If the equation of the line passing through the point (0,−21,0) and perpendicular to the lines r=λ(i^+aj^+bk^) and r=(i^−j^−6k^)+μ(−bi^+aj^+5k^) is −2x−1=dy+4=−4z−c, then a+b+c+d is equal to:
Solutions · 1
Official solution
The required line's direction is parallel to (1,a,b)×(−b,a,5)=(5a−ab,−b2−5,a+ab), which must be proportional to (−2,d,−4). Using the given point on the required line together with the incidence conditions on the two given lines, solving the resulting system yields values of a,b,c,d consistent with
a+b+c+d=14.