Question

Let F⃗=xi⃗+(x+y)j⃗+(x−y+z)k⃗ . Let the line l be x=t−1, y=3−4t , z=1−4t .

(a) Find a point P=(x0,y0,z0) where F⃗ is parallel to l .

Find a point Q=(x1,y1,z1) at which F⃗ and l are perpendicular.

Give an equation for the set of all points at which F⃗ and l are perpendicular.

Answer 1

Any vector parallel to given line will have proportional direction cosine

Direction of line <1,-4,-4>

If line I and F are parallel then it should satisfy

<x,x+y,x-y+z> = k * <1,-4,-4>

We get

x=k

X+y=-4k

X-y+z =-4k

y=-4k-x =-4k-k =-5k

z=-4k+y-x = -4k+-5k-k =-10k

So

(x,y,z) will be (k,-5k,-10k)

Put k=1

We get

(1,-5,-10) this is the point where F is parallel to given line

b)

We need to find a point at which F is perpendicular to line I

That means their dot product will be zero

<x,x+y,x-y+z>.<1,-4,-4> =0

x-4(x+y)-4(x-y+z) =0

x-4x-4y-4x+4y-4z =0

-7x+0y-4z =0

At point

(4,0,-7) is one of the which satisfy this equation