Question

If a, b and c are non-coplanar vectors then,
\[-\overrightarrow{a}+4\overrightarrow{b}-3\overrightarrow{c}\], \[3\overrightarrow{a}+2\overrightarrow{b}-5\overrightarrow{c}\], \[-3\overrightarrow{a}+8\overrightarrow{b}-5\overrightarrow{c}\], \[-3\overrightarrow{a}+2\overrightarrow{b}+\overrightarrow{c}\] are collinear.
(a) True
(b) False

Answer 1

Hint: First of all take the first 3 vectors and prove their determinant to be zero and then take the next 3 vectors and prove their determinant to be zero, which means the first 3 and the last 3 vectors are collinear to each other. So, we can say that all 4 are collinear to each other.

Complete step-by-step answer:

First of all, we are given that \[\overrightarrow{a}\], \[\overrightarrow{b}\] and \[\overrightarrow{c}\] are non-coplanar. Also, we are given 4 vectors formed by combination of \[\overrightarrow{a}\], \[\overrightarrow{b}\] and \[\overrightarrow{c}\] and we have to check if they are collinear or not. First of all, let us consider 4 vectors as,

\[\overrightarrow{P}=-\overrightarrow{a}+4\overrightarrow{b}-3\overrightarrow{c}\]

\[\overrightarrow{Q}=3\overrightarrow{a}+2\overrightarrow{b}-5\overrightarrow{c}\]

\[\overrightarrow{R}=-3\overrightarrow{a}+8\overrightarrow{b}-5\overrightarrow{c}\]

\[\overrightarrow{S}=-3\overrightarrow{a}+2\overrightarrow{b}+\overrightarrow{c}\]

We know that if three vectors are collinear, their scalar product is zero. So, let us check for the vectors \[\overrightarrow{P},\overrightarrow{Q}\text{ and }\overrightarrow{R}\]. If \[\overrightarrow{P},\overrightarrow{Q}\text{ and }\overrightarrow{R}\] are collinear vectors, then we get,

\[\left[ \overrightarrow{P}\text{ }\overrightarrow{Q}\text{ }\overrightarrow{R} \right]=0\]

By substituting the values of \[\overrightarrow{P},\overrightarrow{Q}\text{ and }\overrightarrow{R}\], we get,

\[\left[ \overrightarrow{P}\text{ }\overrightarrow{Q}\text{ }\overrightarrow{R} \right]=\left|

\begin{matrix}

   -1 & 4 & -3 \\

   3 & 2 & -5 \\

   -3 & 8 & -5 \\

\end{matrix} \right|....\left( i \right)\]

We know that

\[\left| \begin{matrix}

   a & b & c \\

   d & e & f \\

   g & h & i \\

\end{matrix} \right|=a\left( ei-hf \right)-b\left( di-gf \right)+c\left( dh-ge \right)\]

So, we get,

\[\left| \begin{matrix}

   -1 & 4 & -3 \\

   3 & 2 & -5 \\

   -3 & 8 & -5 \\

\end{matrix} \right|=-1\left( -10+40 \right)-4\left( -15-15 \right)-3\left( 24+6 \right)\]

\[=-1\left( +30 \right)-4\left( -30 \right)-3\left( -30 \right)\]

\[=-30+120-90\]

\[=-120+120\]

= 0

So, we get, \[\left[ \overrightarrow{P}\text{ }\overrightarrow{Q}\text{ }\overrightarrow{R} \right]=0\]. Hence, \[\overrightarrow{P},\overrightarrow{Q}\text{ and }\overrightarrow{R}\] are collinear vectors.

Now let us check for vectors \[\overrightarrow{Q},\overrightarrow{R}\text{ and }\overrightarrow{S}\]. We know that \[\overrightarrow{Q},\overrightarrow{R}\text{ and }\overrightarrow{S}\] are collinear if

\[\left[ \overrightarrow{Q}\text{ }\overrightarrow{R}\text{ }\overrightarrow{S} \right]=0\]
By substituting the values of \[\overrightarrow{Q},\overrightarrow{R}\text{ and }\overrightarrow{S}\], we get,

\[\left[ \overrightarrow{Q}\text{ }\overrightarrow{R}\text{ }\overrightarrow{S} \right]=\left|

\begin{matrix}

   3 & 2 & -5 \\

   -3 & 8 & -5 \\

   -3 & 2 & 1 \\

\end{matrix} \right|\]

\[=3\left( 8+10 \right)-2\left( -3-15 \right)-5\left( -6+24 \right)\]

\[=3\left( 18 \right)+2\left( 18 \right)-5\left( 18 \right)\]

\[=54+36-90\]

\[=90-90\]

= 0

So, we get, \[\left[ \overrightarrow{Q}\text{ }\overrightarrow{R}\text{ }\overrightarrow{S} \right]=0\]. Hence, \[\overrightarrow{Q},\overrightarrow{R}\text{ and }\overrightarrow{S}\] are collinear vectors.

Now, we have proved that \[\overrightarrow{P},\overrightarrow{Q}\text{ and }\overrightarrow{R}\] and \[\overrightarrow{Q},\overrightarrow{R}\text{ and }\overrightarrow{S}\] are collinear vectors. Hence, we can say that all four vectors \[\overrightarrow{P},\overrightarrow{Q}\text{,}\overrightarrow{R}\text{ and }\overrightarrow{S}\] are collinear. So, the statement given in the question is true.


Note: In this question, some students make this mistake of assuming \[\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\] as points and further assuming \[\overrightarrow{P},\overrightarrow{Q}\text{,}\overrightarrow{R}\text{ and }\overrightarrow{S}\] as points and trying to prove them collinear by equating \[\overrightarrow{PQ}=\lambda \left( \overrightarrow{QR} \right)\] or \[\overrightarrow{QR}=\lambda \left( \overrightarrow{RS} \right)\] which is wrong. Here, it is clearly given that \[\overrightarrow{P},\overrightarrow{Q}\text{,}\overrightarrow{R}\text{ and }\overrightarrow{S}\] and \[\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\] are vectors. So, we can’t use the above expression here as it is the only valid point.