Question

Find the vector equation of the line perpendicular to the vectors `3bb(ulhati)-4bb(ul hatj)+bb(ul hatk)` and `2bb(ul hatj)+3bb(ul hatk)`?

Answer 1

`bb(vecr) = ( (2), (1), (-1) )+ lamda ( (-14), (-9), (6) )`

Explanation:

First consider the vectors `3bb(ulhati)-4bb(ul hatj)+bb(ul hatk)` and `2bb(ul hatj)+3bb(ul hatk)`. A vector, `bb(vecn)`, perpendicular to these vectors is found by forming the cross product:

`bb(vecn) = ( (3), (-4), (1) ) xx ( (0), (2), (3) )`

`\ \ \ \ = | (bb(ul hati), bb(ul hatj), bb(ul hatk)), (3,-4,1), (0,2,3) |`

`\ \ \ \ = | (-4,1), (2,3) | bb(ul hati) - | (3,1), (0,3) | bb(ul hatj) + | (3,-4), (0,2) | bb(ul hatk)`

`\ \ \ \ = (-12-2) bb(ul hati) - (9-0) bb(ul hatj) + (6-0) bb(ul hatk)`
`\ \ \ \ = -14 bb(ul hati) - 9 bb(ul hatj) + 6 bb(ul hatk)`

So now that we have the vector that the line is in the direction of, so the line equation is given by:

`bb(vecr) = bb(vec(OG)) + lamda bb(vecr)`
`\ \ \ \ = ( (2), (1), (-1) )+ lamda ( (-14), (-9), (6) )`

We can confirm this with a 3D diagram: