Question

The distance from the point (2, 2, 2) to the plane 2x-y+3z = 5 is equal to

• A. \(\frac{3\sqrt{7}}{2}\)
• B. \(\frac{\sqrt{3}}{2}\)
• C. \(\frac{3\sqrt{14}}{7}\)
• D. \(\frac{3\sqrt{14}}{14}\)
• E. \(\frac{\sqrt{3}}{3}\)

Answer 1

The Correct Option is D

The distance from a point \( (x_1, y_1, z_1) \) to the plane \( Ax + By + Cz + D = 0 \) is given by the formula: \[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}. \] Here, the point is \( (2, 2, 2) \), and the equation of the plane is \( 2x - y + 3z - 5 = 0 \). Thus, \( A = 2 \), \( B = -1 \), \( C = 3 \), and \( D = -5 \). Substitute the values into the distance formula: \[ d = \frac{|2(2) + (-1)(2) + 3(2) - 5|}{\sqrt{2^2 + (-1)^2 + 3^2}}. \] Simplifying the numerator: \[ 2(2) + (-1)(2) + 3(2) - 5 = 4 - 2 + 6 - 5 = 3. \] Now, simplify the denominator: \[ \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14}. \] Thus, the distance is: \[ d = \frac{|3|}{\sqrt{14}} = \frac{3}{\sqrt{14}}. \] To rationalize the denominator: \[ d = \frac{3}{\sqrt{14}} \times \frac{\sqrt{14}}{\sqrt{14}} = \frac{3\sqrt{14}}{14}. \] Therefore, the distance from the point \( (2, 2, 2) \) to the plane \( 2x - y + 3z = 5 \) is \( \frac{3\sqrt{14}}{14} \).

The correct option is (D) : \(\frac{3\sqrt{14}}{14}\)

Answer 2

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

\(d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}\)

In this case, the point is (2, 2, 2) and the equation of the plane is \(2x - y + 3z = 5\), which can be written as \(2x - y + 3z - 5 = 0\). So, \(x_0 = 2\), \(y_0 = 2\), \(z_0 = 2\), \(A = 2\), \(B = -1\), \(C = 3\), and \(D = -5\).

Substituting these values into the formula, we get:

\(d = \frac{|2(2) - 1(2) + 3(2) - 5|}{\sqrt{2^2 + (-1)^2 + 3^2}} = \frac{|4 - 2 + 6 - 5|}{\sqrt{4 + 1 + 9}} = \frac{|3|}{\sqrt{14}} = \frac{3}{\sqrt{14}}\)

To rationalize the denominator, we multiply by \(\frac{\sqrt{14}}{\sqrt{14}}\):

\(d = \frac{3}{\sqrt{14}} \cdot \frac{\sqrt{14}}{\sqrt{14}} = \frac{3\sqrt{14}}{14}\)

Therefore, the distance from the point (2, 2, 2) to the plane 2x - y + 3z = 5 is \(\frac{3\sqrt{14}}{14}\).