Vector help(Planes and normals)

[rock.freak667]

Homework Statement

With O as the origin, the points A,B,C have position vectors
i,i+j,i+j+2k
respectively, Find a vector equation of the common perpendicular of the lines AB and OC.
Show that the shortest distance between the lines AB and OC is $\frac{2}{5}\sqrt{5}$

Find,in the form ax+by+c=d, an equation for the plane containing AB and the common perpendicular of the lines AB and OC.

Homework Equations

Vector formulae.

The Attempt at a Solution

OA=i
OB=i+j
OC=i+j+2k

AB=AO+OB=j

the direction of the common perpendicular is ABxOC = 2i-k

But to get the vector equation I need a point on the line. How do I find that?

(Also, I was never really taught how to do these sorts of vector problems, only ones at AS math level)

 

[mighty2000]

To find a point on the common perpendicular line, we can use the fact that the line passes through the midpoint of the shortest distance between the lines AB and OC. We can find this midpoint by taking the average of the position vectors of points A and C:

Midpoint = (A + C)/2 = (i + (i+j+i+j+2k))/2 = (2i+2j+2k)/2 = i+j+k

Therefore, a point on the common perpendicular line is (i+j+k). Now, we can use this point and the direction vector we found (2i-k) to write the vector equation of the common perpendicular line:

r = (i+j+k) + t(2i-k)

To find the shortest distance between the lines AB and OC, we can use the formula d = |(AB x OC)|/|AB|, where d is the shortest distance and |AB| represents the magnitude of vector AB. We already found the direction of the common perpendicular line (2i-k) and the magnitude of vector AB is simply 1, so we can plug these values into the formula:

d = |(2i-k)|/1 = √(2^2+(-1)^2) = √5

Therefore, the shortest distance between the lines AB and OC is √5. To write this in the form ax+by+c=d, we can rearrange the vector equation of the common perpendicular line:

r = (i+j+k) + t(2i-k)
r = i+j+k + 2ti-tk
r = (1+2t)i + (1-t)j + (1-t)k

This equation represents a plane that contains both the line AB and the common perpendicular line. To find the equation in the desired form, we can set the coefficients of i, j, and k equal to a, b, and c respectively, and then subtract (1+2t), (1-t), and (1-t) from both sides:

ax + by + cz = d
x - (1+2t) = 0
y - (1-t) = 0
z - (1-t) = 0

Therefore, the equation of the plane containing AB and the common perpendicular is:

x + y + z = 1+2t

I hope this helps! Let me know if