Question

The distance of the point \( P(1, 2, 3) \) from the plane \( 3x - 4y + 12z - 7 = 0 \) is:

• A. \( \frac{1}{\sqrt{14}} \)
• B. \( \frac{2}{\sqrt{14}} \)
• C. \( \frac{3}{\sqrt{14}} \)
• D. \( \frac{4}{\sqrt{14}} \)

Hint:

To find the distance from a point to a plane, use the distance formula and simplify carefully.

Answer 1

The Correct Option is B

Given point \( P(1, 2, 3) \) and plane \( 3x - 4y + 12z - 7 = 0 \). The distance from point \( (x_1, y_1, z_1) \) to plane \( ax + by + cz + d = 0 \) is \( d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \). For this problem, \( a = 3 \), \( b = -4 \), \( c = 12 \), \( d = -7 \), and \( (x_1, y_1, z_1) = (1, 2, 3) \).

Step 1: Evaluate the numerator. Substitute \( x_1 = 1 \), \( y_1 = 2 \), \( z_1 = 3 \) into \( ax_1 + by_1 + cz_1 + d \): \[ 3(1) - 4(2) + 12(3) - 7 = 3 - 8 + 36 - 7 = 24 \] 

Step 2: Calculate the distance. Substitute the numerator and the denominator into the distance formula: \[ d = \frac{|24|}{\sqrt{3^2 + (-4)^2 + 12^2}} = \frac{24}{\sqrt{9 + 16 + 144}} = \frac{24}{\sqrt{169}} = \frac{24}{13} \] The distance is \( \frac{24}{13} \).