Question

The distance of the point whose position vector is \(\mathbf{2\hat{i} + \hat{j} - \hat{k}}\) from the plane vector \(\vec{r} \cdot (\hat{i} - 2\hat{j} + 4\hat{k}) = 4\) is

• A. \(\frac{8}{\sqrt{21}}\)
• B. \(\frac{-8}{\sqrt{21}}\)
• C. \(8\sqrt{21}\)
• D. \(- \frac{8}{21}\)

Answer 1

The Correct Option is A

The formula for the perpendicular distance from a point \((x_0, y_0, z_0)\) to the plane \(Ax + By + Cz + D = 0\) is:

\[ \text{Distance} = \frac{{\left|Ax + By + Cz + D\right|}}{{\sqrt{A^2 + B^2 + C^2}}} \]

In this case, the position vector of the point is given as \(\mathbf{v} = 2\mathbf{i} + \mathbf{j} - \mathbf{k}\), and the plane is defined by the vector equation \(\mathbf{r} \cdot (\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}) = 4\).

Comparing the equation of the plane to the general form \(Ax + By + Cz + D = 0\), we have:

\[ A = 1, \quad B = -2, \quad C = 4, \quad D = -4. \]

Substituting the values into the distance formula, we get:

\[ \text{Distance} = \frac{\left|(1)(2) + (-2)(1) + (4)(-1) + (-4)\right|}{\sqrt{1^2 + (-2)^2 + 4^2}} \]

\[ = \frac{\left|-2 - 2 - 4 - 4\right|}{\sqrt{1 + 4 + 16}} = \frac{\left|-12\right|}{\sqrt{21}} = \frac{12}{\sqrt{21}} = \frac{8}{\sqrt{21}} \]

Therefore, the correct option is (1) \(\frac{8}{\sqrt{21}}\).

Answer 2

Given position vector of the point: \(\vec{v} = 2\hat{i} + \hat{j} - \hat{k}\)
Plane equation: \(\vec{r} \cdot (\hat{i} - 2\hat{j} + 4\hat{k}) = 4\)

This is of the form: \(\vec{r} \cdot \vec{n} = d\), where
Normal vector \(\vec{n} = \hat{i} - 2\hat{j} + 4\hat{k}\) and d = 4.

The formula for distance of point \( \vec{a} \) from plane \( \vec{r} \cdot \vec{n} = d \) is: \[ \text{Distance} = \frac{|\vec{a} \cdot \vec{n} - d|}{|\vec{n}|} \] Substituting: \[ \vec{a} \cdot \vec{n} = (2)(1) + (1)(-2) + (-1)(4) = 2 - 2 - 4 = -4 \] \[ \text{Numerator} = |-4 - 4| = |-8| = 8 \] \[ |\vec{n}| = \sqrt{1^2 + (-2)^2 + 4^2} = \sqrt{1 + 4 + 16} = \sqrt{21} \] \[ \text{Distance} = \frac{8}{\sqrt{21}} \] Correct Answer: \(\frac{8}{\sqrt{21}}\)