Let $\vec{AB} = 3\hat i+ \hat j-\hat k$ and $\vec{AC} = \hat i-\hat j+ 3\hat k$ and a point P on the line segment BC is equidistant from AB and AC, then $\vec{AP}$ is
Answer & explanation
Correct answer: option 3
Clearly, a point equidistant from AB and AC is on
10 the bisector of the ∠BAC.
A vector along the bisector of ∠BAC is
$\frac{\vec{AB}}{|\vec{AB}|}+\frac{\vec{AC}}{|\vec{AC}|}=\frac{1}{\sqrt{11}}(4\hat i+2\hat k)=-\frac{2}{\sqrt{11}}(2\hat i+\hat k)$
Let $\vec{AP}=λ(2\hat i+\hat k)$
$∴\vec{BP}=\vec{AP}-\vec{AB}=λ(2\hat i+\hat k)-(3\hat i+ \hat j-\hat k)$
$⇒\vec{BP}=(2λ-3)\hat i-\hat j+(λ+1)\hat k$
Also, $\vec{BC}=\vec{AC}-\vec{AB}=-2\hat i-2\hat j+4\hat k$
Since $\vec{BP} || \vec{BC}$. Therefore,
$\vec{BP}=t\vec{BC}$
$⇒(2λ-3)\hat i-\hat j+(λ+1) \hat k=t(-2\hat i-2\hat j +4\hat k)$
$⇒2λ-3=-2t,-1=-2t, λ+1=4t$
$⇒λ-1,t=\frac{1}{2}$
$∴\vec{AP}=2\hat i+\hat k$