Answer to Question #91096 in Analytic Geometry for Kamlesh Devi

Question

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?

Answer 1

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.

$$t=w+\alpha v,$$

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

Let's assume

$$w = a$$

and

$$v = a - b$$

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:

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

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

$$l = 1/√3, m = 1/√3, n = 1/√3$$

Knowing the fact that

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

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

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

and

$$x = y = z$$

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

$$u = \theta + \beta <1, 1, 1> = <\beta , \beta , \beta >$$

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

$$<2+\alpha ,-3-4\alpha ,-\alpha >=<\beta ,\beta , \beta >$$

$$\beta = -\alpha , \alpha = -1, \beta = 1$$

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

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