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Section Formula

Let AB=3î+ĵ...

Question

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

2^i -^k

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^i - 2^k

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2^i +^k

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None of these

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Solution

The correct option is D $2^i +^k$
Since $P$ is equidistant from $A C & B C,$ it lies on the angle bisector of $∠ B A C$
Let us also assume that $A$ is the origin.

So, $\to P = x \frac{3^i +^j -^k +^i -^j + 3^k}{\sqrt{11}}$

$\Rightarrow \to P = y 4^i + 2^k$

Hence we can see the options to verify.

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Similar questions

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

and

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

If the point

P

on the line segment

B C

is equdistant from

A B

and

A C,

then

A P

If $\to r \times \to b = \to c \times \to b$ and $\to r . \to a = 0$ where $\to a = 2^i + 3^j -^k, \to b = 3^i -^j +^k$ and $\to c =^i +^j + 3^k$ , then $\to r$ is equal to

Q. If

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and

\to c = 3^i +^j + 2^k

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, if

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and

\to c = 3^i +^j + 2^k .

Q. If A, B, C, D are the points with position vectors

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Let −−→AB=3^i+^j−^k and −−→AC=^i−^j+3^k. If the point P on the line segment BC is equidistant from AB and AC then −−→AP is

The correct option is D 2^i+^kSince P is equidistant from AC & BC, it lies on the angle bisector of ∠BACLet us also assume that A is the origin.So, →P=x3^i+^j−^k+^i−^j+3^k√11⇒→P=y4^i+2^kHence we can see the options to verify.

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