Drone path, radar distance and protected plane
A rescue drone flies along the straight line L given by (x,y,z) = (1,2,3) + λ(2,-1,1), with distances in kilometres and λ a real parameter. A radar station is at P(4,0,1). A protected airspace is the plane π: 2x - y + 2z = 6. The flight controller notices that the point A(1,2,3) satisfies the plane equation and claims that the whole flight path lies inside the protected airspace. The controller must also know how close the drone comes to the radar, because the camera cannot focus on objects nearer than 3 km. The drone cannot leave L during this mission. A second concern is the angle at which the path would cross the boundary of the protected airspace, since a steep crossing is considered riskier. The controller needs a clear geometric report before authorising the flight.
1. The controller claims that because A(1,2,3) lies on π, the whole line L lies in π. Is this correct? Find L ∩ π and describe the intersection.
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The direction of L is d=(2,-1,1) and a normal to π is n=(2,-1,2). Their dot product is d·n = 4+1+2=7, not zero. A line lies in a plane only if its direction is perpendicular to the normal; here it is not. Since A corresponds to λ=0 and satisfies 21-2+23=6, A lies on π. Because d·n ≠ 0, the line crosses the plane at exactly one point, namely A. So the claim is false: the path touches the protected airspace at A and immediately leaves it.
2. Find the point of L closest to P(4,0,1) and the minimum distance. Can the camera focus on the radar there?
Read the worked answer
Let A=(1,2,3) and AP=P-A=(3,-2,-2). Minimise |AP-λd|^2. The projection scalar is λ=(AP·d)/(d·d)=((3)(2)+(-2)(-1)+(-2)(1))/(4+1+1)=(6+2-2)/6=1. The closest point is Q=A+1*d=(3,1,4). The vector from Q to P is (1,-1,-3), whose length is sqrt(1+1+9)=sqrt(11)≈3.317 km. Since 3.317 > 3, the camera can focus at the closest approach. The margin is only about 0.317 km, so the controller should treat it as tight.
3. Compute the angle between L and π. Is the crossing steep?
Read the worked answer
For a line with direction d and a plane with normal n, sin θ = |d·n|/(|d||n|), where θ is the angle between the line and the plane. |d|=sqrt(6), |n|=sqrt(9)=3, and |d·n|=7. Hence sin θ=7/(3sqrt(6))≈0.952, so θ≈72.2°. The line is close to perpendicular to the plane; the angle with the normal is only about 17.8°. This is a steep crossing, so the risk assessment should treat it as such.