Question

If $AB=3i+j-k$ and $AC=i-j+3k$. If the point $P$ on the line segment $BC$ is equidistant from $AB$ and $AC$, then $AP$ is?

• A. $2i–k$
• B. $i–2k$
• C. $2i+k$
• D. None of these

Answer 1

The correct option is C

$2i+k$

Explanation for the correct option:

Step 1. Find the value of $AP$:

Given,

Point $P$ is equidistant from $AB$ and $AC$ is on the bisector of the $\angle BAC$.

$\therefore$vector along the internal bisector of the $\angle BAC=\frac{AB}{\left|AB\right|}+\frac{AC}{\left|AC\right|}$

$$\begin{gathered} =\frac{3i+j+k}{\sqrt{9+1+1}}+\frac{i-j+3k}{\sqrt{1+1+9}} \\ =\frac{3i+j+k}{\sqrt{11}}+\frac{i-j+3k}{\sqrt{11}} \\ =\frac{3i+j+k+i-j+3k}{\sqrt{11}} \\ =\frac{1}{\sqrt{11}}\left(4i+2k\right) \end{gathered}$$

$\therefore AP=t(2i+k)$

And, $BP=AP-AB$

$$\begin{gathered} =t(2i+k)-(3i+j-k) \\ =(2t-3)i-j+(t+1)k \end{gathered}$$

Also, $BC=AC-AB$

$$\begin{gathered} =(i-j+3k)-(3i+j-k) \\ =-2i-2j+4k \end{gathered}$$

Step 2. as per given condition, $BP=sBC$

$\therefore (2t-3)i-j+(t+1)k=s(-2i-2j+4k)$

compare $i,j,k$ Components,

$\Rightarrow$$2t-3=-2s,-1=-2s,t+1=4s$

$\Rightarrow$$s=\frac{1}{2}$and $t=1$

$\mathbf{\therefore }\mathit{A}\mathit{P}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathit{i}\mathbf{}\mathbf{+}\mathbf{}\mathit{k}$

Hence, option ‘C’ is Correct.