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Question: Answered & Verified by Expert
If the points with position vectors $\hat{i}-\hat{j}+\hat{k}, 2 \hat{i}-\hat{k}, \hat{j}+2 \hat{k}$ and $\hat{i}+\hat{j}+\lambda \hat{k}$ are coplanar, then the magnitude of the vector $6 \lambda \hat{i}-3 \hat{j}+6 \hat{k}$ is
MathematicsVector AlgebraTS EAMCETTS EAMCET 2023 (13 May Shift 2)
Options:
  • A $\sqrt{54}$
  • B $\sqrt{46}$
  • C 7
  • D 9
Solution:
2020 Upvotes Verified Answer
The correct answer is: 7
$\begin{aligned} & \text {} \vec{A}=\hat{i}-\hat{j}+\hat{k} \\ & \vec{B}=2 \hat{i}-\hat{k} \\ & \vec{C}=\hat{j}+2 \hat{k} \\ & \vec{D}=\hat{i}+\hat{j}+\lambda \hat{k} \\ & \overrightarrow{A D}=2 \hat{j}+(\lambda-1) \hat{k} \\ & \overline{B D}=-\hat{i}+\hat{j}+(\lambda+1) \hat{k} \\ & \overrightarrow{C D}=\hat{i}+(\lambda-2) \hat{k} \\ & \therefore \quad\left[\begin{array}{lll}\overrightarrow{A D} & \overrightarrow{B D} & \overrightarrow{C D}\end{array}\right]=0 \\ & \Rightarrow\left|\begin{array}{ccc}0 & 2 & \lambda-1 \\ -1 & 1 & \lambda+1 \\ 1 & 0 & \lambda-2\end{array}\right|=0 \\ & \Rightarrow-2(2-\lambda-\lambda-1)+(\lambda-1)(0-1)=0 \\ & \Rightarrow 4 \lambda-2+1-\lambda=0 \\ & \Rightarrow 3 \lambda-1=0 \Rightarrow \lambda=\frac{1}{3} \\ & \therefore \quad 6 \lambda \hat{i}-3 \hat{j}+6 \hat{k}=2 \hat{i}-3 \hat{j}+6 \hat{k} \\ & \text { Magnitude }=\sqrt{2^2+3^2+6^2}=\sqrt{49}=7 \text {. } \\ & \end{aligned}$

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