# Finding Orthogonal Projection of \overrightarrow{x}

• VinnyCee
In summary, the orthogonal projection of the vector [49, 49, 49] onto the subspace spanned by the vectors [2, 3, 6] and [3, -6, 2] is given by the formula proj_v \overrightarrow{x} = \left( \overrightarrow{u_1} \cdot \overrightarrow{x} \right) \overrightarrow{u_1} + \left( \overrightarrow{u_2} \cdot \overrightarrow{x} \right) \overrightarrow{u_2}, and the resulting vector is [28, -36, 28].

#### VinnyCee

Question (5.1, #26 -> Bretscher, O.):

Find the orthogonal projection of $$\left[\begin{array}{c} 49 \\ 49 \\ 49 \end{array}\right]$$ onto the subspace of $$\mathbb{R}^3$$ spanned by $$\left[\begin{array}{c} 2 \\ 3 \\ 6 \end{array}\right]$$ and $$\left[\begin{array}{c} 3 \\ -6 \\ 2 \end{array}\right]$$.

$$\overrightarrow{x} = \left[\begin{array}{c} 49 \\ 49 \\ 49 \end{array}\right]$$

Magnitude -> $$\sqrt{(2)^2 + (3)^2 + (6)^2} = 7\,\,=\,\,\sqrt{(3)^2 + (-6)^2 + (2)^2}$$

$$\overrrightarrow{u_1} = \left[\begin{array}{c} \frac{2}{7} \\ \frac{3}{7} \\ \frac{6}{7}\end{array}\right]$$

$$\overrrightarrow{u_2} = \left[\begin{array}{c} \frac{3}{7} \\ \frac{-6}{7} \\ \frac{2}{7}\end{array}\right]$$

$$proj_v \overrightarrow{x} = \left( \overrightarrow{u_1} \cdot \overrightarrow{x} \right) \overrightarrow{u_1} + \left( \overrightarrow{u_2} \cdot \overrightarrow{x} \right) \overrightarrow{u_2}$$

$$proj_v \overrightarrow{x} = \frac{539}{7} \overrightarrow{u_1} + \frac{7}{7} \overrightarrow{u_2} = 77 \overrightarrow{u_1} + \overrightarrow{u_2}$$

$$proj_v \overrightarrow{x} = \left[\begin{array}{c} \frac{157}{7} \\ \frac{225}{7} \\ \frac{464}{7}\end{array}\right]$$

Does this look correct?

EDIT: IT is incorrect, I now figured it out. Thanks

Last edited:
proj_v \overrightarrow{x} = \left( \overrightarrow{u_1} \cdot \overrightarrow{x} \right) \overrightarrow{u_1} + \left( \overrightarrow{u_2} \cdot \overrightarrow{x} \right) \overrightarrow{u_2}proj_v \overrightarrow{x} = \frac{196}{7} \overrightarrow{u_1} + \frac{343}{7} \overrightarrow{u_2} = 28 \overrightarrow{u_1} + 49 \overrightarrow{u_2}proj_v \overrightarrow{x} = \left[\begin{array}{c} 28 \\ -36 \\ 28 \end{array}\right]

## What is an orthogonal projection?

An orthogonal projection is a method used to find the closest point on a line or plane to a given point. It involves finding the shortest distance between the given point and the line or plane.

## What is the purpose of finding the orthogonal projection of a vector?

The purpose of finding the orthogonal projection of a vector is to break down the given vector into two components: one that lies on a specified line or plane, and another that is perpendicular to that line or plane. This can be useful in solving various geometry and physics problems.

## How do you find the orthogonal projection of a vector onto a line?

To find the orthogonal projection of a vector onto a line, you first need to find the unit vector in the direction of the line. Then, use the dot product to find the scalar projection of the vector onto the unit vector. Finally, multiply the scalar projection by the unit vector to find the orthogonal projection of the vector onto the line.

## What is the difference between orthogonal projection and parallel projection?

The main difference between orthogonal projection and parallel projection is that orthogonal projection preserves the relative lengths of vectors, while parallel projection does not. In other words, orthogonal projection maintains the shape of objects, while parallel projection can distort their shape.

## What are some real-life applications of finding the orthogonal projection of a vector?

One real-life application of finding the orthogonal projection of a vector is in computer graphics, where it is used to create 3D images by projecting objects onto a 2D screen. It is also used in physics and engineering to calculate forces and distances in different directions. Additionally, it is used in navigation and surveying to find the shortest distance between two points on a map or terrain.