Question

A plane passes through the point \( (1, -2, 1) \) and is perpendicular to both the planes \[ 2x - 2y - 2z = 5 \quad \text{and} \quad x - y + 2z = 24 \] Then, the distance of the point \( (1, 2, 2) \) from this plane is:

• A. \( 2\sqrt{2} \)
• B. \( 1 \)
• C. \( \sqrt{2} \)
• D. \( 2 \)

Hint:

To find a plane perpendicular to two given planes, take the cross product of their normal vectors to get the normal of the required plane. Use point-normal form for the equation, and apply the distance formula to calculate perpendicular distance from a point to the plane.

Answer 1

The Correct Option is C

We are given two planes: \[ \pi_1: 2x - 2y - 2z = 5, \quad \pi_2: x - y + 2z = 24 \] Let the required plane \( \pi \) be perpendicular to both \( \pi_1 \) and \( \pi_2 \). Then, its normal vector is perpendicular to the normals of both given planes. Step 1: Find normal vectors of the given planes - Normal to \( \pi_1 \): \( \vec{n_1} = \langle 2, -2, -2 \rangle \) - Normal to \( \pi_2 \): \( \vec{n_2} = \langle 1, -1, 2 \rangle \) The normal to the required plane is the vector perpendicular to both \( \vec{n_1} \) and \( \vec{n_2} \), i.e., \[ \vec{n} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & -2 & -2
1 & -1 & 2 \end{vmatrix} = \hat{i}((-2)(2) - (-2)(-1)) - \hat{j}((2)(2) - (-2)(1)) + \hat{k}((2)(-1) - (-2)(1)) \] \[ = \hat{i}(-4 - 2) - \hat{j}(4 + 2) + \hat{k}(-2 + 2) = -6\hat{i} - 6\hat{j} + 0\hat{k} = \langle -6, -6, 0 \rangle \] So the normal vector to the required plane is \( \vec{n} = \langle -6, -6, 0 \rangle \) Step 2: Equation of the plane The plane passes through point \( (1, -2, 1) \), and has normal \( \langle -6, -6, 0 \rangle \) Using the point-normal form of the plane: \[ -6(x - 1) -6(y + 2) + 0(z - 1) = 0 \Rightarrow -6(x - 1 + y + 2) = 0 \Rightarrow x + y = -1 \] So, required plane is: \[ x + y + 1 = 0 \] Step 3: Distance from point \( (1, 2, 2) \) to this plane Use the distance formula from point to plane: \[ D = \frac{|ax + by + cz + d|}{\sqrt{a^2 + b^2 + c^2}} \] For plane \( x + y + 1 = 0 \), the coefficients are \( a = 1, b = 1, c = 0, d = 1 \) \[ D = \frac{|1 \cdot 1 + 1 \cdot 2 + 0 \cdot 2 + 1|}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{|1 + 2 + 1|}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \] Wait! But this doesn't match the correct answer. Let's recheck the step. Actually: \[ x + y + 1 = 0 \Rightarrow \text{then } d = 1, so equation: \( x + y + 1 = 0 \Rightarrow ax + by + cz + d = x + y + 1 \) Then the correct evaluation for point \( (1, 2, 2) \): \[ D = \frac{|1 + 2 + 1|}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \] Ah! This does match Option A, which is \( 2\sqrt{2} \), not Option C. So final correct choice is: \[ \boxed{\text{Option (A) } 2\sqrt{2}} \]