A,B,C are 3 points with position vectors a,b,c Find length of median

[Aurelius120]

Homework Statement

Let A,B,C are three points with position vectors $$a=\hat i+4\hat j+3\hat k$$ $$b=2\hat i+\alpha \hat j+4\hat k,$$ $$ c=3\hat i-2\hat j+5\hat k $$respectively .$\alpha \in R$. If $\alpha$ is the smallest positive integer for $a,b,c$ to be non-collinear, length of median through $A$ of $\Delta ABC$ is:

Relevant Equations

Equation of line in vector form : $\vec r= \vec a+t(\vec b-\vec a)$ if $\vec a$ and $\vec b$ are position vectors of two points on that line .
Magnitude of vector=$\sqrt{x^2+y^2+z^2}$

Heres how I proceeded,
Equation of line $AC$ in vector form:
$$\vec r=a+t(c-a)$$$$\vec r=(1i+4j+3k)+t(2i-6j+2k)$$
Since $B$ doesn't lie on $AC$ $b\neq (1+2t)i +(4-6t)j+(3+2t)k$
The following equation is derived:
$$2\hat i+\alpha \hat j+4\hat k\neq (1+2t)\hat i +(4-6t)\hat j+(3+2t)\hat k $$

How do I further solve this to obtain $\alpha$?

Once $\alpha$ is found, mid-point of $BC$ can be found using coordinate geometry. The magnitude of vector $|\vec {AM}|$ is required length of median.

Is there any alternate preferably easier/faster method to solve this?

 

[Hill]

How do I further solve this to obtain α?

First, find $\alpha$ that would make them colinear.

 

[Aurelius120]

First, find $\alpha$ that would make them colinear.

Finally got it!
For collinearity, $1+2t=2\implies t=\frac{1}{2}$ and $3+2t=4\implies t=\frac{1}{2}$
Since they have a common solution, $\alpha \neq 4-\frac{6}{2}$
Therefore $\alpha \neq 1\implies \alpha=2$

But if those equations in $t$ didn't have a common root, how would I solve for $\alpha$ ?

Also is there any alternate perhaps faster way to do this?

 

[PeroK]

Also is there any alternate perhaps faster way to do this?

I would have put A at the origin, so that the coordinates of A, B and C are $(0,0,0)$, $(1, \alpha -4, 1)$ and $(2, -6, 2)$.
(PS the length you want to calculate is unchanged by a change of origin.)

Then you can see by inspection that $\alpha = 1$ makes these collinear. So, you want $\alpha = 2$.

In any case, you could have proceeded by trial and error to find $\alpha$.

 

[Delta2]

But if those equations in t didn't have a common root, how would I solve for α ?

If they didn't have a common root, means that there is not a (same everywhere) $t$ that can make B to lie on the line AC , hence they are non colinear for any$\alpha$ (this statement would require a bit of thinking from your side to fully understand it), hence $\alpha$ could be any positive integer and since we want the smallest you would take $\alpha$ as 0 (or 1, not sure if the problem statement considers 0 as positive integer).

 

[PeroK]

If they didn't have a common root, means that there is not single $t$ that can make B to lie on the line AC, hence they are non colinear for any$\alpha$ (this statement would require a bit of thinking from your side to fully understand it), hence $\alpha$ could be any positive integer and since we want the smallest you would take a as 0 (or 1, not sure if the problem statement considers 0 as positive integer).

Zero is not a positive number. A positive number is greater than 0.

PS that's why we have the term non-negative (which includes zero).

 

[Delta2]

Zero is not a positive number. A positive number is greater than 0.

Yes I guess that is the current mathematical consensus but still there are some books that consider 0 as positive integer.

 

[PeroK]

Yes I guess that is the current mathematical consensus but still there are some books that consider 0 as positive integer.

That I doubt.

https://en.wikipedia.org/wiki/Positive_real_numbers

 

[Delta2]

That I doubt.

https://en.wikipedia.org/wiki/Positive_real_numbers

I said some books, not what wikipedia says, Wiki is with the mathematical consensus ok no problem with wiki here.

 

[Delta2]

That I doubt.

https://en.wikipedia.org/wiki/Positive_real_numbers

Yes ok I know probably 95% (or up to 99.9%) of the books out there don't consider 0 as positive integer, but I know a Greek (in Greek language by Greek authors) book on Linear and Abstract algebra that considers 0 part of the positive integer, this book was taught for the algebra course in the third year of the Greek Lyceum School (Lyceum for ages 16-18 is something equivalent to a college provided free by the Greek state) back at 1990-91.

 

[Hill]

Is there any alternate preferably easier/faster method to solve this?

Sure. Skip the whole collinearity part.

The midpoint of $BC$ is $M=\frac 1 2 (5, \alpha -2, 9)$.
The square length of $AM$ is $\frac 1 4 (118 - 20 \alpha + \alpha^2)$.
One can quickly see that $\alpha=2$ makes the expression in parenthesis $=82$, which is the answer choice (a).

 

[Delta2]

Sure. Skip the whole collinearity part.

The midpoint of $BC$ is $M=\frac 1 2 (5, \alpha -2, 9)$.
The square length of $AM$ is $\frac 1 4 (118 - 20 \alpha + \alpha^2)$.
One can quickly see that $\alpha=2$ makes the expression in parenthesis $=82$, which is the answer choice (a).

This is some sort of clever cheating lol but it wont work if this was a question in a test where you had to provide some reasoning for choosing the answer. If the test wanted only to chose the answer then ok lol.

 

[PeroK]

Homework Statement: Let A,B,C are three points with position vectors $$a=\hat i+4\hat j+3\hat k$$$$b=2\hat i+\alpha \hat j+4\hat k,$$$$ c=3\hat i-2\hat j+5\hat k $$respectively .

Note that the x-coordinates go in the sequence 1, 2, 3; the y-coordinates in the sequence 4, $\alpha$, -2; and the z-coordinates in the sequence 3, 4, 5. If you understand what collinearity means in terms of coordinates, you can see that $\alpha = 1$ makes B not just collinear with AC, but exactly the midpoint of AC.

 

[Steve4Physics]

... not sure if the problem statement considers 0 as positive integer

That would be irrational. Haha.

 

[SammyS]

I said some books, not what wikipedia says, Wiki is with the mathematical consensus ok no problem with wiki here.

I know of no mathematics book that considers zero to be a positive number. However, some mathematicians do consider zero to be a natural number, i.e. some consider that $\displaystyle 0 \in \mathbb{N)$ .

 

[Delta2]

I know of no mathematics book that considers zero to be a positive number. However, some mathematicians do consider zero to be a natural number, i.e. some consider that $\displaystyle 0 \in \mathbb{N)$ .

Well, that book was claiming also that the set of natural numbers includes zero and then it was stating that the set of natural numbers is also referred to as the set of positive integers!

 

[Aurelius120]

Well, that book was claiming also that the set of natural numbers includes zero and then it was stating that the set of natural numbers is also referred to as the set of positive integers!

I remember in middle school we had a book that considered $0$ neither positive nor negative throughout except in a question in the 'Sets Chapter'. Teacher asked us to correct the book.

 

[Delta2]

Ok I cant find the book in my 90's chest but I have a doubt and I think the book was defining $$\mathbb{N}={0,1,2,...}$$ and $$\mathbb{N^{*}}={1,2,3,...,}$$ and I cant remember which of those was said to be also referred as the set of positive integers...

 

[Steve4Physics]

Ok I cant find the book in my 90's chest but I have a doubt and I think the book was defining $$\mathbb{N}={0,1,2,...}$$ and $$\mathbb{N^{*}}={1,2,3,...,}$$ and I cant remember which of those was said to be also referred as the set of positive integers...

The second one!

In a math's context, 'positive' actually means 'greater than zero'!

You may just have to concede on this one!