Let the point A be the foot of perpendicular drawn from the point P(a,b,0) on the line 2x−1=1y−2=3z−α. If the midpoint of the line segment PA is (0,43,−41), then the value of a2+b2+α2 is equal to:
Solutions · 1
Official solution
General point on the line: A(1+2r,2+r,α+3r). Using the midpoint conditions with M=(0,43,−41):
\[\frac{2r+1+a}{2}=0 \Rightarrow 2r+a=-1 \quad(i)\]
\[\frac{r+2+b}{2}=\frac34 \Rightarrow r+b=-\frac12 \quad(ii)\]
\[\frac{3r+\alpha}{2}=-\frac14 \Rightarrow 3r+\alpha=-\frac12 \quad(iii)\]
Since A is the foot of the perpendicular, PA⊥(2,1,3). Writing PA=2(M−P) and taking the dot product with (2,1,3) gives, after simplification, a+b=0.
From (i), (ii): a=−1−2r,b=−21−r. Substituting into a+b=0:
\[(-1-2r)+\left(-\frac12-r\right)=0 \implies -\frac32-3r=0 \implies r=-\frac12.\]
Then a=0,b=0, and from (iii): α=−21−3r=1.
\[a^2+b^2+\alpha^2=0+0+1=1.\]