Let the position vectors of the points A, B, C and D be 5i+5j+2 k,i+2j+3k,-2i+ j+4k and -i…

Question

Let the position vectors of the points A, B, C and D be
$5\hat{i}+5\hat{j}+2\lambda \hat{k},\hat{i}+2\hat{j}+3\hat{k},-2\hat{i}+\lambda \hat{j}+4\hat{k}$ and $-\hat{i}+5\hat{j}+6\hat{k}$. Let the set $S={\lambda \in R$ :
the points A, B, C and D are coplanar $}$.

Then $\sum _{\lambda \in S}(\lambda +2{)}^{2}$ is equal to :

Accepted Answer

Given, position vectors of the points $A,B,C$ and $D$ be

$5\hat{i}+5\hat{j}+2\lambda \hat{k},\hat{i}+2\hat{j}+3\hat{k},-2\hat{i}+\lambda \hat{j}+4\hat{k}$ and $-\hat{i}+5\hat{j}+6\hat{k}$

$\begin{gathered} \overset{	o \leftarrow }{AB} =(\hat{i}+2\hat{j}+3\hat{k})-(5\hat{i}+5\hat{j}+2\lambda \hat{k})=-4\hat{i}-3\hat{j}+(3-2\lambda )\hat{k} \ \ \overset{	o \leftarrow }{AC} =(-2\hat{i}+\lambda \hat{j}+4\hat{k})-(5\hat{i}+5\hat{j}+2\lambda \hat{k}) \ \ =-7\hat{i}+(\lambda -5)\hat{j}+(4-2\lambda )\hat{k} $

$ \overset{	o \leftarrow }{AD} =(-\hat{i}+5\hat{j}+6\hat{k})-(5\hat{i}+5\hat{j}+2\lambda \hat{k}) \ \ =-6\hat{i}+(6-2\lambda )\hat{k} \end{gathered}$

Since, points $A,B,C$ and $D$ are coplanar

$\begin{gathered} 	herefore [\overset{	o \leftarrow }{AB}\overset{	o \leftarrow }{AC}\overset{	o \leftarrow }{AD}]=0 \ \Rightarrow |\begin{array}{ccc}-4 \quad  -3 \quad  (3-2\lambda ) \ -7 \quad  (\lambda -5) \quad  (4-2\lambda ) \ -6 \quad  0 \quad  6-2\lambda \end{array}|=0 \end{gathered}$

$\begin{gathered} \Rightarrow -6{(-12+6\lambda )-(13\lambda -15-2{\lambda }^{2})} \ +(6-2\lambda ){-4\lambda +20-21}=0 \ \Rightarrow -6(2{\lambda }^{2}-7\lambda +3)+(6-2\lambda )(-4\lambda -1)=0 \ \Rightarrow -12{\lambda }^{2}+42\lambda -18+8{\lambda }^{2}-22\lambda -6=0 \ \Rightarrow -4{\lambda }^{2}+20\lambda -24=0 \ \Rightarrow {\lambda }^{2}-5\lambda +6=0 \ \Rightarrow (\lambda -2)(\lambda -3)=0 \ \Rightarrow \lambda =2,3 \end{gathered}$

$	herefore \sum _{\lambda \varepsilon S}(\lambda +2{)}^{2}=(2+2{)}^{2}+(3+2{)}^{2}=16+25=41$

Solutions · 1