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: Determinant form
The plane through three points satisfies $\begin{vmatrix} x - 1 & y + 2 & z - 1 \\ 1 & 1 & -4 \\ -1 & 3 & 4\end{vmatrix} = 0$.

Step 2: Expand
$(x - 1)(4 + 12) - (y + 2)(4 - 4) + (z - 1)(3 + 1) = 16(x - 1) + 4(z - 1) = 0$.

Step 3: Simplify
$4(x - 1) + (z - 1) = 0$, so $4x + z = 5$.

Step 4: Distance
From the origin: $\dfrac{5}{\sqrt{4^2 + 1^2}} = \dfrac{5}{\sqrt{17}}$.

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