Let the point, on the line passing through the points P(1,−2,3) and Q(5,−4,7), farther from the origin and at a distance of 9 units from the point P, be \[(\alpha, \beta, \gamma)\]. Then \[\alpha^2+\beta^2+\gamma^2\] is equal to...
Solutions · 1
Official solution
Line through PQ
\[\frac{x-1}{4}=\frac{y+2}{-2}=\frac{z-3}{4}\]
Any point on PQ. be \[R(4 \lambda+1,-2 \lambda-2,4 \lambda+3)\]
PR=9 unit
(PR)2=81
\[(4 \lambda+1-1)^2+(-2 \lambda-2+2)^2+(4 \lambda+3-3)^2=81\]
\[16 \lambda^2+4 \lambda^2+16 \lambda^2=81\]
\[36 \lambda^2=81\]
\[\lambda= \pm \frac{9}{6}= \pm \frac{3}{2}\]
\[\therefore R\] can be (7,−5,9) or (−5,1,−3)
Distance from origin for both points be \[\sqrt{49+25+81}\] and \[\sqrt{25+1+9}=\sqrt{35}\]
\[\therefore\] Distance of (7,−5,9) is farthest from origin
\[\therefore(\alpha, \beta, \gamma)=(7,-5,9)\]
Now 72+(−5)2+92=155