Let Q be the foot of perpendicular from the origin to the plane 4x−3y+z+13=0 and R be the point (−1,1,−6) on the plane. Then length QR is:
Solutions · 1
Official solution
Foot Q lies on line through origin with direction (4,−3,1): Q=(4r,−3r,r); substituting into the plane: 16r+9r+r+13=0⟹26r=−13⟹r=−21.
\[Q=\left(-2,\tfrac32,-\tfrac12\right).\]
\[QR^2=(-1+2)^2+\left(1-\tfrac32\right)^2+\left(-6+\tfrac12\right)^2=1+\tfrac14+\tfrac{121}4=\tfrac{63}2\implies QR=3\sqrt{\tfrac72}.\]