Question

A plane \(\pi\) passes through the points \( (5,1,2) \), \( (3,-4,6) \), and \( (7,0,-1) \). If \( p \) is the perpendicular distance from the origin to the plane \(\pi\) and \([l, m, n]\) are the direction cosines of a normal to the plane \(\pi\), then \([3l + 2m + 5n] =\)

• A. \(3p\)
• B. \(2p\)
• C. \(p\)
• D. \(\frac{p}{2}\)

Hint:

Use vector cross product to easily find a normal vector for planes defined by three points. Normalize vectors when working with direction cosines.

Answer 1

The Correct Option is C

We are given a plane \(\pi\) passing through the points \( (5,1,2) \), \( (3,-4,6) \), and \( (7,0,-1) \). We are to find the perpendicular distance \( p \) from the origin to the plane \(\pi\) and the direction cosines \([l, m, n]\) of a normal to the plane. Finally, we need to compute \([3l + 2m + 5n]\) and determine which option it equals. Step 1: Find the equation of the plane \(\pi\) To find the equation of the plane, we first determine two vectors lying on the plane using the given points: 1. Vector \(\vec{AB} = (3 - 5, -4 - 1, 6 - 2) = (-2, -5, 4)\),
2. Vector \(\vec{AC} = (7 - 5, 0 - 1, -1 - 2) = (2, -1, -3)\).
Next, find the normal vector \(\vec{n}\) to the plane by taking the cross product of \(\vec{AB}\) and \(\vec{AC}\): \[ \vec{n} = \vec{AB} \times \vec{AC} \] Compute the cross product: \[ \vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}
-2 & -5 & 4
2 & -1 & -3 \end{vmatrix} \] \[ \vec{n} = \mathbf{i}((-5)(-3) - (4)(-1)) - \mathbf{j}((-2)(-3) - (4)(2)) + \mathbf{k}((-2)(-1) - (-5)(2)) \] \[ \vec{n} = \mathbf{i}(15 + 4) - \mathbf{j}(6 - 8) + \mathbf{k}(2 + 10) \] \[ \vec{n} = 19\mathbf{i} + 2\mathbf{j} + 12\mathbf{k} \] Thus, the normal vector is \(\vec{n} = (19, 2, 12)\). Step 2: Write the equation of the plane Using the normal vector \(\vec{n} = (19, 2, 12)\) and the point \( (5,1,2) \), the equation of the plane is: \[ 19(x - 5) + 2(y - 1) + 12(z - 2) = 0 \] Simplify: \[ 19x - 95 + 2y - 2 + 12z - 24 = 0 \] \[ 19x + 2y + 12z - 121 = 0 \] Thus, the equation of the plane is: \[ 19x + 2y + 12z = 121 \] Step 3: Find the perpendicular distance \( p \) from the origin to the plane The perpendicular distance from a point \((x_0, y_0, z_0)\) to the plane \(Ax + By + Cz + D = 0\) is given by: \[ p = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \] For the origin \((0, 0, 0)\) and the plane \(19x + 2y + 12z - 121 = 0\), the distance \( p \) is: \[ p = \frac{|19(0) + 2(0) + 12(0) - 121|}{\sqrt{19^2 + 2^2 + 12^2}} \] \[ p = \frac{121}{\sqrt{361 + 4 + 144}} = \frac{121}{\sqrt{509}} \] Step 4: Find the direction cosines \([l, m, n]\) The direction cosines of the normal vector \(\vec{n} = (19, 2, 12)\) are: \[ l = \frac{19}{\sqrt{19^2 + 2^2 + 12^2}} = \frac{19}{\sqrt{509}}, \] \[ m = \frac{2}{\sqrt{509}}, \] \[ n = \frac{12}{\sqrt{509}}. \] Step 5: Compute \([3l + 2m + 5n]\) Substitute the values of \(l\), \(m\), and \(n\): \[ 3l + 2m + 5n = 3\left(\frac{19}{\sqrt{509}}\right) + 2\left(\frac{2}{\sqrt{509}}\right) + 5\left(\frac{12}{\sqrt{509}}\right) \] \[ 3l + 2m + 5n = \frac{57 + 4 + 60}{\sqrt{509}} = \frac{121}{\sqrt{509}} \] From Step 3, we know that \( p = \frac{121}{\sqrt{509}} \). Thus: \[ 3l + 2m + 5n = p \] Final Answer: \(\boxed{3}\)