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Equation of Perpendicular from a Point on a Line

Show that the...

Question

Show that the four points A, B, C, D with position vectors $\to a$ , $\to b$ , $\to c$ , $\to d$ respectively such that 3 $\to a$ - 2 $\to b$ + 5 $\to c$ - 6 $\to d$ = $\to 0$ , are coplaner. Also, find the position vector of the point of intersection of the line segment AC and BD.

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Solution

$3 \to a - 2 \to b + 5 \to c - 6 \to d = 0$
$\Rightarrow 3 \to a - 3 \to b + \to b - \to c + 6 \to c - 6 \to d = 0$
$\Rightarrow 3 (\to a - \to b) + (\to b - \to c) + 6 (\to c - \to d) = 0$
$\Rightarrow 3 \to B A + \to C B + 6 \to D C = 0 \to (1)$
We know, three points are always coplanar.
$∴$ Let $A, B$ and $C$ lie on a plane, hence we have to prove $D$ lies on the same plane.
If $D$ lies on the same plane, then
$\to D C . (\to B A \times \to B C) = 0$
$\Rightarrow (\frac{- \to B A - \to C B}{6}) . (\to B A - \to B C)$
$\Rightarrow (\frac{- \to B A + \to B C}{6}) . (\to B A - \to B C)$
$\Rightarrow \frac{1}{6} [- \to B A . (\to B A - \to B C) + \to B C . (\to B A - \to B C)] \to (2)$
$\to B A \times \to B C$ is perpendicular to $\to B A$
$\Rightarrow \to B A . (\to B A \times \to B C) = 0$
Similarly, $\to B C . (\to B A \times \to B C) = 0$
$∴ (2) = \to D C . (\to B A \times \to B C) = 0$
Hence, $A, B, C a n d D$ are coplanar.
$W e K n o w, 3 \to a - 2 \to b + 5 \to c - 6 \to d = 0$
$\Rightarrow 3 \to a + 5 \to c = 2 \to b + 6 \to d$
Dividing both sides by $8,$ we get
$\Rightarrow \frac{3 \to a + 5 \to c}{8} = \frac{2 \to b + 6 \to d}{8}$
Hence by using section formula, we can say that there exists a point of intersection between $\to A C a n d \to B D .$ divides line $A C$ in $5 : 3$ and line $\to B D$ in $6 : 2$ or $3 : 1.$
Hence, point of intersection is $\frac{3 \to a + 5 \to c}{8} o r \frac{2 \to b + 6 \to d}{8} .$

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