Question

Find the distance between the point \(2\hat{i} + \hat{j} - \hat{k}\) and the plane \(\vec{r} \cdot (\hat{i} - 2\hat{j} + 4\hat{k}) = 9\).

• A. \(\frac{13}{\sqrt{21}}\)
• B. \(\frac{9}{\sqrt{21}}\)
• C. \(\frac{13}{21}\)
• D. \(\frac{21}{\sqrt{13}}\)

Hint:

For point–plane distance problems, always treat the plane as \( \vec{r}\cdot \vec{n} - d = 0 \) to avoid sign mistakes in exams.

Answer 1

The Correct Option is A

Concept: The shortest distance from a point to a plane is along the perpendicular (normal) to the plane. For a plane \[ \vec{r}\cdot \vec{n} = d \] and a point with position vector \(\vec{a}\), the distance is: \[ D = \frac{|\vec{a}\cdot \vec{n} - d|}{|\vec{n}|} \]

Step 1: Identifying given values.
Point: \[ \vec{a} = 2\hat{i} + \hat{j} - \hat{k} \] Plane: \[ \vec{r} \cdot (\hat{i} - 2\hat{j} + 4\hat{k}) = 9 \] So, \[ \vec{n} = \hat{i} - 2\hat{j} + 4\hat{k}, \quad d = 9 \]

Step 2: Compute dot product \(\vec{a}\cdot \vec{n}\).
\[ \vec{a}\cdot \vec{n} = (2)(1) + (1)(-2) + (-1)(4) \] \[ = 2 - 2 - 4 = -4 \]

Step 3: Magnitude of normal vector.
\[ |\vec{n}| = \sqrt{1^2 + (-2)^2 + 4^2} \] \[ |\vec{n}| = \sqrt{1 + 4 + 16} = \sqrt{21} \]

Step 4: Apply distance formula.
\[ D = \frac{|\vec{a}\cdot \vec{n} - d|}{|\vec{n}|} \] \[ D = \frac{|-4 - 9|}{\sqrt{21}} \] \[ D = \frac{|-13|}{\sqrt{21}} = \frac{13}{\sqrt{21}} \]