Find Intersection Point of Vectors | Problem Solved

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To find the intersection point of two vectors given their initial points and equations, set up parametric equations for each vector and solve the resulting equations for the parameters. In three dimensions, lines may be skew and not intersect, while in two dimensions, only two equations are needed. If the vectors do not intersect, the minimum distance can be calculated by subtracting one vector from the other and taking the magnitude of the resultant vector. However, if two parallel vectors have the same direction and magnitude, the distance between them is zero, which highlights the need for careful consideration of their initial points. Understanding these concepts is crucial for accurately determining intersections and distances between vectors.
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i have just joined ur sit e
can u help me out
if we have the initial point of two vectors and equations of these vectors , how can we find the intersection point of these vectors
 
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?? Vectors are "free floating"- only specific representations have initial points. And even those may not intersect- they may not reach far enough.
If you mean you have two lines, one passing through (x_0,y_0,z_0) in the same direction as vector Ai+ Bj+ Ck, the other passing through (x_1, y_1, z_1), in the same direction as vector Ui+ Vj+ Wk, then the first line can be written in parametric equations x= x_0+ At, y= y_0+ Bt, z= z_0+ Ct and the second line can be written in parametric equations x= x_1+ Ut, y= y_1+ Vt, z= z_1+ Wt. Set "x=x", "y= y", "z= z", and solve the resulting three equations for s and t.
Of course, in general, three equations won't give just two unknown values- in three dimensions most lines are "skew" and don't intersect. If you are working in 2-dimensions, (x, y), just ignore the z equations and you have two equations to solve for s and t.
 
thanks sir for ur answer.
u got me right.

here is one other problem.
given initial points of two vector with their equation
what will be the minimum distance between them if they don't intersect.

i had read one of the answer written in ur site , that subtract one vector from the other and taking the magnitude of that resultant vector will give us result.

now suppose if our one vector start at (1,0,0) with equation 2i +0j + 0k i.e 2i. Other has (2,0,0) as initial point with eq 2i( as i think two parallel vectors with equal magnitude have same equation).then according to above solution ,the answer is zero. pease if possible give formula.


thanks
vinay
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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