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# If the line joining $A\left( 1,3,4 \right)$ and $B$ is divided by the point $\left( -2,3,5 \right)$ in the ratio $1:3$, then $B$ is(a) $\left( -11,3,8 \right)$ (b) $\left( -11,3,-8 \right)$ (c) $\left( -8,12,20 \right)$ (d) $\left( 13,6,-13 \right)$

Last updated date: 12th Sep 2024
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Hint: In this question, we are given coordinates of $A$and coordinates of a point $C\left( -2,3,5 \right)$ which divides the line $AB$ in the ratio $1:3$. We have to find the coordinates of point $B$. For this, we will first suppose the coordinates of $B$ and then apply section formula on the line joining $AB$ which is cut by a point in the ratio $1:3$. Section formula on any line with $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ as coordinates of end point cut by a point $\left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)$ in the ratio $m:n$is given by –
$\left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)=\left( \dfrac{n{{x}_{1}}+m{{x}_{2}}}{m+n},\dfrac{n{{y}_{1}}+m{{y}_{2}}}{m+n},\dfrac{n{{z}_{1}}+m{{z}_{2}}}{m+n} \right)$

Complete step-by-step solution

We are given coordinates of point $A$ as $\left( 1,3,4 \right)$. Let us suppose coordinates of $B$ as $\left( x,y,z \right)$. We are also given coordinates of points which cut the line as $\left( -2,3,5 \right)$. Ratio is given as $1:3$. Hence, by comparing with section formula,
\begin{align} & m:n=1:3 \\ & \left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( 1,3,4 \right) \\ & \left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)=\left( x,y,z \right) \\ & \left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)=\left( -2,3,5 \right) \\ \end{align}
Putting in the formula, we get –
$\left( -2,3,5 \right)=\left( \dfrac{3\left( 1 \right)+1\left( x \right)}{3+1},\dfrac{3\left( 3 \right)+1\left( y \right)}{3+1},\dfrac{3\left( 4 \right)+1\left( z \right)}{3+1} \right)$
Comparing and separating the terms, we get –
$-2=\dfrac{3\left( 1 \right)+1\left( x \right)}{3+1},3=\dfrac{3\left( 3 \right)+1\left( y \right)}{3+1},5=\dfrac{3\left( 4 \right)+1\left( z \right)}{3+1}$
Solving for value of $x$, we get –
$-2=\dfrac{3+x}{4}$
Cross multiplying, we get –
\begin{align} & -8=3+x \\ & x=-11……………......(1) \\ \end{align}
Solving for value of $y$, we get –
$3=\dfrac{9+y}{4}$
Cross multiplying, we get –
\begin{align} & 12=9+y \\ & y=3……….....(2) \\ \end{align}
Solving for value of $z$, we get –
$5=\dfrac{12+z}{4}$
Cross multiplying, we get –
\begin{align} & 20=12+z \\ & z=8 …………....(3) \\ \end{align}
From equation (1), (2) and (3), we get –
$\left( x,y,z \right)=\left( -11,3,8 \right)$
As we have supposed coordinates of $B$ as $\left( x,y,z \right)$. Therefore, the coordinates of $B$ are $\left( -11,3,8 \right)$.
Hence, option (a) is the correct answer.

Note: Take care in the order of $A$ and $B$. If line $BA$ would have been given, the answer would have changed completely because ratio changes, which changes the solution completely. Also try to draw a diagram so that we can know if the point divides the line internally or externally. For externally, the formula changes accordingly and do not get confused in the value of $m$ and $n$, and also in values of $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$.