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. $\text{Distance} = \frac{{\left|Ax + By + Cz + D\right|}}{{\sqrt{A^2 + B^2 + C^2}}}$
  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:
  $A = 1, B = -2, C = 4$, and$D = -4.$
  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) $\frac{8}{\sqrt{21}}$
• B. $\frac{-8}{\sqrt{21}}$
• C. $8\sqrt{21}$
• D. $- \frac{8}{21}$

Answer 1

Answer: (A)

$\text{Distance} = \frac{{\left|Ax + By + Cz + D\right|}}{{\sqrt{A^2 + B^2 + C^2}}}$
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:
$A = 1, B = -2, C = 4$, and$D = -4.$
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) $\frac{8}{\sqrt{21}}$