1. Tardigrade
  2. Question
  3. Mathematics
  4. The points A(3,2,0), B(5,3,2) and C(0,2,4) are the vertices of a triangle. Find the distance of the point A from the point in which the bisector of ∠BAC meets [BC].

Q. The points $A(3,2,0)$, $B(5,3,2)$ and $C(0,2,4)$ are the vertices of a triangle. Find the distance of the point $A$ from the point in which the bisector of $∠BAC$ meets $[BC]$.

Introduction to Three Dimensional Geometry Report Error

Solution:

Let $D$ be the points at which bisector of $∠BAC$ meets $[BC]$, then $D$ divides $[BC]$ internally in the ratio $c : b$ where $c = |AB|$ and $b = |AC|$.
image
Now $c = \left|AB\right| = \sqrt{\left(5-3\right)^{2}+\left(3-2\right)^{2}+\left(2-0\right)^{2}} = 3$ units
and $b = \left|AC\right| = \sqrt{\left(0-3\right)^{2}+\left(2-2\right)^{2}+\left(4-0\right)^{2}}=\sqrt{25}=5$ units
$\therefore D$ divides $\left[BC\right]$ in the ratio $3 : 5$
Hence, $D \equiv \left(\frac{3\times0+5\times 5}{3+5}, \frac{3\times2+5\times3}{3+5}, \frac{3\times4+5\times2}{3+5}\right)$,
i.e., $D \equiv \left(\frac{25}{8}, \frac{21}{8}, \frac{22}{8}\right)$.
Now, $\left|AD\right|= \sqrt{\left(\frac{25}{8}-3\right)^{2}+\left(\frac{21}{8}-2\right)^{2}+\left(\frac{22}{8}-0\right)^{2}}$
$= \sqrt{\frac{1}{64}+\frac{25}{64}+\frac{484}{64}}$
$=\sqrt{\frac{510}{64}}=\frac{1}{8}\sqrt{510}$

Questions from Introduction to Three Dimensional Geometry

1. The point which divides the line joining the points (1, 3, 4)and (4, 3, 1) internally in the ratio $2 : 1$, is

KEAM 2012

2. The line $\frac{x-1}{3} = \frac{y-1}{-1} = \frac{z+1}{0}$ and $ \frac{x-4}{2} = \frac{y+0}{0} = \frac{z+1}{3} $

3. The equation of the plane passing through the points $(2, 1, -1), (1, 1, 1), (3, 3, 0)$ is

4. The perpendicular distance of the point $P(6,7,8)$ from $XY$-plane is

KCET 2017

5. The point on the x-axis equidistant from the points $ (4,3,1) $ and $ (-2,\,-6,\,-2) $ is

J & K CET 2009

Mathematics Most Viewed Questions

1. The solution of $\frac{dy}{dx} = \frac{y}{x}+\tan \frac{y}{x}$ is

WBJEE 2011 Differential Equations

2. The solution of the differential equation $\frac{dy}{dx} = (x +y)^2$ is

COMEDK 2009 Differential Equations

3. $\int\frac{1}{\sin x\, \cos x}$ dx is equal to

KEAM 2016 Integrals

4. If $\begin{bmatrix}1&- \tan\theta \\ \tan \theta&1\end{bmatrix}\begin{bmatrix}1&\tan \theta \\ - \tan \theta &1\end{bmatrix}^{-1} = \begin{bmatrix}a&-b\\ b&a\end{bmatrix}$ then

COMEDK 2009 Matrices

5. The value of $ \int{\frac{{{x}^{2}}+1}{{{x}^{4}}-{{x}^{2}}+1}}dx $ is

KEAM 2007 Integrals

Latest Updates