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Properties of Isosceles and Equilateral Triangles

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Question

Show that the four points P, Q, R, S with position vectors $\to p$ , $\to q$ , $\to r$ , $\to s$ respectively such that 5 $\to p$ - 2 $\to q$ + 6 $\to r$ - 9 $\to s$ = $\to 0$ , are coplaner. Also, find the position vector of the point of intersection of the line segment PR and QS.

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Solution

$5 \to p - 2 \to q + 6 \to r - 9 \to s = 0$
$\Rightarrow 5 \to p - 5 \to q + 3 \to q - 3 \to r + 9 \to r - 9 \to s = 0$
$\Rightarrow 5 (\to p - \to q) + 3 (\to q - \to r) + 9 (\to r - \to s) = 0$
$\Rightarrow 5 \to Q P + 3 \to R Q + 9 \to S R = 0$
If $P, Q, R$ and $S$ are coplanar,
We know $P, Q, R$ are coplanar,
$∵$ $3$ points always lie in the same plane and hence we have to only check if $S$ lies in this plane or not.
If $S$ lies in the same plane then
$\to S R . (\to Q P \times \to Q R) = 0 \to (1)$
$\Rightarrow \frac{- 5 \to Q P - 3 \to R Q}{9} . (\to Q P \times \to Q R)$
$\Rightarrow \frac{- 1}{9} [5 \to Q P . (\to Q P \times \to Q R) - 3 \to R Q . (\to Q P \times (\to - R Q))]$
$\Rightarrow \frac{- 1}{9} [5 \to Q P . (\to Q P \times \to Q R) + 3 \to R Q . (\to Q P \times \to R Q)]$
$\to Q P . (\to Q P \times \to Q R) = 0$
and $\to R Q . (\to Q P \times \to R Q) = 0$
( $∵$ Dot product of perpendicular vector =0)
$\Rightarrow (1) = \to S R . (\to Q P \times \to Q R) = 0$
$∴$ $P, Q, R$ and $S$ are coplanar
Now, $5 \to p + 6 \to r = 2 \to q + 9 \to s$
Dividing by $11$ , we get
$\Rightarrow \frac{5 \to p + 6 \to r}{11} = \frac{2 \to q + 9 \to s}{11}$
By section formula, we can say that point of intersection of $\to P R a n d \to Q S$ divides $\to P R$ segment in the ratio $9 : 2$
$∴$ Point of intersection is $\frac{5 \to p + 6 \to r}{11}$ or $\frac{2 \to q + 9 \to s}{11}$

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