Question

Consider the set of eight vectors \[V = \left\{ {a\hat i + b\hat j + c\hat k;a,b,c \in \left\{ { - 1,1} \right\}} \right\}\]. Three non-coplanar vectors are chosen from \[V\] in \[{2^p}\] ways. Then \[P\] is
A. 2
B. 3
C. 4
D. 5

Answer 1

Hint: Here we will proceed by writing all the set of eight given vectors. Then find the number of ways of selecting two collinear vectors and then the number of ways of selecting three coplanar vectors. The number of ways of selecting three non-coplanar vectors is equal to the difference of number of ways of selecting three vectors and the number of ways of selecting three coplanar vectors.

 Complete step by step answer:
Given the set of eight vectors \[V = \left\{ {a\hat i + b\hat j + c\hat k;a,b,c \in \left\{ { - 1,1} \right\}} \right\}\]
So, the vectors are \[\vec a\left( {1,1,1} \right),\vec b\left( { - 1,1,1} \right),\vec c\left( {1, - 1,1} \right),\vec d\left( { - 1, - 1,1} \right)\] and the rest of the vectors are \[ - \vec a\left( { - 1, - 1, - 1} \right), - \vec b\left( {1, - 1, - 1} \right), - \vec c\left( { - 1,1, - 1} \right), - \vec d\left( {1,1, - 1} \right)\]
Now, the number of ways of selecting three vectors from eight vectors \[ = {}^8{C_3} = \dfrac{{8 \times 7 \times 6}}{{1 \times 2 \times 3}} = 56\]
Let us consider the number of ways of selecting three coplanar vectors.
We know that the vectors which are negative to each other in magnitude and the opposite to each other in direction are collinear.
So, the vectors \[\vec a\& - \vec a;\vec b\& - \vec b;\vec c\& - \vec c\] and \[\vec d\& - \vec d\] are collinear vectors.
Hence, the number of ways two select two collinear vectors = 4
We know that any other vector in a plane with two collinear are said to be in coplanar.
So, now we have to select any other vector for two collinear vectors of out set of eight vectors.
Hence, the number of ways of selecting any other vector with two collinear vectors = 6
By multiplicative rule of combinations, we have
The number of ways of selecting three coplanar vectors \[ = 4 \times 6 = 24\]
Now, number of ways of selecting three non-coplanar vectors = number of ways of selecting three vectors – number of ways of selecting three coplanar vectors
Thus, number of ways of selecting 3 non-coplanar vectors out of set of 8 vectors \[ = 56 - 24 = 32\]
But given that it is equal to \[{2^p}\] ways. So, we have
\[
   \Rightarrow {2^p} = 32 \\
   \Rightarrow {2^p} = {2^5} \\
\]
Equating the powers as bases are equal, we have
\[\therefore p = 5\]
Thus, the correct option is D. 5

Note:
In this problem we have used multiplicative principle of combinations i.e., if there are \[x\]number of ways of selecting one thing and\[y\] number of ways of selecting another, then the total number of ways of selecting both the things is given in \[xy\] number of ways.