Question

Find the distance from the point to the given plane.

\((1, - 2,4),\quad 3x + 2y + 6z = 5\)

Short answer

The distance between the point \((1, - 2,4)\) and the plane \(3x + 2y + 6z = 5\) is \(\frac{{18}}{7}\).

Step-by-step solution

Formula used

The expression to find the distance from the point \(P\left( {{x_1},{y_1},{z_1}} \right)\) to the plane.

\(D = \frac{{\left| {a{x_1} + b{y_1} + c{z_1} + d} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\) ……. (1)

Here,

\({x_1},{y_1}\), and \({z_1}\) are the coordinates of the point \(P\left( {{x_1},{y_1},{z_1}} \right)\), which is \((1, - 2,4)\),

a, b, and \(c\) are the normal vector numbers to the plane and

\(d\)is the constant parameter.

The expression to find constant parameter \((d)\).

\(a{x_0} + b{y_0} + c{z_0} + d = 0\) ……. (2) Here,

\({x_0},{y_0}\), and \({z_0}\) are the coordinates of any point \(P\left( {{x_0},{y_0},{z_0}} \right)\) in the plane.

Calculate the normal vector and constant parameter from the equation

The equation of the plane as follows.

\(3x + 2y + 6z = 5\)

Write the normal vector from the equation of the plane.

\({\rm{n}} = \langle 3,2,6\rangle \)

Rewrite the equation of the plane in the form of equation (2)

\(3x + 2y + 6z - 5 = 0{\rm{ }}\) ……. (3)

Compare equation (2) and (3), we get

The value of constant parameter \((d)\) is \( - 5\).

Calculate the distance

Substituting the values in equation (1)

\(\begin{array}{l}D = \frac{{|(3)(1) + (2)( - 2) + (6)(4) + ( - 5)|}}{{\sqrt {{3^2} + {2^2} + {6^2}} }}\\ = \frac{{|3 - 4 + 24 - 5|}}{{\sqrt {9 + 4 + 36} }}\\ = \frac{{18}}{{\sqrt {49} }}\\ = \frac{{18}}{7}\end{array}\)

Thus, the distance between the point \((1, - 2,4)\) and the plane \(3x + 2y + 6z = 5\) is \(\frac{{18}}{7}\).