Question

What are the co-ordinates of P which divides AB externally in the ratio $1:2$ if $A\left( {1,2, - 1} \right)$ and $B\left( { - 1,0,1} \right)$ are given:
A. $\dfrac{1}{3}\left( {1,4, - 1} \right)$
B. $\left( {3,4, - 3} \right)$
C. $\dfrac{1}{3}\left( {3,4, - 3} \right)$
D. None of these

Answer 1

Hint: In order to solve this type of question, we will use the section formula for the point dividing the line in some ratio externally i.e., $\left( {x,y,z} \right) = \left( {\dfrac{{{m_1}{x_2} - {m_2}{x_1}}}{{{m_1} - {m_2}}},\dfrac{{{m_1}{y_2} - {m_2}{y_1}}}{{{m_1} - {m_2}}},\dfrac{{{m_1}{z_2} - {m_2}{z_1}}}{{{m_1} - {m_2}}}} \right)$ to find the coordinates of that point. Here, $\left( {{x_1},{y_1},{z_1}} \right) = \left( {1,2, - 1} \right)$ and $\left( {{x_2},{y_2},{z_2}} \right) = \left( { - 1,0,1} \right)$. Also, ${m_1} = 1$ and ${m_2} = 2.$

Formula used:
$\left( {x,y,z} \right) = \left( {\dfrac{{{m_1}{x_2} - {m_2}{x_1}}}{{{m_1} - {m_2}}},\dfrac{{{m_1}{y_2} - {m_2}{y_1}}}{{{m_1} - {m_2}}},\dfrac{{{m_1}{z_2} - {m_2}{z_1}}}{{{m_1} - {m_2}}}} \right)$

Complete step by step solution:
We are given,
$A\left( {1,2, - 1} \right)$ and $B\left( { - 1,0,1} \right)$
Let P divides AB externally in the ratio $1:2$ then applying the section formula for the point dividing the line in some ratio externally and substituting the values in it.
$P = \left( {\dfrac{{1 \times \left( { - 1} \right) - 2 \times 1}}{{1 - 2}},\dfrac{{1 \times 0 - 2 \times 2}}{{1 - 2}},\dfrac{{1 \times 1 - 2 \times \left( { - 1} \right)}}{{1 - 2}}} \right)$
Solve it to get the required coordinates of the point P.
$P = \left( {\dfrac{{ - 1 - 2}}{{ - 1}},\dfrac{{0 - 4}}{{ - 1}},\dfrac{{1 + 2}}{{ - 1}}} \right)$
$P = \left( {3, - 4, - 3} \right)$
$\therefore $The correct option is B.

Note: The key to solve this type of question is to be sure that you are using the section formula for the point dividing the line in some ratio externally and not internally. Otherwise, it may lead to an incorrect answer. It is also called external division.