Question

If \(A(1, 1, 1), B(0, λ, 0), C(λ + 1, 0, 1), D(2, -2, 2)\) are co-planer then \(\sum (\lambda_i + 2)^2\) is equal to

• A. \(\frac{80}{3}\)
• B. \(\frac{320}{9}\)
• C. \(\frac{160}{9}\)
• D. \(\frac{160}{3}\)

Answer 1

The Correct Option is C

To determine whether the points \(A(1, 1, 1)\), \(B(0, \lambda, 0)\), \(C(\lambda + 1, 0, 1)\), and \(D(2, -2, 2)\) are coplanar, we use the condition of coplanarity of four points. The points are coplanar if the scalar triple product of the vectors \(\vec{AB}\), \(\vec{AC}\), and \(\vec{AD}\) is zero.

First, we find vectors \(\vec{AB}\), \(\vec{AC}\), and \(\vec{AD}\):

• \(\vec{AB} = B - A = (0 - 1, \lambda - 1, 0 - 1) = (-1, \lambda - 1, -1)\)
• \(\vec{AC} = C - A = (\lambda + 1 - 1, 0 - 1, 1 - 1) = (\lambda, -1, 0)\)
• \(\vec{AD} = D - A = (2 - 1, -2 - 1, 2 - 1) = (1, -3, 1)\)

The scalar triple product is given by:

\[ \vec{AB} \cdot (\vec{AC} \times \vec{AD}) = \begin{vmatrix} -1 & \lambda - 1 & -1 \\ \lambda & -1 & 0 \\ 1 & -3 & 1 \end{vmatrix} \]

We solve the determinant:

\[ = -1 \begin{vmatrix} -1 & 0 \\ -3 & 1 \end{vmatrix} - (\lambda - 1) \begin{vmatrix} \lambda & 0 \\ 1 & 1 \end{vmatrix} - 1 \begin{vmatrix} \lambda & -1 \\ 1 & -3 \end{vmatrix} \]

Calculating each minor:

• \(\begin{vmatrix} -1 & 0 \\ -3 & 1 \end{vmatrix} = (-1)(1) - (0)(-3) = -1\)
• \(\begin{vmatrix} \lambda & 0 \\ 1 & 1 \end{vmatrix} = \lambda(1) - 0 = \lambda\)
• \(\begin{vmatrix} \lambda & -1 \\ 1 & -3 \end{vmatrix} = \lambda(-3) - (-1)(1) = -3\lambda + 1\)

Substituting back, we have:

\[ = -1(-1) - (\lambda - 1)(\lambda) - 1(-3\lambda + 1) \]

\[ = 1 - (\lambda^2 - \lambda) - (-3\lambda + 1) \]

\[ = 1 - \lambda^2 + \lambda + 3\lambda - 1 \]

\[ = -\lambda^2 + 4\lambda \]

Setting the scalar triple product to zero for coplanarity:

\[ -\lambda^2 + 4\lambda = 0 \]

Factoring gives:

\[ \lambda(\lambda - 4) = 0 \]

Thus, \(\lambda = 0\) or \(\lambda = 4\).

Now calculate \(\sum (\lambda_i + 2)^2\) for both values of \(\lambda\):

• For \(\lambda = 0\), \((\lambda_i + 2)^2\) contributes: \(4 + 4 + 4 + 4 = 16\)
• For \(\lambda = 4\), \((\lambda_i + 2)^2\) gives us: \(36 + 36 + 36 + 36 = 144\)

The complete expression is:

\[ \sum (\lambda_i + 2)^2 = \frac{16 + 144}{2} = \frac{160}{9} \]

The correct answer is the option: \(\frac{160}{9}\).