Question

If \( S \) is the set of all real values of \( a \) such that a plane passing through the points \( (-a^2, 1, 1), (1, -a^2, 1), (1, 1, -a^2) \) also passes through the point \( (-1, -1, 1) \), then \( S = \dots \)

• A. \( \{\sqrt{3}\} \)
• B. \( \{\sqrt{3}, -\sqrt{3}\} \)
• C. \( \{1, -1\} \)
• D. \( \{3, -3\} \)

Hint:

To find the equation of a plane passing through three points, substitute the coordinates into the general equation of a plane and solve for the coefficients.

Answer 1

The Correct Option is B

We are given three points: \( (-a^2, 1, 1) \), \( (1, -a^2, 1) \), and \( (1, 1, -a^2) \), and the condition that the plane passing through these points also passes through the point \( (-1, -1, 1) \). Step 1: To find the equation of the plane passing through the three given points, we can use the general form of the plane equation: \[ Ax + By + Cz + D = 0 \] Substitute the coordinates of the given points into the equation to find the values of \( A \), \( B \), \( C \), and \( D \). Step 2: Substitute the coordinates of the point \( (-1, -1, 1) \) into the equation of the plane to find the condition for \( a \). Step 3: Solving this equation, we find that the values of \( a \) that satisfy the condition are \( \pm \sqrt{3} \). Thus, the set \( S \) is: \[ S = \{\sqrt{3}, -\sqrt{3}\} \] % Final Answer The set \( S = \{\sqrt{3}, -\sqrt{3}\} \).

Answer 2

Step 1: Understand the problem
We are given three points that lie on a plane:
\( A = (-a^2, 1, 1) \),
\( B = (1, -a^2, 1) \),
\( C = (1, 1, -a^2) \)
We are told that the point \( D = (-1, -1, 1) \) also lies on the same plane.

Step 2: Use vector approach to find normal
Let’s find two direction vectors on the plane:
\( \vec{AB} = B - A = (1 + a^2, -a^2 - 1, 0) \)
\( \vec{AC} = C - A = (1 + a^2, 0, -a^2 - 1) \)

Step 3: Find the normal vector using cross product
\( \vec{n} = \vec{AB} \times \vec{AC} \)
Using determinant method:
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 + a^2 & -1 - a^2 & 0 \\ 1 + a^2 & 0 & -1 - a^2 \\ \end{vmatrix} = \hat{i}((−1 − a^2)(−1 − a^2) − 0) - \hat{j}((1 + a^2)(−1 − a^2) − 0) + \hat{k}((1 + a^2)(0) − (1 + a^2)(−1 − a^2)) \]
Simplifying:
\( \vec{n} = \hat{i}(1 + a^2)^2 - \hat{j}(−(1 + a^2)^2) + \hat{k}((1 + a^2)^2) \)
So \( \vec{n} = (1 + a^2)^2 (\hat{i} + \hat{j} + \hat{k}) \)

Step 4: General equation of the plane
Now use point-normal form of the plane with point \( A \):
Normal vector: \( (1, 1, 1) \)
Point: \( (-a^2, 1, 1) \)
So plane equation:
\( 1(x + a^2) + 1(y - 1) + 1(z - 1) = 0 \)
→ \( x + y + z + a^2 - 2 = 0 \)

Step 5: Check if point D lies on the plane
Substitute \( (-1, -1, 1) \) into the plane:
\( -1 + (-1) + 1 + a^2 - 2 = 0 \Rightarrow -1 + a^2 - 2 = 0 \Rightarrow a^2 = 3 \)
So, \( a = \pm \sqrt{3} \)

Final Answer:
\( S = \{\sqrt{3}, -\sqrt{3}\} \)