Question

The parametric equation of a line \(L\) in a coordinate system with origin \(O\) is $$ x=2 t+1, \quad y=t+1, \quad z=-2 t+2 $$ A point \(P\) has coordinates \((3,-2,1)\). (a) Using the projection operators, find the length of the projection of \(\overline{O P}\) on the line \(L\). (b) Find the vector whose beginning is \(P\) and ends perpendicularly on \(L\). (c) From this vector calculate the distance from \(P\) to \(L\).

Short answer

The length of the projection is 1, the perpendicular vector from the point to the line is (1,-1,3), and the distance from the point to the line is \( \sqrt{11}\)

Step-by-step solution

Compute Vector OP

Extract the coordinates of the origin \(O\) which is \((0,0,0)\) and the point \(P\) which is \((3,-2,1)\). Compute the vector \(\overrightarrow{OP}\) by subtracting the coordinates of \(O\) from those of \(P\). That is: \(\overrightarrow{OP} = P - O\). Resulting in: \(\overrightarrow{OP} = (3,-2,1)\)

Compute Projection Operator

The parametric equations of the line \(L\) can be rewritten in vector form as \(L(t) = (2t+1)\hat{i} + (t+1)\hat{j} - 2t\hat{k}= t \cdot (2\hat{i} + \hat{j} -2\hat{k}) + \hat{i} + \hat{j} + 2\hat{k}\). Here, the vector direction of the line is \( \hat{d} = 2\hat{i} + \hat{j} -2\hat{k}\). The projection of \(\overrightarrow{OP}\) on line \(L\) is given by \( \overrightarrow{Proj} = \frac{\overrightarrow{OP} \cdot \hat{d}}{\hat{d} \cdot \hat{d}}\). Computing this gives \( \overrightarrow{Proj} = 1\). The projection length is hence 1.

Compute the Perpendicular Vector

The perpendicular vector from point \(P\) to the line \(L\), denoted as \(\overrightarrow{u}\), is given by \(\overrightarrow{u} = \overrightarrow{OP} - \overrightarrow{Proj} \cdot \hat{d}\). Substituting the values gives: \(\overrightarrow{u} = (1,-1,3)\).

Compute the Distance

The distance from point \(P\) to the line \(L\) is the magnitude of the vector \(\overrightarrow{u}\). This is computed as: \( distance = ||\overrightarrow{u}|| = \sqrt{1^2 + (-1)^2 + 3^2} = \sqrt{11}\)

Key concepts

Projection Operators

Projection operators tend to be one of those mathematical concepts that can seem daunting at first, but with a bit of a breakdown, they're quite intuitive. When we talk about projection in a geometrical sense, we are typically referring to casting a shadow of one element onto another. In terms of vectors in three-dimensional space, projection operators allow us to take one vector and project it onto another vector or line.

To understand the projection operator, consider two vectors, \( \mathbf{a} \) and \( \mathbf{b} \). The projection operator is used to project \( \mathbf{a} \) onto \( \mathbf{b} \), resulting in a new vector that lies entirely on \( \mathbf{b} \) and points in the same (or opposite, depending on the direction) direction. Mathematically, the projected vector, often denoted as \( \mathbf{proj}_\mathbf{b}\mathbf{a} \), is calculated using the formula:\[ \mathbf{proj}_\mathbf{b}\mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \mathbf{b} \]Here, \( \cdot \) denotes the dot product of the two vectors. This operation scales the vector \( \mathbf{b} \) so that when you add enough of these units to it (or subtract, if necessary), you land exactly on the shadow of \( \mathbf{a} \) onto \( \mathbf{b} \) along the direction of \( \mathbf{b} \).

The beauty of projection operators is that they encapsulate all this scaling and direction within a neat mathematical formula. When you apply this to our textbook problem, the projection operator helps determine the length of the projection of a vector from the origin to point P on the given line L.

Vector Projection

Moving on from the theory of projection operators, let's delve into vector projection—visualizing how we can 'throw' one vector onto another. Vector projection is the result you obtain when using a projection operator. It is the vector that represents the shadow cast when one vector is projected onto another.

In the context of the exercise at hand, vector projection allows us to find the component of the vector \( \overrightarrow{OP} \) that lies along the direction of the line \( L \). This component is effectively the shortest route from our point P to the line—if you were to slide P straight down to L, the path it would take is our vector projection.

Mathematically, we've determined the vector direction of line L as \( \hat{d} = 2\hat{i} + \hat{j} -2\hat{k} \). The projection involves calculating the extent to which \( \overrightarrow{OP} \) aligns with this direction, which is done through the dot product as mentioned earlier. Specifically, we calculate the projection of \( \overrightarrow{OP} \) onto \( L \) by the formula involving the dot product with \( \hat{d} \). After performing the corresponding calculations, we get the length of this projection, which helps interpret the positional relationship between point P and line L.

Distance from Point to Line

Lastly, in the realm of three-dimensional geometry, discerning the distance from a point to a line is a practical application of vector arithmetic. The shortest distance between a point and a line is along the line perpendicular to the given line that also passes through the point in question. To determine this distance mathematically, we harness the power of vector projection.

After finding the vector projection, as demonstrated in the previous section, the remaining component of the vector \( \overrightarrow{OP} \) that is perpendicular to the line is the key to determining the distance from the point P to the line L. This component, which we've called the perpendicular vector \( \overrightarrow{u} \), is calculated by subtracting the vector projection from the original vector \( \overrightarrow{OP} \). In algebraic form, \( \overrightarrow{u} = \overrightarrow{OP} - (\overrightarrow{Proj} \cdot \hat{d}) \).Once obtained, the magnitude of \( \overrightarrow{u} \) is essentially the distance from point P to the line L. It's found using the Euclidean norm (or magnitude formula), yielding a distance of \( \sqrt{11} \) in our problem set. To summarize: the shortest distance is the length of the perpendicular dropped from the point to the line, and with vector projection, we find both the direction and magnitude of this critical line segment.