Intersection of straight lines (vectors)

[Foggy]

I'm getting myself muddled on this one. I know that the vector dot product will be zero for perpendicular vectors. I've worked out the point on line L and its direction vector but I'm going round in circles. Any help appreciated!

 

[pka]

I'm getting myself muddled on this one. I know that the vector dot product will be zero for perpendicular vectors. I've worked out the point on line L and its direction vector but I'm going round in circles. Any help appreciated!
View attachment 28531

Why not post what you have done? We will check the work.
In terms of [imath]\lambda[/imath] what is [imath]\overline{AP}~?[/imath]

 

[Foggy]

I've done a sketch to make the facts clearer
I've put line L into parametric form
I can't work out how to get a direction vector or mu for line AP

 

[Foggy]

When I couldn't work out how to use a.b=0, I then went on to try equating the x,y and z from each line. However I dont know what to do in this method when there are the extra variables d1,d2,d3 from the unknown direction vector.

equating x: 9+lambda = 4 + d1 mu
equating y: 13 + 4 lambda = 16 + d2 mu
equating z: -3 - 2 lambda = -3 + d3 mu

 

[pka]

If [imath]L(\lambda)=(9+\lambda,~13+4\lambda,~-3-2\lambda)[/imath] then [imath](\exists t)[L(t)=P=(9+t,13+4t,-3-2t)[/imath]
In terms of that [imath]t[/imath] can you express [imath]\overrightarrow {PA}~?[/imath]
If you can then find [imath]\overrightarrow {PA} \cdot <1,4,-2>=0~?[/imath]

 

[Foggy]

Great, finally got it. Thank you