Question

If the points $P(-{\lambda }^2,\lambda ,0)$ and $Q({\lambda }^2,\frac{7\lambda }{3},7)$ lies on the same sides of the plane $x-3y+2z=4.$ Then the integral value(s) of $\lambda$ can be ?

• A. $-2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (B), (C)

$4$

Given Points:
$P(-{\lambda }^2,\lambda ,0)$ and $Q({\lambda }^2,\frac{7\lambda }{3},7)$
and plane $x-3y+2z=4.$
As points lies on the same sides ,
$\Rightarrow (a{x}_1+b{x}_1+c{z}_1-d)\cdot (a{x}_2+b{x}_2+c{z}_2-d)>0\Rightarrow (-{\lambda }^2-3\lambda -4)\cdot ({\lambda }^2-7\lambda +10)>0\cdots \left(i\right)$
As, $-{\lambda }^2-3\lambda -4<0\forall \lambda \in R$
$\Rightarrow {\lambda }^2-7\lambda +10<0$ [From $\left(i\right)$]
$\Rightarrow (\lambda -2)(\lambda -5)<0\Rightarrow \lambda \in (2,5)$
So, integral values are : $3,4.$