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=?
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:
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 $\displaystyle -\frac{2}{3}$
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.