Question

Find the distance between the point and the line given by the set of parametric equations. (4,-1,5)\(; \quad x=3, \quad y=1+3 t, \quad z=1+t\)

Short answer

The distance between the point and the line is given by \(d = \sqrt{13.1}\)

Step-by-step solution

Determine the Coordinates of the Point on the Line

From the parametric equations, \( x=3, y=1+3 t, z=1+t \), for \( t=0 \), the line passes through the point A=(3,1,1).

Calculate Vector AP

Subtract the position vector of point A from the position vector of point P: \(\vec{AP} = P - A = (4,-1,5) - (3,1,1) = (1, -2, 4)\)

Find Directional Vector v

The directional vector \( \vec{v} \) of the line can be read directly from the parametric equation: \( \vec{v} = (0, 3, 1) \).

Calculate Cross Product of AP and v

By using the definition for cross product, calculate \( \vec{AP} \times \vec{v} = (1, -2, 4) \times (0, 3, 1) = (-11, -1, 3)\). For magnitude we get \(|\vec{AP} \times \vec{v}| = \sqrt{(-11)^2 + (-1)^2 + 3^2} = \sqrt{131}\)

Calculate the Magnitude of v

The magnitude of directional vector \( \vec{v} \) is \(|\vec{v}| = \sqrt{0^2 + 3^2 + 1^2} = \sqrt{10}\)

Calculate the Distance

Now use the formula \(d= \frac{| \vec{AP} \times \vec{v} |}{| \vec{v} |}\) to find the distance: \(d = \frac{\sqrt{131}}{\sqrt{10}} = \sqrt{13.1}\)

Key concepts

Parametric Equations

Parametric equations are a powerful way to describe the motion of a point in space. They define a set of equations where each coordinate of a point is expressed as a function of one or more parameters (often time). For example, in the given exercise, the parametric equations
\( x = 3, y = 1 + 3t, z = 1 + t \)
describe a line in three-dimensional space. Here, \( t \) is our parameter, and as it changes, the values of \( x, y, \) and \( z \) change accordingly, tracing out the path of a line. The beauty of parametric equations lies in their ability to capture complex motions and curves in a relatively simple form.

Cross Product

The cross product is a crucial mathematical operation used when dealing with vectors in three-dimensional space. It takes two vectors and returns a third vector that is perpendicular to both of the original vectors. The magnitude of this new vector is proportional to the area of the parallelogram spanned by the original two vectors. In the exercise, the cross product is used to calculate
\( \vec{AP} \times \vec{v} \)
where \( \vec{AP} \) is the vector from point A on the line to point P, and \( \vec{v} \) is the directional vector of the line. The cross product of these vectors gives us a vector that is useful in finding the shortest distance between point P and the line.

Vector Magnitude

Vector magnitude, also known as the vector's length or norm, is a measure of how long a vector is. Any vector, such as the directional vector \( \vec{v} \) in the problem, can be quantified in terms of its length. To compute a vector's magnitude, one must take the square root of the sum of the squares of its components. This is often symbolized as
\( |\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2} \)
where \( v_x, v_y, \) and \( v_z \) are the vector's components along the x, y, and z axes, respectively. In our example, the magnitude of the directional vector \( \vec{v} \) and the cross product \( \vec{AP} \times \vec{v} \) helps us determine the final distance between the point and line.

Directional Vector

A directional vector, often notated as \( \vec{v} \), gives direction and magnitude and thus describes the orientation of a line in space. It is derived from the coefficients of the parameter in the parametric equations. For instance, the directional vector of the line in our exercise is
\( \vec{v} = (0, 3, 1) \)
which results from the coefficients of \( t \) in the parametric equations. Crucially, the directional vector is central to finding the cross product with another vector and is essential in calculating distances to points not on the line, as demonstrated in the steps for solving the depicted exercise.