Let $- - \to A B = 3^i +^j -^k$ and $- - \to A C =^i -^j + 3^k$ and a point $P$ on the line segment $B C$ be equidistant from $A B$ and $A C$ , the $- - \to A P$ is

2^i -^k

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^i - 2^k

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2^i +^k

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^i + 2^k

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Solution

The correct option is C $2^i +^k$
Clearly, a point equidistant from $A B$ and $A C$ is on the bisector of $∠ B A C$ is $\frac{\to A B}{∣ ∣ \to A B ∣ ∣} + \frac{\to A C}{∣ ∣ \to A C ∣ ∣} = \frac{1}{\sqrt{11}} (4^i + 2^k) = \frac{2}{\sqrt{11}} (2^i +^k)$

Let $\to A B = λ (2^i +^k)$
$∴ \to B P = \to A P - \to A B = λ (2^i +^k) - (3^i +^j -^k)$
$∴ \to B P = (2 λ - 3)^i -^j + (λ + 1)^k$

Also $\to B C = \to A C - \to A B = (- 2^i - 2^j + 4^k)$
$∵ \to B P ∥ \to B C ∴ \to B P = t \to B C$
$\Rightarrow (2 λ - 3)^i -^j + (λ + 1)^k = t (- 2^i - 2^j + 4^k)$
$\Rightarrow (2 λ - 3) = - 2 t, - 1 = - 2 t, λ + 1 = 4 t$
$\Rightarrow λ = 1, t = \frac{1}{2} \Rightarrow \to A P = 2^i +^k$

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Similar questions

- - \to A B = 3^i +^j -^k

- - \to A C =^i -^j + 3^k

- - \to A P

A B = 3^i +^j -^k

A C =^i -^j + 3^k .

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Find $\to a . (\to b \times \to c)$ , if $\to a = 2^i +^j + 3^k, \to b = -^i + 2^j +^k$ and $\to c = 3^i +^j + 2^k$ .

\to a = 2^i +^j + 3^k, \to b = -^i + 2^j +^k

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\to a \cdot (\to b \times \to c)

\to a . (\to b \times \to c)

\to a = 2^i +^j + 3^k, \to b = -^i + 2^j +^k a n d \to c = 3^i +^j + 2^k

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