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,
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−bThen 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,−3−4α,−α>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=θ+β<1,1,1>=<β,β,β> And the point we need is the intersection point of those lines
<2+α,−3−4α,−α>=<β,β,β>
β=−α,α=−1,β=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) .
Comments
Leave a comment