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In how many ways can three vectors be chosen from the set of vectors (S ={ai+bj+ck : a,b ,c belong to {1,-1,0 } } such that they are not coplanar?

Note by Keshav Tiwari
2 years, 8 months ago

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The question is easier to solve if we count all the possible configurations of the three vectors and then subtract the number of vectors that are coplanar. By the way, my solution ignores the $$0$$ vector because if we were to include it every set of $$3$$ vectors would be coplanar. Also, your question doesn't specify whether the vectors are distinct or not, but my solution assumes that they're distinct.

$\text{Number of Possible Vectors}=26\times25\times24=15600$

For each value of $$a$$, $$b$$ and $$c$$ we can choose from a set of $$3$$ elements, namely the set $$\{-1,0,1\}$$. This means there are $$3^3$$ possible vectors for the first vector, minus the zero vector gives us $$3^3-1=26$$. Since the vectors are distinct, we have one less possible vector to choose from for the second vector, and two less for the third, therefore our answer must be $$26\times25\times24=15600$$ possible vectors to choose from.

Now how many are coplanar? First, let's reduce the problem to finding out how many there are in just one plane, starting with the $$xy$$-plane. This plane has $$8$$ vectors in it because we set $$c=0$$, there are $$3^2$$ choices for $$a$$ and $$b$$ and we ignore the case $$a=b=0$$ so $$3^2-1=8$$ options. How many ways can we pick three vectors from a set of $$8$$? Easy, $${8\choose3}=56$$.

So how many planes can we make with three coplanar vectors? First note that all of the planes must go through the origin. It's a lot simpler if you can visualize a 3d graph of all the vectors, I found the best way to do so was to pick up a Rubik's cube, where each piece represents a vector. Using this technique I counted $$9$$ possible planes, so the total number of combinations of vectors such that all three are coplanar must be the number of planes available times the number of combinations of coplanar vectors in that plane.

$9{8\choose3}=9(56)=504$

Finally, if there are $$15600$$ possible combinations of vectors available to choose from, and $$504$$ of those combinations result in coplanar vectors, then our final answer must be the difference of the two numbers.

$\text{Total Number of Vectors}-\text{Sets of Three Coplanar Vectors} = 15600-504=\boxed{15096}$

- 2 years, 7 months ago

ahaa ! Thankyou !

- 2 years, 7 months ago

- 2 years, 7 months ago

Can you tell me what have you tried? Do you know the condition for three vectors to be not coplanar?

- 2 years, 7 months ago

Sure , I thought of using 3 D rather than matrices. Actually I tried this problem after solving another which said a,b,c belong to {-1,1} . Clearly there were 8 possible vectors in total . I made groups of 2 each (with opposite signs eg i+j+k and -i-j-k make a group) , so i got 4 groups (with each having two vectors) . If we choose three vectors (from different groups ) they will be non coplanar( as they represent the body diagonal of a cube) . Hence total no. of non coplanar vectors were 3C1*(2C1)^3 =32. I trieed a similar approach without any success .

- 2 years, 7 months ago

For three vectors to be coplanar their determinant should be zero.

$$(\begin{matrix} a1 & b1 & c1 \\ a2 & b2 & c2 \\ a3 & b3 & c3 \end{matrix})$$.

Where ai+bj+ck are vectors.

- 2 years, 7 months ago