Find All Unit Vectors In The Plane Determined By Vectors U U U And V V V That Are Perpendicular To The Vector W.

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Introduction

In vector spaces, finding unit vectors that are perpendicular to a given vector is a crucial concept in various mathematical and scientific applications. Given two vectors $u$ and $v$, we can determine the plane that contains both vectors. However, to find unit vectors in this plane that are perpendicular to a third vector $w$, we need to apply specific mathematical techniques. In this article, we will explore the process of finding all unit vectors in the plane determined by vectors $u$ and $v$ that are perpendicular to the vector $w$.

Understanding the Problem

To begin with, let's understand the problem at hand. We are given two vectors $u=(0,1,1)$ and $v=(2,-1,3)$, and a third vector $w=(5,7,-4)$. Our goal is to find all unit vectors in the plane determined by $u$ and $v$ that are perpendicular to the vector $w$. This means we need to find vectors that lie in the plane containing $u$ and $v$ and are orthogonal to $w$.

Finding the Plane Determined by $u$ and $v$

The plane determined by two vectors $u$ and $v$ can be found by taking the cross product of the two vectors. The cross product of two vectors $u=(u_1,u_2,u_3)$ and $v=(v_1,v_2,v_3)$ is given by:

$$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ \end{vmatrix}$$

where $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ are the unit vectors along the $x$, $y$, and $z$ axes, respectively.

For our given vectors $u=(0,1,1)$ and $v=(2,-1,3)$, the cross product is:

$$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & 1 \\ 2 & -1 & 3 \\ \end{vmatrix} = (3-(-1))\mathbf{i} - (0-2)\mathbf{j} + (-1-2)\mathbf{k} = 4\mathbf{i} + 2\mathbf{j} - 3\mathbf{k} = (4, 2, -3)$$

Therefore, the plane determined by $u$ and $v$ is spanned by the vector $(4, 2, -3)$.

Finding Unit Vectors Perpendicular to $w$

To find unit vectors in the plane determined by $u$ and $v$ that are perpendicular to the vector $w$, we need to find vectors that are orthogonal to both $(4, 2, -3)$ and $w=(5,7,-4)$. This can be achieved by taking the cross product of the two vectors.

The cross product of $(4, 2, -3)$ and $w=(5,7,-4)$ is:

$$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 2 & -3 \\ 5 & 7 & -4 \\ \end{vmatrix} = (-8-(-21))\mathbf{i} - (-16-(-15))\mathbf{j} + (28-10)\mathbf{k} = 13\mathbf{i} - \mathbf{j} + 18\mathbf{k} = (13, -1, 18)$$

However, this vector is not a unit vector. To find the unit vector, we need to normalize the vector by dividing it by its magnitude.

Normalizing the Vector

The magnitude of the vector $(13, -1, 18)$ is:

$$\sqrt{13^2 + (-1)^2 + 18^2} = \sqrt{169 + 1 + 324} = \sqrt{494}$$

Therefore, the unit vector is:

$$\frac{(13, -1, 18)}{\sqrt{494}} = \left(\frac{13}{\sqrt{494}}, \frac{-1}{\sqrt{494}}, \frac{18}{\sqrt{494}}\right)$$

Conclusion

In this article, we have explored the process of finding all unit vectors in the plane determined by vectors $u$ and $v$ that are perpendicular to the vector $w$. We have found the plane determined by $u$ and $v$ by taking the cross product of the two vectors, and then found the unit vectors in this plane that are perpendicular to $w$ by taking the cross product of the two vectors. Finally, we have normalized the vector to obtain the unit vector.

References

• [1] Hoffman, K., & Kunze, R. (1971). Linear Algebra. Prentice-Hall.
• [2] Strang, G. (1988). Linear Algebra and Its Applications. Academic Press.

Additional Information

• The cross product of two vectors $u$ and $v$ is a vector that is orthogonal to both $u$ and $v$.
• The magnitude of a vector is the length of the vector.
• A unit vector is a vector with a magnitude of 1.
  

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Introduction

In our previous article, we explored the process of finding all unit vectors in the plane determined by vectors $u$ and $v$ that are perpendicular to the vector $w$. In this article, we will answer some frequently asked questions related to this topic.

Q: What is the significance of finding unit vectors in the plane determined by vectors $u$ and $v$ that are perpendicular to the vector $w$?

A: Finding unit vectors in the plane determined by vectors $u$ and $v$ that are perpendicular to the vector $w$ is significant in various mathematical and scientific applications. For example, it can be used to find the normal vector to a plane, which is essential in geometry and physics.

Q: How do I find the plane determined by vectors $u$ and $v$?

A: To find the plane determined by vectors $u$ and $v$, you need to take the cross product of the two vectors. The cross product of two vectors $u=(u_1,u_2,u_3)$ and $v=(v_1,v_2,v_3)$ is given by:

$$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ \end{vmatrix}$$

Q: How do I find unit vectors in the plane determined by vectors $u$ and $v$ that are perpendicular to the vector $w$?

A: To find unit vectors in the plane determined by vectors $u$ and $v$ that are perpendicular to the vector $w$, you need to take the cross product of the two vectors. The cross product of $(4, 2, -3)$ and $w=(5,7,-4)$ is:

$$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 2 & -3 \\ 5 & 7 & -4 \\ \end{vmatrix} = (-8-(-21))\mathbf{i} - (-16-(-15))\mathbf{j} + (28-10)\mathbf{k} = 13\mathbf{i} - \mathbf{j} + 18\mathbf{k} = (13, -1, 18)$$

Q: How do I normalize the vector to obtain the unit vector?

A: To normalize the vector, you need to divide it by its magnitude. The magnitude of the vector $(13, -1, 18)$ is:

$$\sqrt{13^2 + (-1)^2 + 18^2} = \sqrt{169 + 1 + 324} = \sqrt{494}$$

Therefore, the unit vector is:

$$\frac{(13, -1, 18)}{\sqrt{494}} = \left(\frac{13}{\sqrt{494}}, \frac{-1}{\sqrt{494}}, \frac{18}{\sqrt{494}}\right)$$

Q: What are some common applications of finding unit vectors in the plane determined by vectors $u$ and $v$ that are perpendicular to the vector $w$?

A: Some common applications of finding unit vectors the plane determined by vectors $u$ and $v$ that are perpendicular to the vector $w$ include:

• Finding the normal vector to a plane
• Finding the area of a triangle
• Finding the distance between two points
• Finding the equation of a plane

Q: What are some common mistakes to avoid when finding unit vectors in the plane determined by vectors $u$ and $v$ that are perpendicular to the vector $w$?

A: Some common mistakes to avoid when finding unit vectors in the plane determined by vectors $u$ and $v$ that are perpendicular to the vector $w$ include:

• Not normalizing the vector
• Not taking the cross product of the two vectors
• Not using the correct formula for the cross product
• Not checking the magnitude of the vector

Conclusion

In this article, we have answered some frequently asked questions related to finding unit vectors in the plane determined by vectors $u$ and $v$ that are perpendicular to the vector $w$. We hope that this article has been helpful in clarifying any doubts you may have had on this topic.

References

• [1] Hoffman, K., & Kunze, R. (1971). Linear Algebra. Prentice-Hall.
• [2] Strang, G. (1988). Linear Algebra and Its Applications. Academic Press.

Additional Information

• The cross product of two vectors $u$ and $v$ is a vector that is orthogonal to both $u$ and $v$.
• The magnitude of a vector is the length of the vector.
• A unit vector is a vector with a magnitude of 1.