Question

Let $(\lambda ,2,1)$ be a point on the plane which passes through the point $(4,–2,2)$. If the plane is perpendicular to the line joining the points $(–2,–21,29)$ and$(–1,–16,23)$, then ${\left(\frac{\lambda }{11}\right)}^2-\left(\frac{4\lambda }{11}\right)-4$is equal to

Answer 1

Step 1: Illustrating the figure and finding the value for $\lambda$

Let $A=(\lambda ,2,1)$ be a point on the plane which passes through the point $B=(4,–2,2)$

Let $P=(-2,-21,29)\text{and}Q=(-1,-16,23)$

$$\begin{gathered} AB\perp PQ \\ \therefore \overrightarrow{AB}.\overrightarrow{PQ}=0 \\ \left[\right(4–\lambda )i–4j+k].[i+5j–6k]=0 \\ 4–\lambda –20–6=0 \\ \therefore \lambda =–22 \end{gathered}$$

Step 2: Finding the value for the given equation

Let's find $\frac{\lambda }{11}$

$\left(\frac{\lambda }{11}\right)=–2$

Therefore,

$$\begin{gathered} {\left(\frac{\lambda }{11}\right)}^2-\left(\frac{4\lambda }{11}\right)-4 \\ ={\left(-2\right)}^2-4\left(-2\right)-4 \\ =4+8-4 \\ =8 \end{gathered}$$

Thus, ${\left(\frac{\lambda }{11}\right)}^2-\left(\frac{4\lambda }{11}\right)-4=8$