If the plane 2x-y+2z+3=0 has the distances 13 and 23 units from the planes 4x-2y+4z+=0 and…

Question

If the plane $2x-y+2z+3=0$ has the distances $\dfrac13$ and $\dfrac23$ units from the planes $4x-2y+4z+\lambda=0$ and $2x-y+2z+\mu=0$, respectively, then the maximum value of $\lambda+\mu$ is equal to:

Accepted Answer

$4x-2y+4z+\lambda=0$ simplifies to $2x-y+2z+\tfrac\lambda2=0$, parallel to $2x-y+2z+3=0$:
$$\frac{\left|3-\tfrac\lambda2\right|}3=\frac13\implies \left|3-\tfrac\lambda2\right|=1\implies \lambda=8\text{ or }4 \implies \lambda_{\max}=8.$$
$$\frac{|\mu-3|}3=\frac23\implies|\mu-3|=2\implies \mu=5\text{ or }1\implies\mu_{\max}=5.$$
$$(\lambda+\mu)_{\max}=8+5=13.$$

Solutions · 1