Question

If $a,b,c$ ae non-coplanar vectors, then which of the following points are collinear whose position vectors are given by:
This question has multiple correct options
A. $a - 2b + 3c,2a + 3b - 4c, - 7b + 10c$
B. $3a - 4b + 3c, - 4a + 5b - 6c,4a - 7b + 6c$
C. $2a + 5b - 4c,a + 4b - 3c,4a + 7b - 6c$
D. $6a - b - 2c,2a + 3b + 2c, - a - 9b + 7c$

Answer 1

In this question, first of all we consider the position vectors and apply the condition of collinearity to find whether the given points are in collinear or not. So, use this concept to reach the solution of the problem.

Complete step-by-step answer :
A. let $\overrightarrow P ,\overrightarrow Q ,\overrightarrow R$ be the position vectors which are given by $\overrightarrow P = \overrightarrow a - 2\overrightarrow b + 3\overrightarrow c ,\overrightarrow Q = 2\overrightarrow a + 3\overrightarrow b - 4\overrightarrow c$ and $\overrightarrow R = - 7\overrightarrow b + 10\overrightarrow c$.
We know that, the condition of collinearity to be the position vectors $\overrightarrow P = \overrightarrow a + \overrightarrow b + \overrightarrow c ,\overrightarrow Q = \overrightarrow d + \overrightarrow e + \overrightarrow f$ and $\overrightarrow R = \overrightarrow g + \overrightarrow h + \overrightarrow i$ are in collinear is \left| {\begin{array}{*{20}{c}} a&b;&c; \\\ d&e;&f; \\\ g&h;&i; \end{array}} \right| = 0
So, the condition for the position vectors $\overrightarrow P ,\overrightarrow Q ,\overrightarrow R$ to be in collinear is

\Rightarrow \left| {\begin{array}{*{20}{c}} 1&{ - 2}&3 \\\ 2&3&{ - 4} \\\ 0&{ - 7}&{10} \end{array}} \right| = 0 \\\ \Rightarrow 1\left\\{ {\left( 3 \right)\left( {10} \right) - \left( { - 4} \right)\left( { - 7} \right)} \right\\} - \left( { - 2} \right)\left\\{ {\left( 2 \right)\left( {10} \right) - \left( { - 4} \right)\left( 0 \right)} \right\\} + 3\left\\{ {\left( 2 \right)\left( { - 7} \right) - \left( 3 \right)\left( 0 \right)} \right\\} = 0 \\\ \Rightarrow 1\left\\{ {30 - 28} \right\\} + 2\left\\{ {20 - 0} \right\\} + 3\left\\{ { - 14 - 0} \right\\} = 0 \\\ \Rightarrow 1\left( 2 \right) + 2\left( {20} \right) + 3\left( { - 14} \right) = 0 \\\ \Rightarrow 2 + 40 - 42 = 0 \\\ \therefore 0 = 0 \\\

Therefore, the position vectors $\overrightarrow P = \overrightarrow a - 2\overrightarrow b + 3\overrightarrow c ,\overrightarrow Q = 2\overrightarrow a + 3\overrightarrow b - 4\overrightarrow c$ and $\overrightarrow R = - 7\overrightarrow b + 10\overrightarrow c$ are in collinear.
Thus, the point A. $a - 2b + 3c,2a + 3b - 4c, - 7b + 10c$ is collinear.
B. let $\overrightarrow P ,\overrightarrow Q ,\overrightarrow R$ be the position vectors which are given by $\overrightarrow P = 3\overrightarrow a - 4\overrightarrow b + 3\overrightarrow c ,\overrightarrow Q = - 4\overrightarrow a + 5\overrightarrow b - 6\overrightarrow c$ and $\overrightarrow R = 4\overrightarrow a - 7\overrightarrow b + 6\overrightarrow c$.
The condition for the position vectors $\overrightarrow P ,\overrightarrow Q ,\overrightarrow R$ to be in collinear is

\Rightarrow \left| {\begin{array}{*{20}{c}} 3&{ - 4}&3 \\\ { - 4}&5&{ - 6} \\\ 4&{ - 7}&6 \end{array}} \right| = 0 \\\ \Rightarrow 3\left\\{ {\left( 5 \right)\left( 6 \right) - \left( { - 7} \right)\left( { - 6} \right)} \right\\} - \left( { - 4} \right)\left\\{ {\left( { - 4} \right)\left( 6 \right) - \left( { - 6} \right)\left( 4 \right)} \right\\} + 3\left\\{ {\left( { - 4} \right)\left( { - 7} \right) - \left( 5 \right)\left( 4 \right)} \right\\} = 0 \\\ \Rightarrow 3\left\\{ {30 - 42} \right\\} + 4\left\\{ { - 24 + 24} \right\\} + 3\left\\{ {28 - 20} \right\\} = 0 \\\ \Rightarrow 3\left( { - 12} \right) + 4\left( 0 \right) + 3\left( 8 \right) = 0 \\\ \Rightarrow - 36 - 0 + 24 = 0 \\\ \Rightarrow - 12 = 0 \\\

But $- 12 \ne 0$. So, the position vectors $\overrightarrow P = 3\overrightarrow a - 4\overrightarrow b + 3\overrightarrow c ,\overrightarrow Q = - 4\overrightarrow a + 5\overrightarrow b - 6\overrightarrow c$ and $\overrightarrow R = 4\overrightarrow a - 7\overrightarrow b + 6\overrightarrow c$ are not in collinear.
Thus, the point B. $3a - 4b + 3c, - 4a + 5b - 6c,4a - 7b + 6c$ is not collinear.
C. let $\overrightarrow P ,\overrightarrow Q ,\overrightarrow R$ be the position vectors which are given by $\overrightarrow P = 2\overrightarrow a + 5\overrightarrow b - 4\overrightarrow c ,\overrightarrow Q = \overrightarrow a + 4\overrightarrow b - 3\overrightarrow c$ and $\overrightarrow R = 4\overrightarrow a + 7\overrightarrow b - 6\overrightarrow c$.
The condition for the position vectors $\overrightarrow P ,\overrightarrow Q ,\overrightarrow R$ to be in collinear is

\Rightarrow \left| {\begin{array}{*{20}{c}} 2&5&{ - 4} \\\ 1&4&{ - 3} \\\ 4&7&{ - 6} \end{array}} \right| = 0 \\\ \Rightarrow 2\left\\{ {\left( 4 \right)\left( { - 6} \right) - \left( { - 3} \right)\left( 7 \right)} \right\\} - 5\left\\{ {\left( 1 \right)\left( { - 6} \right) - \left( { - 3} \right)\left( 4 \right)} \right\\} + \left( { - 4} \right)\left\\{ {\left( 1 \right)\left( 7 \right) - \left( 4 \right)\left( 4 \right)} \right\\} = 0 \\\ \Rightarrow 2\left\\{ { - 24 + 21} \right\\} - 5\left\\{ { - 6 + 12} \right\\} - 4\left\\{ {7 - 16} \right\\} = 0 \\\ \Rightarrow 2\left( { - 3} \right) - 5\left( 6 \right) - 4\left( { - 9} \right) = 0 \\\ \Rightarrow - 6 - 30 + 36 = 0 \\\ \therefore 0 = 0 \\\

So, the position vectors $\overrightarrow P = 2\overrightarrow a + 5\overrightarrow b - 4\overrightarrow c ,\overrightarrow Q = \overrightarrow a + 4\overrightarrow b - 3\overrightarrow c$ and $\overrightarrow R = 4\overrightarrow a + 7\overrightarrow b - 6\overrightarrow c$ are in colinear.
Thus, the point C. $2a + 5b - 4c,a + 4b - 3c,4a + 7b - 6c$ is collinear.
D. let $\overrightarrow P ,\overrightarrow Q ,\overrightarrow R$ be the position vectors which are given by $\overrightarrow P = 6\overrightarrow a - \overrightarrow b - 2\overrightarrow c ,\overrightarrow Q = 2\overrightarrow a + 3\overrightarrow b + 2\overrightarrow c$ and $\overrightarrow R = - \overrightarrow a - 9\overrightarrow b + 7\overrightarrow c$.

\Rightarrow \left| {\begin{array}{*{20}{c}} 6&{ - 1}&{ - 2} \\\ 2&3&2 \\\ { - 1}&{ - 9}&7 \end{array}} \right| = 0 \\\ \Rightarrow 6\left\\{ {\left( 3 \right)\left( 7 \right) - \left( 2 \right)\left( { - 9} \right)} \right\\} - \left( { - 1} \right)\left\\{ {\left( 2 \right)\left( 7 \right) - \left( 2 \right)\left( { - 1} \right)} \right\\} + \left( { - 2} \right)\left\\{ {\left( 2 \right)\left( { - 9} \right) - \left( 3 \right)\left( { - 1} \right)} \right\\} = 0 \\\ \Rightarrow 6\left\\{ {21 + 18} \right\\} + 1\left\\{ {14 + 2} \right\\} - 2\left\\{ { - 18 + 3} \right\\} = 0 \\\ \Rightarrow 6\left( {39} \right) + 1\left( {16} \right) - 2\left( { - 15} \right) = 0 \\\ \Rightarrow 234 + 16 + 30 = 0 \\\ \Rightarrow 280 = 0 \\\

But, $280 \ne 0$. So, the position vectors $\overrightarrow P = 6\overrightarrow a - \overrightarrow b - 2\overrightarrow c ,\overrightarrow Q = 2\overrightarrow a + 3\overrightarrow b + 2\overrightarrow c$ and $\overrightarrow R = - \overrightarrow a - 9\overrightarrow b + 7\overrightarrow c$are not in collinear.
Thus, point D. $6a - b - 2c,2a + 3b + 2c, - a - 9b + 7c$is not collinear.

Note :Three points with position vectors $\overrightarrow A ,\overrightarrow B ,\overrightarrow C$ are collinear if and only if the vectors $\left( {\overrightarrow B - \overrightarrow A } \right)$ and $\left( {\overrightarrow C - \overrightarrow A } \right)$ are in parallel. In other words, to prove collinearity, we would need to show $\left( {\overrightarrow B - \overrightarrow A } \right) = \lambda \left( {\overrightarrow C - \overrightarrow A } \right)$ for some constant $\lambda$. By this method also we can prove the collinearity.