# Thread: Very Urgent: (distances of vectors)

1. ## Very Urgent: (distances of vectors)

1) Find the coordinates of the foot of the perpendicular from Q(3,2,4) to the line r= (-6,-7,-3) +t(5,3,4)

2) The common perpendicular of two skew lines with the direction vectors d1 and d2 is the line that intersects both the skew lines and has direction vector n= d1 x d2. Find the points of intersection fo the common perpendicular with each of the lines (x,y,z)=(0,-1,0) + s(1,2,1) and
(x,y,z)= (-2,2,0) + t(2,-1,2)

These were assigned tonight and this content will be on tomorrows test I have no idea what to do. We've been studying distance between vectors,planes and points, but I can't do these at all.

Thanks!

2. Originally Posted by white_cap
1) Find the coordinates of the foot of the perpendicular from Q(3,2,4) to the line r= (-6,-7,-3) +t(5,3,4)

...
1. Create a plane perpendicular to the line containing Q:

$[5, 3, 4] \cdot (r - [3, 2, 4]) = 0$

2. Calculate the point of intersection between the line and the plane:

$[5, 3, 4] \cdot ([-6,-7,-3] +t[5,3,4] - [3, 2, 4]) = 0$

$[5, 3, 4] \cdot ([-9,-9,-7] +t[5,3,4] ) = 0$

$-100+50t = 0~\implies~ t = 2$

3. Plug in this value into the equation of the line. You'll get F(4, -1, 5)

3. Originally Posted by white_cap
...

2) The common perpendicular of two skew lines with the direction vectors d1 and d2 is the line that intersects both the skew lines and has direction vector n= d1 x d2. Find the points of intersection fo the common perpendicular with each of the lines (x,y,z)=(0,-1,0) + s(1,2,1) and
(x,y,z)= (-2,2,0) + t(2,-1,2)

...
1. Calculate the normal vector to the 2 direction vectors:

$\vec n = [1, 2, 1] \times [2, -1, 2] = [5, 0, -5]$

2. Calculate the equation of a line through a point of $l_1$ in the direction of $\vec n$:

$l_n: [x, y, z] = \underbrace{[0,-1,0] + s \cdot [1,2,1]}_{\text{ arbitrary point on the first line }} + r \cdot [5, 0, -5]$

3. Calculate the point of intersection $l_n \cap l_2$:

$[0,-1,0] + s \cdot [1,2,1] + r \cdot [5, 0, -5] = [-2,2,0] + t \cdot [2,-1,2]$

4. You have now a system of simultaneous equations in 3 variables. I don't know how you solve such a system but you should get:

$r=-\frac15 ~\wedge~ s=1~\wedge~t=1$

5. Plug in s into the equation of $l_1$ to get A(1, 1, 1)
and plug in t into the equation of $l_2$ to get B(0, 1, 2)