First, I suspect you are missing parentheses in your equations. It is very easy to forget them when working with typeset fractions which contain implied parentheses in the numerator and denominator.
I assume the equations are:
(x-1)/k = (y-2)/2 = (z+1)/(k-1).....(1)
(x+3)/(-2) = z/1, y=-1 ..........(2)
If that's the case, solution follows.
First find the direction vector of the line 1:
We look for the slope of each of variable x, y and z to form the direction vector, namely:
V1=<1/k, 1/2, 1/(k-1)>
1/k is obtained from (x-1)/k=x/k-1/k. The slope of x is therefore 1/k. You would do the same for the slopes of y and z to complete the vector V1.
Then follows the vector V2 of line 2, which is actually a line lying in the plane y=-1, therefore y does not vary, so the y-component of V2=0.
Working with x, and z similarly as line 1, we have
V2=<1/(-2), 0, 1/1>
We now proceed to find the conditions under which V1 and V2 are perpendicular. It turns out that two vectors are perpendicular when the dot product is zero. So we calculate
V1.V2 = 1/(k-1)-1/(2k) = 0
after some simplifications.
To find k, we only have to solve for k in the above equation to get k=-1.
Determine the value of k for which the direction vectors of the lines
x-1/k = y-2/2 = z+1/k-1
and
x+3/-2 = z/1, y=-1
are perpendicular.
2 answers
great, thank you!
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