Question

What's the line through (2,1,0) and perpendicular to vectors i+j and also perpendicular to j+k? Is it <1+t,1,0> where t is the independent variable?

Answer 1

to me, first off, take cross product between <1,1,0> x <0,1,1> to get the normal vector of the plane content the required line. I got \(\vec n\)=<1,-1,1>
then, apply formula of the equation of the line whose direction vector is \(\vec n\)
so that the equation of the line is
x = 2 +t
y= 1-t
z = t

Answer 2

Can you elaborate a little more on forming the planar equation? that's my current problem.

Answer 3

do you got me from the first step ( find out the normal vector of the plane which perpendicular to both vectors <i, j, 0> and <0, j, k>?

Answer 4

Yes, right after getting the normal vector can you explain that?

Answer 5

so, you have the point , you have direction vector, you can write the equation of the line pass through that point as I did. ( that's formula)

Answer 6

Sorry, I wasn't being specific enough, but that's alright, I see what I did wrong thanks.

Answer 7

the formula of the line past through the point \((x_0,y_0, z_0)\) with the direction <a, b , c> is
x =\(x_0\) + at
y= \(y_0\)+ bt
z =\(z_0\) + ct