SATHEE: Vectors 5 Question 10

Vectors 5 Question 10

10. 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)

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Answer:

Correct Answer: 10. $p = r = \frac{1}{\sqrt{1 + 2 \cos θ}}, q = \frac{- 2 \cos θ}{\sqrt{1 + 2 \cos θ}}$

Solution:

Let the position vectors of $A, B, C$ be $\vec{a}, \vec{b}$ and $\vec{c}$ respectively and that of $P, Q, R$ be $\vec{p}, \vec{q}$ and $\vec{r}$ , respectively. Let $\vec{h}$ be the position vector of the orthocentre $H$ of the $△ P Q R$ . We have, $H P ⊥ Q R$ . Equation of straight line passing through $A$ and perpendicular to $Q R$ i.e. parallel to $\vec{H P} = \vec{P} - \vec{h}$ is

$\vec{r} = \vec{a} + t_{1} (\vec{p} - \vec{h})$

where, $t_{1}$ is a parameter.

Similarly, equation of straight line through $B$ and

perpendicular to $R P$ is $\vec{r} = \vec{b} + t_{2} (\vec{q} - \vec{h})$

Again, equation of straight line through $C$ and

perpendicular to $P Q$ is $\vec{r} = \vec{c} + t_{3} (\vec{r} - \vec{h})$

If the lines (i), (ii) and (iii) are concurrent, then there exists a point $D$ with position vector $\vec{d}$ which lies on all of them, that is for some values of $t_{1}, t_{2}$ and $t_{3}$ ,

which implies that

$\begin{aligned} \frac{1}{t_{1}} \vec{d} = \frac{1}{t_{1}} \vec{a} + \vec{p} - \vec{h} \\ \frac{1}{t_{2}} \vec{d} = \frac{1}{t_{2}} \vec{b} + \vec{q} - \vec{h} \\ \frac{1}{t_{3}} \vec{d} = \frac{1}{t_{3}} \vec{c} + \vec{r} - \vec{h} \end{aligned}$

From Eqs. (iv) and (v),

$(\frac{1}{t_{1}} - \frac{1}{t_{2}}) \vec{d} = \frac{1}{t_{1}} \vec{a} - \frac{1}{t_{2}} \vec{b} + \vec{p} - \vec{q}$

and from Eqs. (v) and (vi),

$(\frac{1}{t_{2}} - \frac{1}{t_{3}}) \vec{d} = \frac{1}{t_{2}} \vec{b} - \frac{1}{t_{3}} \vec{c} + \vec{q} - \vec{r}$

Eliminating $\vec{d}$ from Eqs. (vii) and (viii), we get

$\begin{aligned} (\frac{1}{t_{2}} - \frac{1}{t_{3}}) [\frac{1}{t_{1}} \vec{a} - \frac{1}{t_{2}} \vec{b} + \vec{p} - \vec{q}] \\ = (\frac{1}{t_{1}} - \frac{1}{t_{2}}) [\frac{1}{t_{2}} \vec{b} - \frac{1}{t_{3}} \vec{c} + \vec{q} - \vec{r}] \\ \Rightarrow (t_{3} - t_{2}) [t_{2} \vec{a} - t_{1} \vec{b} + t_{1} t_{2} (\vec{p} - \vec{q})] \\ = (t_{2} - t_{1}) [t_{3} \vec{b} - t_{2} \vec{c} + t_{2} t_{3} (\vec{q} - \vec{r})] \\ [multiplying both sides by t_{1} t_{2} t_{3}] \\ \Rightarrow t_{2} (t_{3} - t_{2}) \vec{a} + t_{2} (t_{1} - t_{3}) \vec{b} \\ + t_{2} (t_{2} - t_{1}) \vec{c} + t_{1} t_{2} (t_{3} - t_{2}) \vec{p} \\ + t_{2}^{2} (t_{1} - t_{3}) \vec{q} + t_{2} t_{3} (t_{2} - t_{1}) \vec{r} = \vec{0} \end{aligned}$

Thus, lines (i), (ii), and (iii) are concurrent is equivalent to say that there exist scalars $t_{1}, t_{2}$ and $t_{3}$ such that

$\begin{aligned} (t_{2} - t_{3}) \vec{a} + (t_{3} - t_{1}) \vec{b} & + (t_{1} - t_{2}) \vec{c} + t_{1} (t_{2} - t_{3}) \vec{p} \\ + t_{2} (t_{3} - t_{1}) \vec{q} + t_{3} (t_{1} - t_{2}) \vec{r} = \vec{0} \end{aligned}$

On dividing by $t_{1} t_{2} t_{3}$ , we get

$\begin{aligned} (λ_{2} - λ_{3}) \vec{p} + (λ_{3} - λ_{1}) \vec{q} + (λ_{1} - λ_{2}) \vec{r} \\ + λ_{1} (λ_{2} - λ_{3}) \vec{a} + λ_{2} (λ_{3} - λ_{1}) \vec{b} + λ_{3} (λ_{1} - λ_{2}) \vec{c} = 0 \end{aligned}$

where, $λ_{i} = \frac{1}{t_{i}}$ 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).

Vectors 5 Question 10

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Vectors 5 Question 10

10. Let ABC and PQR be any two triangles in the same plane. Assume that the perpendiculars from the points A,B,C to the sides QR,RP,PQ respectively are concurrent. Using vector methods or otherwise, prove that the perpendiculars from P,Q,R to BC,CA,AB respectively are also concurrent.

Answer:

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