Distance of the point Pp- from the line r-a-b- is-

The correct option is A | (a⃗ p⃗)+((p⃗ a⃗)·b⃗)b⃗|b⃗|^2| Let Q(q⃗) be the foot of perpendicular drawn from P(p⃗) to the line r⃗ = a⃗ + λb⃗ ...

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Distance of t...

Question

Distance of the point $P (\to p)$ from the line $\to r = \to a + λ \to b$ is-

∣ (\to a - \to p) + \frac{((\to p - \to a) \cdot \to b) \to b}{∣ \to b ∣^{2}} ∣

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∣ (\to b - \to p) + \frac{((\to p - \to a) \cdot \to b) \to b}{∣ \to b ∣^{2}} ∣

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∣ (\to a - \to p) + \frac{((\to p - \to b) \cdot \to b) \to b}{∣ \to b ∣^{2}} ∣

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None of these.

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\to a, \to b, \to c

\to p, \to q, \to r

(\to a \times \to p) + (\to b \times \to q) + (\to c \times \to r)

\to a, \to b, \to c

\to p, \to q, \to r

\to p = \frac{\to b \times \to c}{[\begin{matrix} \to b & \to c & \to a \end{matrix}]}

\to q = \frac{\to c \times \to a}{[\begin{matrix} \to c & \to a & \to b \end{matrix}]}

\to r = \frac{\to a \times \to b}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]}

(\to a + \to b) . \to p + (\to b + \to c) . \to q + (\to c + \to a) . \to r

If $\to A$ + $\to B$ = $\to P$ and $\to A$ $\times$ $\to B$ = $\to Q$ then

\to a \times {\to b \times (\to c \times \to a) + (\to p \times \to q)} =

| \to a | = | \to b | = | \to c | = 1

\to a + \to b + \to c = \to 0

\to a . \to b + \to b . \to c + \to c . \to a = \frac{- 3}{2}

(\to p + \to q)^{2} = | \to p |^{2} + | \to q |^{2} + 2 \to p . \to q

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