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  • Let the position vectors of the points A, B, C and D be <img alt="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}" src="https://entrancecorne

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{\mathrm{i}}+5 \hat{\mathrm{j}}+6 \hat{\mathrm{k}} . Let the set  \mathrm{S}=\{\lambda \in \mathbb{R}$ : the points $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and $D$ are coplanar $\}$. Then $\sum_{\lambda \in \mathbb{S}}(\lambda+2)^2is equal to :

Option: 1

\frac{37}{2}


Option: 2

13


Option: 3

25


Option: 4

41


Answers (1)

best_answer

A, B, C, D are coplanar
\Rightarrow[\overrightarrow{\mathrm{AB}} \overrightarrow{\mathrm{AC}} \overrightarrow{\mathrm{AD}}]= 0 \quad \Rightarrow \begin{bmatrix} -4& -3 & 3-2\lambda\\ -7&\lambda-5 &4-2\lambda \\ -6& 0& 6-2\lambda \end{bmatrix}= 0

\Rightarrow-6\left[6 \lambda-12+2 \lambda^2+15-13 \lambda\right]+(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 \quad \Rightarrow \lambda^2-5 \lambda+6=0

(\lambda-3)(\lambda-2)=0 \sum_{\lambda=2}^{\lambda=2}
Now \sum_{\lambda \in \mathrm{s}}(\lambda+2)^2=16+25=41


 

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manish

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