}}) $$
where, $t _1$ is a parameter.
Similarly, the equation of a straight line through $B$ and
perpendicular to $R P$ is $\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{b}}+t_2(\overrightarrow{\mathbf{q}}-\overrightarrow{\mathbf{h}})$
Again, equation of straight line through $C$ and $D$
perpendicular to $P Q$ is $\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{c}}+t _3(\overrightarrow{\mathbf{h}}-\overrightarrow{\mathbf{r}})$
If the lines (i), (ii) and (iii) are concurrent, then there exists a point $D$ with position vector $\overrightarrow{\mathbf{d}}$ which lies on all of them, that is for some values of $t _1, t _2$ and $t _3$,
which implies that the process is reversible
$$ \begin{aligned} & \frac{1}{t _1} \overrightarrow{\mathbf{d}}=\frac{1}{t _1} \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{p}}-\overrightarrow{\mathbf{h}} \\ & \frac{1}{t _2} \overrightarrow{\mathbf{d}}=\frac{1}{t _2} \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{q}}-\overrightarrow{\mathbf{h}} \\ & \frac{1}{t _3} \overrightarrow{\mathbf{d}}=\frac{1}{t _3} \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{r}}-\overrightarrow{\mathbf{h}} \end{aligned} $$
From Eqs. (iv) and (v),
$$ \left(\frac{1}{t _1}-\frac{1}{t _2}\right) \overrightarrow{\mathbf{d}}=\frac{1}{t _1} \overrightarrow{\mathbf{a}}-\frac{1}{t _2} \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{p}}-\overrightarrow{\mathbf{q}} $$
and from Eqs. (5) and (6),
$$ \left(\frac{1}{t _2}-\frac{1}{t _3}\right) \overrightarrow{\mathbf{d}}=\frac{1}{t _2} \overrightarrow{\mathbf{b}}-\frac{1}{t _3} \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{q}}-\overrightarrow{\mathbf{r}} $$
Eliminating $\overrightarrow{\mathbf{d}}$ from Eqs. (vii) and (viii), we get
$$ \begin{aligned} & \left(\frac{1}{t _2}-\frac{1}{t _3}\right)\left[\frac{1}{t _1} \overrightarrow{\mathbf{a}}-\frac{1}{t _2} \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{p}}-\overrightarrow{\mathbf{q}}\right] \\ & =\left(\frac{1}{t _1}-\frac{1}{t _2}\right)\left[\frac{1}{t _2} \overrightarrow{\mathbf{b}}-\frac{1}{t _3} \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{q}}-\overrightarrow{\mathbf{r}}\right] \\ & \Rightarrow\left(t _3-t _2\right)\left[t _2 \overrightarrow{\mathbf{a}}-t _1 \overrightarrow{\mathbf{b}}+t _1 t _2(\overrightarrow{\mathbf{p}}-\overrightarrow{\mathbf{q}})\right] \\ & =\left(t _2-t _1\right)\left[t _3 \overrightarrow{\mathbf{b}}-t _2 \overrightarrow{\mathbf{c}}+t _2 t _3(\overrightarrow{\mathbf{q}}-\overrightarrow{\mathbf{r}})\right] \\ & \text { [multiplying both sides by } t _1 t _2 t _3 \text { ] } \\ & \Rightarrow \quad t _2\left(t _3-t _2\right) \overrightarrow{\mathbf{a}}+t _2\left(t _1-t _3\right) \overrightarrow{\mathbf{b}} \\ & +t _2\left(t _2-t _1\right) \overrightarrow{\mathbf{c}}+t _1 t _2\left(t _3-t _2\right) \overrightarrow{\mathbf{p}} \\ & +t _2^{2}\left(t _1-t _3\right) \overrightarrow{\mathbf{q}}+t _2 t _3\left(t _2-t _1\right) \overrightarrow{\mathbf{r}}=\overrightarrow{0} \end{aligned} $$
Thus, lines (i), (ii), and (iii) are concurrent if and only if there exist scalars $t_1, t_2$ and $t_3$ such that
$$ \begin{aligned} \left(t _2-t _3\right) \overrightarrow{\mathbf{a}}+\left(t _3-t _1\right) \overrightarrow{\mathbf{b}} & +\left(t _1-t _2\right) \overrightarrow{\mathbf{c}}+t _1\left(t _2-t _3\right) \overrightarrow{\mathbf{p}} \\ & +t _2\left(t _3-t _1\right) \overrightarrow{\mathbf{q}}+t _3\left(t _1-t _2\right) \overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{0}} \end{aligned} $$
On dividing by $t _1 t _2 t _3$, we get
$$ \begin{alignedat} & \left(\lambda _2-\lambda _3\right) \overrightarrow{\mathbf{p}}+\left(\lambda _3-\lambda _1\right) \overrightarrow{\mathbf{q}}+\left(\lambda _1-\lambda _2\right) \overrightarrow{\mathbf{r}} \\ & \quad+\lambda _1\left(\lambda _2-\lambda _3\right) \overrightarrow{\mathbf{a}}+\lambda _2\left(\lambda _3-\lambda _1\right) \overrightarrow{\mathbf{b}}+\lambda _3\left(\lambda _1-\lambda _2\right) \overrightarrow{\mathbf{c}}=0 \end{aligned} $$
where, $\lambda _i=\frac{1}{t _i}$ for $i=1,2,3$
So, this is the condition that the lines from $P$ perpendicular to $B C$, from $Q$, perpendicular to $C A$ and from $R$ perpendicular to $A B$ are concurrent (by changing $A B C$ and $P Q R$ simultaneously).
BETA
