Please help!

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?

#Geometry #Matrices #JEE

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$ 2 \times 3 $
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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.

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

Garrett Clarke - 5 years, 10 months ago

ahaa ! Thankyou !

Keshav Tiwari - 5 years, 10 months ago

Your welcome! Interesting problem!

Garrett Clarke - 5 years, 10 months ago

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

Pranshu Gaba - 5 years, 10 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.

Nelson Mandela - 5 years, 10 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 .

Keshav Tiwari - 5 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	\( 2 \times 3 \)
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$