Question

The distance of the point $(7,-3,-4)$ from the plane passing through the points $(2,-3,1),(-1,1,-2)$ and $(3,-4,2)$ is :

• A. $5 \sqrt{2}$
• B. $4 \sqrt{2}$
• C. 4
• D. 5

Answer 1

The Correct Option is A

To find the distance of the point \((7, -3, -4)\) from the plane passing through the points \((2, -3, 1)\), \((-1, 1, -2)\), and \((3, -4, 2)\), we must first determine the equation of the plane. Let's go through the solution step-by-step.

1. Identify the points given in the plane: \((2, -3, 1)\), \((-1, 1, -2)\), and \((3, -4, 2)\).
2. Find two vectors lying on the plane using these points. Let's denote vector \(\vec{A}\) as the vector from point \((2, -3, 1)\) to point \((-1, 1, -2)\), and vector \(\vec{B}\) as the vector from point \((2, -3, 1)\) to point \((3, -4, 2)\).
   • \(\vec{A} = (-1 - 2, 1 + 3, -2 - 1) = (-3, 4, -3)\)
   • \(\vec{B} = (3 - 2, -4 + 3, 2 - 1) = (1, -1, 1)\)
3. Calculate the normal vector \(\vec{n}\) to the plane by taking the cross product of \(\vec{A}\) and \(\vec{B}\).
   • \(\vec{n} = \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 4 & -3 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(4 \cdot 1 - (-1) \cdot (-3)) - \hat{j}(-3 \cdot 1 - 1 \cdot (-3)) + \hat{k}(-3 \cdot (-1) - 4 \cdot 1)\)
   • Simplifying gives: \(\vec{n} = \hat{i} (4 - 3) - \hat{j}(-3 + 3) + \hat{k} (3 - 4)\)
   • Simplified further, \(\vec{n} = \hat{i}(1) + \hat{j}(0) + \hat{k}(-1) = (1, 0, -1)\).
4. The equation of the plane can now be derived using the normal vector \(\vec{n} = (1, 0, -1)\) and point \((2, -3, 1)\) on the plane: \[ 1(x - 2) + 0(y + 3) - 1(z - 1) = 0 \] This simplifies to: \[ x - z = 1 \]
5. To find the distance of point \((7, -3, -4)\) from the plane \(x - z = 1\), use the point-to-plane distance formula:
   • Distance \(d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}\) where \(A = 1\), \(B = 0\), \(C = -1\), and \(D = -1\), and \((x_1, y_1, z_1) = (7, -3, -4)\).
   • Plugging in the values: \(d = \frac{|1\cdot7 + 0\cdot(-3) - 1\cdot(-4) -1|}{\sqrt{1^2 + 0^2 + (-1)^2}} = \frac{|7 + 4 - 1|}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2}\)

Therefore, the distance of the point \((7, -3, -4)\) from the given plane is \(5\sqrt{2}\). The correct answer is

$5 \sqrt{2}$