The plane 2x-y+z=4 intersects the line segment joining the points A(a,-2,4) and B(2,b,-3) …

Question

The plane $2x-y+z=4$ intersects the line segment joining the points $A(a,-2,4)$ and $B(2,b,-3)$ at the point $C$ in the ratio $2:1$, and the distance of the point $C$ from the origin is $\sqrt5$. If $ab<0$ and $P$ is the point $(a-b,\ b,\ 2b-a)$, then $CP^2$ is equal to:

Accepted Answer

$AC:CB=2:1 \Rightarrow C=\left(\dfrac{a+4}{3},\ \dfrac{2b-2}{3},\ \dfrac{-2}{3}\right)$.

$C$ on the plane: $\dfrac{2a+8}{3}-\dfrac{2b-2}{3}-\dfrac23=4 \implies a-b=2.\quad(1)$

$OC=\sqrt5$: $\left(\dfrac{a+4}{3}\right)^2+\left(\dfrac{2b-2}{3}\right)^2+\dfrac49=5.\quad(2)$

Substituting $a=b+2$ into (2): $(b+6)^2+(2b-2)^2=41 \implies 5b^2+4b-1=0 \implies b=-1\ \text{or}\ \tfrac15$.

Correspondingly $a=1$ or $\tfrac{11}{5}$. Since $ab<0$: $(a,b)=(1,-1)$.
$$C=\left(\frac53,-\frac43,-\frac23\right),\qquad P=(2,-1,-3).$$
$$CP^2=\frac19+\frac19+\frac{49}{9}=\frac{51}{9}=\frac{17}{3}.$$

Solutions · 1