Problem 20
Question
Find the distance between the point and the plane (see figure). The distance \(D\) between a point \(\left(x_{0}, y_{0}, z_{0}\right)\) and the plane \(a x+b y+c z+d=0\) is $$ D=\frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}} $$ $$ (-2,1,0), 2 x+5 y-z=20 $$
Step-by-Step Solution
Verified Answer
The distance between the point and the plane is approximately 3.47 units.
1Step 1: Identifying the variables
The plane equation is \(2x + 5y -z = 20\), so a = 2, b = 5, c = -1, and d = -20. The point \((-2,1,0)\) gives us \(x_0 = -2\), \(y_0 = 1\), and \(z_0 = 0\).
2Step 2: Substitute the values into the equation
Now you substitute the obtained variables into the formula for the distance: \[D = \frac{| 2(-2) + 5(1) -1(0) - 20 |}{\sqrt{2^2 + 5^2 + (-1)^2}}\].
3Step 3: Calculate the result
Work out the absolute value calculation in the numerator and the square roots in the denominator. This gives \[D = \frac{| -4 + 5 - 20|}{\sqrt{4 + 25 + 1}} = \frac{|-19|}{\sqrt{30}} = \frac{19}{\sqrt{30}}\].
4Step 4: Simplify the result
To simplify the result, the square root of 30 can be approximated up to two decimal places, which is about 5.48. So the final answer is \[D = \frac{19}{5.48} ≈ 3.47\].
Key Concepts
Understanding the Plane EquationSubstituting Variables AccuratelyCalculating the Absolute Value
Understanding the Plane Equation
Before delving into complex calculations, it is crucial to understand the equation of a plane. In three dimensions, the general form of the equation of a plane is \(ax + by + cz + d = 0\), where \(a\), \(b\), and \(c\) are coefficients that define the plane's orientation in space, and \(d\) is the displacement from the origin.
Visualizing the plane, imagine it as a flat surface extending infinitely in all directions. The coefficients \(a\), \(b\), and \(c\) can be thought of as pointing in the direction perpendicular to the plane's surface. This is what is known as the normal vector of the plane. The value of \(d\) determines the plane's position relative to the origin; if \(d\) is zero, the plane passes through the origin. Otherwise, it is offset.
When tasked with finding the distance between a point and a plane, understanding this equation becomes fundamental. It sets the stage for us to plug in values and use algebra to find the required distance, as per the given formula.
Visualizing the plane, imagine it as a flat surface extending infinitely in all directions. The coefficients \(a\), \(b\), and \(c\) can be thought of as pointing in the direction perpendicular to the plane's surface. This is what is known as the normal vector of the plane. The value of \(d\) determines the plane's position relative to the origin; if \(d\) is zero, the plane passes through the origin. Otherwise, it is offset.
When tasked with finding the distance between a point and a plane, understanding this equation becomes fundamental. It sets the stage for us to plug in values and use algebra to find the required distance, as per the given formula.
Substituting Variables Accurately
With the equation of a plane in hand, our next step involves substituting specific values for the variables \(x_0\), \(y_0\), and \(z_0\), which represent the coordinates of a point in space. This process requires attention to detail, as errors in substitution can lead to incorrect solutions.
In our problem, we have the coefficients \(a = 2\), \(b = 5\), \(c = -1\), and \(d = -20\) from the plane equation, and the point of interest is \( (-2, 1, 0)\), giving us \(x_0 = -2\), \(y_0 = 1\), and \(z_0 = 0\). Substitution merely means replacing the general variables in the plane equation with these specific values.
The purpose of this step is to convert a general expression into one that calculates the precise distance for this unique point and plane. Once the values are substituted correctly, computation is straightforward and brings us closer to the final result.
In our problem, we have the coefficients \(a = 2\), \(b = 5\), \(c = -1\), and \(d = -20\) from the plane equation, and the point of interest is \( (-2, 1, 0)\), giving us \(x_0 = -2\), \(y_0 = 1\), and \(z_0 = 0\). Substitution merely means replacing the general variables in the plane equation with these specific values.
The purpose of this step is to convert a general expression into one that calculates the precise distance for this unique point and plane. Once the values are substituted correctly, computation is straightforward and brings us closer to the final result.
Calculating the Absolute Value
One often underemphasized step is the absolute value calculation. It's essential because the distance can never be negative; it's a measure of magnitude. The absolute value ensures this by converting possible negative results into positive values.
In the distance formula, \(D=\frac{|ax_{0}+ by_{0}+ cz_{0}+ d|}{\sqrt{a^{2}+b^{2}+c^{2}}}\), the numerator involves taking the absolute value of a linear expression that includes the substituted point and the plane's coefficients. The absolute value bars act like a safeguard, capturing the correct 'distance' even when the algebraic sum is negative.
During computation, as we carry out the operations within the absolute value, \(|-4 + 5 - 20|\) becomes \(|-19|\), which simplifies to 19. This is a key step, as ignoring the absolute value may lead to the incorrect conclusion that distance can be negative, which contradicts its very definition. Therefore, the absolute value calculation is not just a mechanical step but is rooted in the conceptual understanding of distance as a positive quantity or zero.
In the distance formula, \(D=\frac{|ax_{0}+ by_{0}+ cz_{0}+ d|}{\sqrt{a^{2}+b^{2}+c^{2}}}\), the numerator involves taking the absolute value of a linear expression that includes the substituted point and the plane's coefficients. The absolute value bars act like a safeguard, capturing the correct 'distance' even when the algebraic sum is negative.
During computation, as we carry out the operations within the absolute value, \(|-4 + 5 - 20|\) becomes \(|-19|\), which simplifies to 19. This is a key step, as ignoring the absolute value may lead to the incorrect conclusion that distance can be negative, which contradicts its very definition. Therefore, the absolute value calculation is not just a mechanical step but is rooted in the conceptual understanding of distance as a positive quantity or zero.
Other exercises in this chapter
Problem 20
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Find the coordinates of the midpoint of the line segment joining the two points. $$ (0,-2,5),(4,2,7) $$
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Determine whether there is a relative maximum, a relative minimum, a saddle point, or insufficient information to determine the nature of the function \(f(x, y)
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