# Finding perpendicular unit vector using inner product

• zecuria
In summary, using the given inner product, a unit vector perpendicular to the vector (1,1) can be found by setting the inner product of the perpendicular vector with (1,1) equal to 0 and solving for one of its components. Then, the resulting vector can be made a unit vector by dividing by its length, which is determined by the inner product.
zecuria

## Homework Statement

You are given that with x = (x1,x2), y = (y1,y2), the formula

(x,y) = [x1 x2] [2 1;1 2] [y1;y2] (where ; represents a new row).

is a inner product for the vectors in R2

Using this inner product, find a unit vector perpendicular to the vector (1,1)

## The Attempt at a Solution

I don't particularly understand what a perpendicular unit vector has to do with the inner product, I can find a perpendicular unit vector in the normal dot product way however I don't know how to find one using the inner product, any help would be much appreciated,

Two vectors are perpendicular if their inner product is equal to zero. The dot product is a common example of an inner product, but there are many others. Two vectors might be perpendicular when using one inner product, but perhaps not using another inner product. So just try taking the inner product of (1,1) with a vector (x1, x2) and setting the result equal to zero to come up with some dependency of x1 in terms of x2. Then you can just choose a value for x2 and TADA! you have a perpendicular vector. After that you can scale it to unit length (don't forget that you have a new inner product here! You should completely forget about the dot product during this problem).

Last edited:
zecuria said:

## Homework Statement

You are given that with x = (x1,x2), y = (y1,y2), the formula

(x,y) = [x1 x2] [2 1;1 2] [y1;y2] (where ; represents a new row).

is a inner product for the vectors in R2

Using this inner product, find a unit vector perpendicular to the vector (1,1)

## The Attempt at a Solution

I don't particularly understand what a perpendicular unit vector has to do with the inner product, I can find a perpendicular unit vector in the normal dot product way however I don't know how to find one using the inner product, any help would be much appreciated,

Just work things out. Supposed p=(1,1) and the perpendicular unit vector is u=(u1,u2). Then, in terms of your inner product, to have u perpendicular to p means (u,p)=0, and u being a unit vector means (u,u)=1. Work both of those out in components and try to solve the resulting equations.

zecuria said:

## Homework Statement

You are given that with x = (x1,x2), y = (y1,y2), the formula

(x,y) = [x1 x2] [2 1;1 2] [y1;y2] (where ; represents a new row).

is a inner product for the vectors in R2
So the inner product is
$$\begin{bmatrix}x_1 & x_2 \end{bmatrix}\begin{bmatrix}2 & 1 \\ 1 & 2 \end{bmatrix}\begin{bmatrix}y_1\\ y_2\end{bmatrix}= 2(x_1y_1+x_2y_2)+ (x_1y_2+ x_2y_1)$$

Using this inner product, find a unit vector perpendicular to the vector (1,1)

## The Attempt at a Solution

I don't particularly understand what a perpendicular unit vector has to do with the inner product,
Really? The definition of "perpendicular" in terms of an inner product is that the inner product of two perpendicular vectors is 0.

I can find a perpendicular unit vector in the normal dot product way however I don't know how to find one using the inner product, any help would be much appreciated,
An "inner product" is just a generalization of "dot product". Here, with $(x_1, y_1)= (1, 1)$ the inner product with (x, y) is 2((1)x+ (1)y)+ ((1)x+ (1)y)= 3x+ 3y= 0 which is the same as saying y= -x. Any vector perpendicular to (1, 1) is of the form (x, -x). Now you want to make that a "unit vector" by dividing by its length. Of course, you determine the length using that inner product: the length of a vector v is $\sqrt{(v, v)}$. The inner product of (x, -x) with itself is $2((x)(x)+ (-x)(-x))+ ((x)(-x)+ (-x)(x))= 2(x^2+ x^2)+ (-x^2- x^2)= 2x^2$. Set that equal to 1 and solve for x.

Last edited by a moderator:

## What is the inner product?

The inner product is a mathematical operation that takes two vectors and produces a scalar value. It is also known as the dot product and is calculated by multiplying the corresponding components of the two vectors and then summing the results.

## How do you find the perpendicular unit vector using the inner product?

To find the perpendicular unit vector, you first need to take the inner product of the two given vectors. Then, you can divide the result by the magnitude of one of the vectors. Finally, multiply the quotient by the other vector and subtract it from the original vector. The resulting vector will be perpendicular to the original vector and will also have a magnitude of 1, making it a unit vector.

## Why is the inner product used to find the perpendicular unit vector?

The inner product is used because it gives us a scalar value that can be used to calculate the perpendicular unit vector. By dividing the inner product by the magnitude of one of the vectors, we can ensure that the resulting vector has a magnitude of 1, making it a unit vector. Additionally, taking the inner product helps us find the angle between the two vectors, which is necessary for finding the perpendicular vector.

## Can the inner product be used with any type of vector?

Yes, the inner product can be used with any type of vector as long as the vectors have the same number of dimensions. This means that it can be used with both two-dimensional and three-dimensional vectors, as well as with vectors in higher dimensions.

## Is there a geometric interpretation of the inner product in finding the perpendicular unit vector?

Yes, the inner product can be interpreted geometrically as the projection of one vector onto another. This means that the inner product gives us the length of the shadow cast by one vector onto another, which can be used to find the angle between the two vectors and ultimately, the perpendicular unit vector.

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