Question

The perpendicular distance from the origin to the plane containing the points \((1,-2,1),(2,-1,-3)\) and \((0,1,5)\) is...(in units)

• A. \(\frac{1}{\sqrt{17}}\)
• B. \(\frac{3}{\sqrt{26}}\)
• C. \(\frac{5}{\sqrt{17}}\)
• D. \(\frac{7}{\sqrt{26}}\)

Hint:

Find the normal as the cross product of two vectors in the plane, then use the distance formula.

Answer 1

The Correct Option is C

Step 1: Find two vectors in the plane
Take \(P = (1, -2, 1)\). Then \(\overrightarrow{PQ} = (1, 1, -4)\) and \(\overrightarrow{PR} = (-1, 3, 4)\).

Step 2: Normal vector
\[ \overrightarrow{PQ}\times\overrightarrow{PR} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -4 \\ -1 & 3 & 4\end{vmatrix} = (16, 0, 4) \]
So the normal direction is \((4, 0, 1)\).

Step 3: Plane equation
\(4x + 0y + z = d\). Using \(P\): \(4 + 1 = 5\), so the plane is \(4x + z = 5\). Check \(Q\): \(8 - 3 = 5\). Check \(R\): \(0 + 5 = 5\).

Step 4: Distance
\[ \text{distance} = \frac{|5|}{\sqrt{16 + 0 + 1}} = \frac{5}{\sqrt{17}} \]
Option (C).

Final Answer:
The distance is 5/sqrt(17). This is option (C). \[ \boxed{\text{(C) }\frac{5}{\sqrt{17}}} \]