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Derivative from First Principle

If a,b, c a...

Question

If $¯ ¯ ¯ a, ¯ ¯ b, ¯ ¯ c$ are position vectors of three non-collinear points $A, B, C$ respectively, then the shortest distance of $A$ from $B C$ is

\frac{| \to A B \times (\to A B \times \to A C) |}{| \to B C |^{2}}

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\frac{| \to A C \times (\to A C \times \to B C) |}{| \to A C |^{2}}

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\frac{| \to A B \times (\to A B \times \to A C) |}{| \to A B |^{2}}

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\frac{| \to B C \times (\to A B \times \to B C) |}{| \to B C |^{2}}

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Solution

The correct option is D $\frac{| \to B C \times (\to A B \times \to B C) |}{| \to B C |^{2}}$
Shortest distance of $A$ on the line $B C$ will be at foot of perpendicular from $A$ on $A (\to a)$
$D (\to d) = \to b + λ (\to b - \to c)$
For some $^{'} A^{'}, D (\to d)$ will be foot of perpendicular
$- - \to A D$ is perpendicular to $- - \to B C$
$\Rightarrow (\to b - \to c) . (\to a - \to b - λ (\to b - \to c)) = 0$
$\Rightarrow (\to b - \to c) . ((\to a - \to b) - λ (\to b - \to c)) = 0$
$λ = \frac{(\to a - \to b) . (\to b - \to c)}{{| \to b - \to c |}^{2}}$
$D (\to d) = \to b + \frac{(\to a - \to b) . (\to b - \to c)}{{| \to b - \to c |}^{2}} (\to b - \to c)$
$- - \to A D = \to a - ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \to b + \frac{(\to a - \to b) . (\to b - \to c)}{{| \to b - \to c |}^{2}} (\to b - \to c) ⎞ ⎟ ⎟ ⎟ ⎟ ⎠$
$= (\to a - \to b) - ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{(\to a - \to b) . (\to b - \to c)}{{| \to b - \to c |}^{2}} (\to b - \to c) ⎞ ⎟ ⎟ ⎟ ⎟ ⎠$
$= \frac{(\to a - \to b) {| \to b - \to c |}^{2} - (\to a - \to b) . (\to b - \to c) (\to b - \to c)}{{| \to b - \to c |}^{2}}$
$= \frac{(\to b - \to c) \times ((\to a - \to b) \times (\to b - \to c))}{{| \to b - \to c |}^{2}}$
$= \frac{- - \to B C \times (- - \to A B \times - - \to B C)}{{| - - \to B C |}^{2}}$
$| - - \to A D | = ∣ ∣ ∣ ∣ ∣ ∣ \frac{- - \to B C \times (- - \to A B \times - - \to B C)}{{| - - \to B C |}^{2}} ∣ ∣ ∣ ∣ ∣ ∣$

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