If the plane 2x -- y + 2z + 3 = 0 has the distances [1 3] and [2 3] units from the planes …

Question

If the plane 2x -- y + 2z + 3 = 0 has the distances
$${1 \over 3}$$
 and
$${2 \over 3}$$
 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

Distance formula

(i) $${{\left| {\lambda  - 6} \right|} \over {\sqrt {16 + 4 + 16} }} = \left| {{{\lambda  - 6} \over 6}} \right| = {1 \over 3}$$

$$ \Rightarrow $$ $$\left| {\lambda  - 6} \right| = 2$$

$$ \Rightarrow $$ $$\lambda  = 8,4$$

(ii) $${{\left| {\mu  - 3} \right|} \over {\sqrt {4 + 4 + 1} }} = {2 \over 3}$$

$$ \Rightarrow $$ $$\left| {\mu  - 3} \right| = 2$$

$$ \Rightarrow $$ $$\mu  = 5,1$$

$$ \therefore $$ $${\left( {\mu  + \lambda } \right)_{\max }} = 13$$

Solutions · 1