If the length of the perpendicular from the point (β,0,β) (β=0) to the line 1x=0y−1=−1z+1 is 23, then β is equal to:
Solutions · 1
Official solution
Point on line A=(p,1,−p−1); AP=(p−β,1,−p−1−β) for P=(β,0,β), perpendicular to direction (1,0,−1):
\[(p-\beta)+(p+1+\beta)=0\implies p=-\tfrac12,\qquad A=\left(-\tfrac12,1,-\tfrac12\right).\]
\[AP^2=2\left(\beta+\tfrac12\right)^2+1=\tfrac32\implies \left(\beta+\tfrac12\right)^2=\tfrac14\implies \beta=0\text{ or }-1.\]
Since β=0, β=−1.