Recover plane-distance and coupled-motion parameters

Use the completed vector foundations

This transfer follows spatial projection, line/plane equations and vector calculus. Distance uses the normal projection; collision uses position at one common time; perpendicular velocity uses a separate dot product. Keep these roles distinct

Derive the unsigned distance and the foot

For a nonzero normal, use the signed residual to find the distance and perpendicular foot. Substitution shows that the foot lies on the plane and the displacement from the point to the foot is normal to it

Work both inverse-distance branches

The normal length is 7. A distance of 2 gives two real allowed parameters, one on each side of the plane

Give the smaller parameter for this point at distance 2 from the plane

Give the larger parameter for this point at distance 2 from the plane

How many distinct parameter values put this point on the plane?

Keep the sign when finding the foot

The signed residual is negative, so finding the foot adds a positive multiple of the normal. The foot lies on the plane, and its displacement from the point has length 2

Give the z-coordinate of the perpendicular foot from the point to the plane

Couple position and velocity conditions

The particles share nonnegative time. Require collision and perpendicular velocities at the stated time. Position equality gives the offsets; the velocity dot product gives the remaining condition

Give the smaller value satisfying the perpendicular-velocity condition

Give the horizontal offset for this branch of the collision equation

Give the shared y-offset when these positions agree at the stated time

Which verification establishes a collision at T=2?

Check acceleration without changing branches

The accelerations are(0,2) and(0,−18), so their magnitudes are2 and18 for both branches. For k=6,u=−36,v=40 both particles are at(12,4) at T=2; for k=−6,u=36 they meet at(−12,4). Each branch must retain its matched offset

Justify distance branches and degeneracies, ungraded

Derive the distance and foot formulas. For P(λ)=A+λV, explain the numbers of solutions when n·V≠0 for positive or zero target distance. Then explain what changes if n·V=0, rather than dividing by zero

Model: use a normal decomposition and check the slope

Choose any Q on the plane. The normal component of P−Q has signed length(n·P−d)/|n|, giving the distance formula and the stated foot. If n·V≠0, each signed distance equation is linear with one solution; positive distance gives two distinct values, while distance 0 gives one. If n·V=0, the distance is constant along the line: every λ qualifies when that constant equals the target, otherwise none qualifies

Recover and verify every coupled branch, ungraded

The particles collide with perpendicular velocities at the stated time. Independently solve for all parameters, derive both accelerations, verify collision and orthogonality for each branch, and prove there is no other collision time for nonnegative time

Model: solve together and substitute each result

AtT=2, position equality gives u=−6k,v=40. The dot product4k²−144=0 gives k=±6, yielding(k,u,v)=(6,−36,40) or(−6,36,40). Direct substitution gives common positions(12,4) or(−12,4), while velocity dot products vanish. Accelerations are(0,2),(0,−18), with magnitudes2,18. The y-position equation is t²=−9t²+40, so 10t²=40. Only t=2 is allowed by t≥0. Thus no second future collision has been omitted