Angle Between a Plane and a Line: Solving for the Complement

Today, we will delve into the topic of determining the angle between a plane and a line, specifically using the given example and leveraging vector algebra. We'll explore the direction ratios of the normal to the plane, the direction ratios of the line passing through points, and how to calculate the angle between these elements. We will also discuss the complementary angle between the plane and the line to arrive at the final result.

Understanding the Given Plane and Line

The plane in question is defined by the equation 2x 3y - 5z 10. This equation allows us to determine the normal vector to the plane, which serves as a critical step in solving the problem.

The direction ratios of the normal to the plane are derived from the coefficients of x, y, and z. Therefore, the normal vector (vec{A}) is:

(vec{A} 2hat{i} 3hat{j} - 5hat{k})

Direction Vectors of the Line

The line passing through the points (2, 3, -1) and (1, 2, 1) can be represented by its direction ratios:

The direction ratios of the line are the differences in the coordinates of these points:

The direction vector (vec{B}) is:

(vec{B} hat{i} - hat{j} - 2hat{k})

Calculating the Angle Between the Vectors

To find the angle (theta) between the vectors (vec{A}) and (vec{B}), we use the dot product formula:

Formula and Calculation

(theta arccos left(frac{vec{A} cdot vec{B}}{|vec{A}| |vec{B}|}right))

First, we calculate the dot product (vec{A} cdot vec{B}):

(vec{A} cdot vec{B} (2)(1) (3)(-1) (-5)(-2) 2 - 3 10 9)

Next, we calculate the magnitudes (|vec{A}|) and (|vec{B}|):

(|vec{A}| sqrt{2^2 3^2 (-5)^2} sqrt{4 9 25} sqrt{38})

(|vec{B}| sqrt{1^2 (-1)^2 (-2)^2} sqrt{1 1 4} sqrt{6})

Substituting these values into the formula gives:

(theta arccos left(frac{9}{sqrt{38} cdot sqrt{6}}right) arccos left(frac{9}{sqrt{228}}right) arccos left(frac{9}{2sqrt{57}}right))

Evaluating this using a calculator:

(theta approx 6.587^o)

The Complementary Angle

In the original problem, the angle between the normal to the plane and the line has been calculated. However, what the question actually asks is the angle between the plane and the line, which is the complement of the angle between the normal and the line.

Therefore, to find the angle between the plane and the line, we subtract the angle between the normal and the line from 90 degrees:

Calculation of the Complementary Angle

(theta 90^o - 6.587^o 83.413^o)

This is the final answer to the problem.

Conclusion

Understanding the relationship between the normal vector to a plane, the direction vector of a line, and their angles with each other is crucial in solving geometric problems. By calculating the angle between the normal and the line and then finding the complementary angle, we can accurately determine the angle between a plane and a line.