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

The Correct Option 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) \(\frac{8}{\sqrt{21}}\)