Question

Find the distance between the point and the plane. $$ \begin{array}{l} (2,8,4) \\ 2 x+y+z=5 \end{array} $$

Short answer

The distance between the point (2,8,4) and the plane \(2x + y + z = 5\) is \(\frac{11*sqrt{6}}{6}\).

Step-by-step solution

Identify the coefficients of the plane equation

The equation of the plane is given as 2x + y + z = 5. From this equation, it can be noticed that the coefficients for x, y, z are 2, 1, and 1 respectively. These coefficients form the normal vector of the plane, \(N = (2, 1, 1)\).

Calculate the dot product of the normal vector and the given point

The given point is P = (2, 8, 4). Compute the dot product of this point with the normal vector. The dot product is given as \(N.P = 2*2 + 1*8 + 1*4 = 4 + 8 + 4 = 16\).

Calculate the magnitude of the normal vector

The magnitude of a vector \(N = (a, b, c)\) is given by \(\sqrt{{a}^2 + {b}^2 + {c}^2 }\). Thus, the magnitude of N = \(\sqrt{{2}^2 + {1}^2 + {1}^2} = \sqrt{6}\).

Calculating the perpendicular distance

The perpendicular distance 'd' from a point to the plane is given by \(\frac{{|N.P + D|}}{{||N||}}\) where N is the normal vector to the plane, P is the given point, and D is the constant value from the equation of the plane. Here, \(D = -5\) (because the equation of the plane is \(2x + y + z - 5 = 0\)). So, \(d = \frac{{|16 - 5|}}{{\sqrt{6}}} = \frac{11}{\sqrt{6}} = \frac{11*sqrt{6}}{6}\).

Key concepts

Plane Equation

When studying three-dimensional geometry, the plane equation is an essential tool for describing the position of a flat surface within the space. It takes the form of \( Ax + By + Cz + D = 0 \), where \(A, B, C\) are the coefficients that correspond to the normal vector of the plane, and \(D\) is a constant term that influences the plane's position relative to the origin.

In the context of our exercise, the plane equation given is \(2x + y + z = 5\). To use it for calculating distances, we can rewrite the equation in its standard form \(2x + y + z - 5 = 0\), where we can see that \(A = 2\), \(B = 1\), \(C = 1\), and \(D = -5\). These components are vital as they directly contribute to finding the perpendicular distance from a point to the plane.

Dot Product

The dot product, also known as the scalar product, is a way to multiply two vectors that results in a scalar value. Mathematically, it's calculated as \( A \cdot B = A_xB_x + A_yB_y + A_zB_z \), where \(A_x, A_y, A_z\) and \(B_x, B_y, B_z\) are the components of vectors \(A\) and \(B\), respectively.

For our example, consider the normal vector \(N = (2, 1, 1)\) and point \(P = (2, 8, 4)\). The dot product is then \(N \cdot P = 2 \times 2 + 1 \times 8 + 1 \times 4\), which simplifies to 16. It's a crucial step as this scalar will be used to compute the perpendicular distance from the point to the plane in conjunction with the plane equation.

Normal Vector

In three-dimensional space, a normal vector is a vector that is perpendicular to a surface. For a plane, it tells us the direction in which the plane 'stands up' from its surface. The normal vector's components are taken directly from the coefficients of the plane equation \( Ax + By + Cz = D \), meaning for the plane \(2x + y + z = 5\), the normal vector is \(N = (2, 1, 1)\).

Knowing the normal vector is essential because it defines the orientation of the plane and is necessary for calculating distances from points to the plane. Its magnitude, or length, plays a role in finding the perpendicular distance, emphasizing its importance in the calculation process.

Perpendicular Distance

The perpendicular distance from a point to a plane is the shortest distance between them, which is always along the direction of the normal vector to the plane. It is given by the formula \(d = \frac{|N \cdot P + D|}{||N||}\), where \(|N \cdot P + D|\) is the absolute value of the dot product of the normal vector \(N\) and point \(P\), plus the plane's constant \(D\), and \(||N||\) is the magnitude of \(N\).

In the problem at hand, we have already computed \(N \(cdot\) P = 16\) and know that \(D = -5\). We also calculated the magnitude of \(N\) to be \(\sqrt{6}\). Putting these values into the formula gives us the distance \(d = \frac{11}{\sqrt{6}} = \frac{11\sqrt{6}}{6}\), which is the perpendicular distance from the given point (2, 8, 4) to the plane defined by \(2x + y + z = 5\).