Vector Algebra
Coplanar Vectors — Sum of Squares
nta_pyq_2023_apr
Grade 12

Question:

Let position vectors of $A,B,C,D$ 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 $S=\{\lambda\in\mathbb{R}: A,B,C,D\text{ coplanar}\}$. Then $\sum_{\lambda\in S}(\lambda+2)^2$ is equal to
25
\dfrac{37}{2}
14
41

Step-by-Step Solution

Key Concept: $[\overrightarrow{AB}\ \overrightarrow{AC}\ \overrightarrow{AD}]=0$. Expand the scalar triple product determinant and solve for $\lambda$.
$\lambda=3,2$. Sum $=41$.
Correct Answer: 4

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