Problem 32 Find the distance from \((3,-1,2... [FREE SOLUTION] (2024)

Short Answer

distance formula

The distance formula is a crucial concept in coordinate geometry. It helps in measuring the shortest distance between a point and a plane or between two points. For a point \( (x_1, y_1, z_1) \) and a plane expressed as \( Ax + By + Cz + D = 0 \), the formula is: \[ \text{Distance} = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]. This formula is derived using the perpendicular distance from the point to the plane.

plane equation

Understanding the plane equation is essential in solving distance problems in coordinate geometry. A plane in three-dimensional space can be represented by the general form \( Ax + By + Cz + D = 0 \). Here, \( A, B, \) and \( C \) are the coefficients of \( x, y, \) and \( z \) respectively, and \( D \) is the constant term. For example, the plane equation from the given exercise is \( 5x - y - z = 4 \). It can be rewritten as \[ 5x - y - z - 4 = 0 \]. This standard form helps in applying the distance formula easily.

numerical calculation

Numerical calculation involves substituting known values into a formula and simplifying the result. For our distance problem, we identified the plane equation's coefficients \( A = 5, B = -1, C = -1, \) and constant term \( D = -4 \). The point is \( (3, -1, 2) \). Substitute these values into the distance formula: \[ \text{Distance} = \frac{|5(3) + (-1)(-1) + (-1)(2) - 4|}{\sqrt{5^2 + (-1)^2 + (-1)^2}} \]. First, calculate the numerator: \[ |15 + 1 - 2 - 4| = 10 \]. Then, simplify the denominator: \[ \sqrt{25 + 1 + 1} = \sqrt{27} = 3\sqrt{3} \]. Combining these, we get the distance: \[ \text{Distance} = \frac{10}{3\sqrt{3}} \]. Finally, rationalize to get: \[ \frac{10\sqrt{3}}{9} \].

coordinate geometry

Coordinate geometry studies the geometric properties and spatial relations represented by coordinates. Problems involving points and planes in three-dimensional space often use coordinate geometry techniques to find solutions. A typical problem includes finding the distance from a point to a plane. Using the given plane equation and the point coordinates, we apply the distance formula, performing steps to substitute values, calculate numerators and denominators, simplify, and get the final distance. This approach underscores the elegance of coordinate geometry in solving spatial problems effectively.