Question

\(5\^{i}+5\hat{j}+2λ\hat{k},\hat{i}+2\hat{j}+3\hat{k},-2\hat{i}+λ\hat{j}+4\hat{k}\) and \(-\hat{i}+5\hat{j}+6\hat{k}.\) Let the set S = {λ∈ \(\mathbb{R}\) : The points A, B, C and D are coplanar}. Then \(\displaystyle\sum_{λ∈S}^{}(λ+2)^2\) is equal to

• A. 13
• B. 25
• C. 41
• D. \(\frac{37}{2}\)

Answer 1

The Correct Option is C

Coplanar Points Calculation 

The points A, B, C, and D are coplanar if the scalar triple product of the vectors $\overrightarrow{AB}$, $\overrightarrow{AC}$, and $\overrightarrow{AD}$ is zero. Let's find these vectors:

$\overrightarrow{AB} = i + 2j + 3k − (5i + 5j + 2λk) = −4i − 3j + (3 − 2λ)k$

$\overrightarrow{AC} = −2i + λj + 4k − (5i + 5j + 2λk) = −7i + (λ − 5)j + (4 − 2λ)k$

$\overrightarrow{AD} = −i + 5j + 6k − (5i + 5j + 2λk) = −6i + 0j + (6 − 2λ)k$

The scalar triple product is given by the determinant:

$\begin{vmatrix} -4 & -3 & 3 - 2λ \\ -7 & λ - 5 & 4 - 2λ \\ -6 & 0 & 6 - 2λ \end{vmatrix} = 0$

Expanding this determinant:

$-4λ^2 + 20λ − 24 = 0$

Dividing by -4:

$λ^2 − 5λ + 6 = 0$

Factoring the quadratic equation:

$(λ − 2)(λ − 3) = 0$

Thus, the solutions are $λ = 2$ and $λ = 3$. Therefore, $S = \{2, 3\}$.

Now we compute the sum:

$\sum_{λ \in S} (λ + 2)^2 = (2 + 2)^2 + (3 + 2)^2 = 16 + 25 = 41$

Conclusion: The value of $\sum_{λ \in S} (λ + 2)^2$ is 41.