# Thread: coliniar vectors

1. ## coliniar vectors

xOy. we have points M(1,n) , N(n,3) and P(2n,5) , n is natural. vectors MN and MP are coliniar. Find out n=?

2. ## Re: coliniar vectors

Originally Posted by archie13
xOy. we have points M(1,n) , N(n,3) and P(2n,5) , n is natural. vectors MN and MP are coliniar. Find out n=?
To get you started: If two vectors are linear then the dot product between them is either 1 or -1. Can you finish from here?

-Dan

3. ## Re: coliniar vectors

Thanks but i dont know what dot product means?

4. ## Re: coliniar vectors

A dot product of 2 vectors is defined as
a*b = |a|*|b|*cos(θ) where θ (theta) is the angle between those 2 vectors, and |a| and |b| is the magnitude of those vector.

Here's my solution, but we encourage you to first try for yourself.

Spoiler:
Translate the MN and MP vectors to the origin by subtracting M from N and M from P.
Now we only have to deal with 2 points (which we can treat as vectors with their start in the coordinate origin).
Let's call these 2 new points Y and Z vectors with their coordinates (n-1, 3-n) and (2n-1, 5-n) respectively.
Now we know that for 2 vectors to be collinear their direction must be same/opposite.
We can express this as an equality of ratios of their x y coordinates
(n-1)/(3-n)=(2n-1)/(5-n)
We know that the 2 ratios are equal when their cross products are equal (i.e. multiplying each side with the denominators)
So we do that and we get this: (n-1)*(5-n)=(2n-1)*(3-n) , multiply this and rearrange everything on one side and simplify and you've got yourself a nice simple quadratic equation which will give you 2 solutions. n=2 v n=-1, plug those in and check if they are correct.
Edit: Since n is natural, we exclude n=-1 from the solution

5. ## Re: coliniar vectors

Originally Posted by topsquark
To get you started: If two vectors are linear then the dot product between them is either 1 or -1. Can you finish from here?

-Dan
This is not true! Two unit vectors are collinear if and only if their dot product is -1 or 1 but that only applies to unit vectors. The vector from M to N is (n-1,3- n) and the vector from M to P is (2n- 1, 5- n). Those are NOT necessarily unit vectors. In order that they be "collinear" one must be a multiple or the other.

One way to check that is to divide the first vector by (2n-1)/(n-1) to get the vector that clearly is collinear to (n- 1, 3- n), (2n- 1, ((3- n)(2n-1))/(n-1)). For (n-1, 3- n) itself to be collinear to (2n-1, 5- n) we must have ((3- n)(2n- 1))/(n- 1)= 5- n. Multiply both sides by n- 1 to get (3- n)(2n- 1)= (n- 1)(5- n). That gives $\displaystyle 7n- 4n^2- 3= 6n- n^2- 5$, $\displaystyle 3n^2- n- 2= 0$. n must be $\displaystyle 1$ or $\displaystyle -\frac{2}{3}$

6. ## Re: coliniar vectors

Originally Posted by HallsofIvy
This is not true! Two unit vectors are collinear if and only if their dot product is -1 or 1 but that only applies to unit vectors. The vector from M to N is (n-1,3- n) and the vector from M to P is (2n- 1, 5- n). Those are NOT necessarily unit vectors. In order that they be "collinear" one must be a multiple or the other.
Whoops! I had meant to say that the cosine of the angle between them is 1 or -1.

Thanks for the catch.

-Dan

7. ## Re: coliniar vectors

Equivalently, two vectors are collinear if and only if there cross product is 0. Strictly speaking, the cross product can only be calculated for three dimensional vectors but we can take the third component to be 0.

Given points M= (1, n, 0), N= (n, 3, 0), P= (2n, 5), then MN= <n- 1, 3- n, 0> and MP= <2n-1, 5- n, 0>. Their cross product is $\displaystyle \left | \begin{array}{ccc} i & j & k \\ n- 1 & 3- n & 0 \\ 2n- 1 & 5- n & 0\end{array}\right|= 0i+ 0j+ [(n- 1)(5- n)- (3- n)(2n- 1)]k= [6n- 5- n^2- 7n+ 3+ 2n^2]k= [n^2- n- 2]k$.

We must have $\displaystyle n^2- n- 2= 0$. Solve that equation for n. There are two solutions but only one is a natural number.

8. ## Re: coliniar vectors

Originally Posted by HallsofIvy
Equivalently, two vectors are collinear if and only if there cross product is 0. Strictly speaking, the cross product can only be calculated for three dimensional vectors but we can take the third component to be 0.

Given points M= (1, n, 0), N= (n, 3, 0), P= (2n, 5), then MN= and MP= . Their cross product is $\displaystyle \left | \begin{array}{ccc} i & j & k \\ n- 1 & 3- n & 0 \\ 2n- 1 & 5- n & 0\end{array}\right|= 0i+ 0j+ [(n- 1)(5- n)- (3- n)(2n- 1)]k= [6n- 5- n^2- 7n+ 3+ 2n^2]k= [n^2- n- 2]k$.

We must have $\displaystyle n^2- n- 2= 0$. Solve that equation for n. There are two solutions but only one is a natural number.
Thanks a lot! n=2

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