Answer to Question #91096 in Analytic Geometry for Kamlesh Devi

Question #91096
Find the vector equation of the line passing through 2i - 3j and i + j + k. Which point on this line has direction cosines, 1/√3, 1/√3, 1/√3 and why?
1
Expert's answer
2019-06-24T10:19:57-0400

Assume we have two vectors (a and b) with coordinates <2, -3, 0> and <1, 1, 1> in the (i, j, k) basis.

Firstly, we need to find a vector equation of the line passing through those points.



t=w+αv,t=w+\alpha v,


where w is some vector with the end on that line and v is a vector parallel to it.

Let's assume

w=aw = a

and

v=abv = a - b

Then w has the coordinates <2, -3, 0> and v has <2 - 1, -3 - 1, 0 - 1> = <1, -4, -1>.

Vector equation of the line:


t=<2,3,0>+α<1,4,1>=<2+α,34α,α>t = <2,-3,0> + \alpha <1,-4,-1> = <2+\alpha ,-3-4\alpha,-\alpha>

Next, find a vector equation of another line that has given direction cosines


l=1/3,m=1/3,n=1/3l = 1/√3, m = 1/√3, n = 1/√3

Knowing the fact that


l=x/r,m=y/r,n=z/r,l = x/|r|, m = y/|r|, n = z/|r|,

where |r| is a length of some vector on that line and x, y, z are its coordinates, we have


x=r/3,y=r/3,z=r/3x = |r|/√3, y = |r|/√3, z = |r|/√3

and

x=y=zx = y = z

So, vector equation of the line with given direction cosines is


u=θ+β<1,1,1>=<β,β,β>u = \theta + \beta <1, 1, 1> = <\beta , \beta , \beta >

And the point we need is the intersection point of those lines


<2+α,34α,α>=<β,β,β><2+\alpha ,-3-4\alpha ,-\alpha >=<\beta ,\beta , \beta >

β=α,α=1,β=1\beta = -\alpha , \alpha = -1, \beta = 1

So, the point we need has the coordinates (1, 1, 1).

Answer. t=<2,3,0>+α<1,4,1>;A=(1,1,1)t = <2,-3,0> + \alpha <1,-4,-1>; A = (1,1,1) .


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