Step 1 : Understanding the Question:
The problem asks us to find the shortest (perpendicular) distance between a point in three-dimensional space and a given plane. The point is specified by its position vector, and the plane is given in its vector form. This is a fundamental concept in 3D coordinate geometry and vector algebra, as it involves projecting a vector onto the normal of the plane to find the offset distance.
Step 2 : Key Formulas and approach:
For a point defined by the position vector $\vec{a}$ and a plane defined by the equation $\vec{r} \cdot \vec{n} = d$, where $\vec{n}$ is the normal vector to the plane, the perpendicular distance $D$ is given by the formula:
\[ D = \frac{|\vec{a} \cdot \vec{n} - d|}{|\vec{n}|} \]
Our approach will be:
1. Identify the components of the point vector $\vec{a}$.
2. Identify the normal vector $\vec{n}$ and the constant $d$ from the plane's equation.
3. Compute the dot product $\vec{a} \cdot \vec{n}$.
4. Calculate the magnitude of the normal vector $|\vec{n}|$.
5. Substitute these into the formula to find the final distance.
Step 3 : Detailed Explanation:
Given point vector: $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$.
Plane equation: $\vec{r} \cdot (\hat{i} - 2\hat{j} + 4\hat{k}) = 9$.
From the plane equation, we extract the normal vector $\vec{n} = \hat{i} - 2\hat{j} + 4\hat{k}$ and the scalar $d = 9$.
Calculate the dot product $\vec{a} \cdot \vec{n}$: $(2)(1) + (1)(-2) + (-1)(4) = 2 - 2 - 4 = -4$.
Calculate the magnitude of $\vec{n}$: $|\vec{n}| = \sqrt{1^2 + (-2)^2 + 4^2} = \sqrt{1 + 4 + 16} = \sqrt{21}$.
Now, substitute these values into the distance formula: $D = \frac{|-4 - 9|}{\sqrt{21}}$.
Simplify the numerator: $|-13| = 13$.
Thus, the distance is $D = \frac{13}{\sqrt{21}}$.
Step 4 : Final Answer:
The distance from the point to the plane is $\frac{13}{\sqrt{21}}$.