Question

The distance of the point \((-1, 2, -2)\) from the line of intersection of the planes \(2x + 3y + 2z = 0\) and \(x - 2y + z = 0\) is :

• A. \(\frac{1}{\sqrt{2}}\)
• B. \(\frac{5}{2}\)
• C. \(\frac{\sqrt{34}}{2}\)
• D. \(\frac{\sqrt{42}}{2}\)

Hint:

When both plane equations have no constant term, the line of intersection always passes through the origin \((0, 0, 0)\). This simplifies finding a point on the line significantly.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The line of intersection of two planes is a line perpendicular to the normal vectors of both planes. We first find the direction of this line and a point on it, and then calculate the perpendicular distance from the given point to this line.
Step 2: Key Formula or Approach:
1. The direction vector \(\vec{b}\) of the line is given by \(\vec{n}_1 \times \vec{n}_2\), where \(\vec{n}_1\) and \(\vec{n}_2\) are the normals to the given planes.
2. The distance \(d\) of a point \(P\) from a line passing through point \(A\) with direction \(\vec{b}\) is: \[ d = \frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|} \]
Step 3: Detailed Explanation:
The planes are \(P_1: 2x + 3y + 2z = 0\) and \(P_2: x - 2y + z = 0\).
The normal vectors are \(\vec{n}_1 = (2, 3, 2)\) and \(\vec{n}_2 = (1, -2, 1)\).
Direction of line \(\vec{b} = \vec{n}_1 \times \vec{n}_2\): \[ \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & 3 & 2
1 & -2 & 1 \end{vmatrix} = \hat{i}(3 - (-4)) - \hat{j}(2 - 2) + \hat{k}(-4 - 3) = 7\hat{i} - 0\hat{j} - 7\hat{k} \] We can take the simplified direction vector as \(\vec{b} = \hat{i} - \hat{k} = (1, 0, -1)\).
Since both planes pass through the origin (constant terms are 0), the origin \(A(0, 0, 0)\) lies on the line.
Let the given point be \(P(-1, 2, -2)\). Then \(\vec{AP} = (-1, 2, -2)\).
Calculate \(\vec{AP} \times \vec{b}\): \[ \vec{AP} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-1 & 2 & -2
1 & 0 & -1 \end{vmatrix} = \hat{i}(-2 - 0) - \hat{j}(1 - (-2)) + \hat{k}(0 - 2) = -2\hat{i} - 3\hat{j} - 2\hat{k} \] Magnitude \(|\vec{AP} \times \vec{b}| = \sqrt{(-2)^2 + (-3)^2 + (-2)^2} = \sqrt{4 + 9 + 4} = \sqrt{17}\).
Magnitude \(|\vec{b}| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{2}\).
The distance is \(d = \frac{\sqrt{17}}{\sqrt{2}} = \sqrt{\frac{17 \times 2}{2 \times 2}} = \frac{\sqrt{34}}{2}\).
Step 4: Final Answer:
The distance is \(\frac{\sqrt{34}}{2}\).