Math Problem Statement

Find a unit vector that is orthogonal to both u and v.

u

 = 

1, −3, 5 

v

 = 

4, 2, 3

Solution

To find a unit vector that is orthogonal to both u\mathbf{u} and v\mathbf{v}, we need to first calculate the cross product u×v\mathbf{u} \times \mathbf{v}, which will give us a vector orthogonal to both u\mathbf{u} and v\mathbf{v}. Then, we will normalize this vector to obtain a unit vector.

Step 1: Find the cross product u×v\mathbf{u} \times \mathbf{v}

Given:

u=i+(3)j+5k\mathbf{u} = \mathbf{i} + (-3)\mathbf{j} + 5\mathbf{k} v=4i+2j+3k\mathbf{v} = 4\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}

The cross product u×v\mathbf{u} \times \mathbf{v} is:

u×v=ijk135423\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -3 & 5 \\ 4 & 2 & 3 \end{vmatrix}

Expanding the determinant:

u×v=i((3)×35×2)j(1×35×4)+k(1×2(3)×4)\mathbf{u} \times \mathbf{v} = \mathbf{i} \left((-3) \times 3 - 5 \times 2 \right) - \mathbf{j} \left(1 \times 3 - 5 \times 4 \right) + \mathbf{k} \left(1 \times 2 - (-3) \times 4 \right)

u×v=i(910)j(320)+k(2+12)\mathbf{u} \times \mathbf{v} = \mathbf{i}(-9 - 10) - \mathbf{j}(3 - 20) + \mathbf{k}(2 + 12)

u×v=i(19)j(17)+k(14)\mathbf{u} \times \mathbf{v} = \mathbf{i}(-19) - \mathbf{j}(-17) + \mathbf{k}(14)

u×v=19i+17j+14k\mathbf{u} \times \mathbf{v} = -19\mathbf{i} + 17\mathbf{j} + 14\mathbf{k}

So, the vector orthogonal to both u\mathbf{u} and v\mathbf{v} is:

w=19i+17j+14k\mathbf{w} = -19\mathbf{i} + 17\mathbf{j} + 14\mathbf{k}

Step 2: Normalize the vector to obtain a unit vector

To normalize w\mathbf{w}, we need to divide it by its magnitude w\|\mathbf{w}\|, which is given by:

w=(19)2+172+142\|\mathbf{w}\| = \sqrt{(-19)^2 + 17^2 + 14^2}

w=361+289+196\|\mathbf{w}\| = \sqrt{361 + 289 + 196}

w=846\|\mathbf{w}\| = \sqrt{846}

The unit vector w^\mathbf{\hat{w}} is:

w^=1846(19i+17j+14k)\mathbf{\hat{w}} = \frac{1}{\sqrt{846}} \left(-19\mathbf{i} + 17\mathbf{j} + 14\mathbf{k}\right)

Or, simplifying the components:

w^=(19846)i+(17846)j+(14846)k\mathbf{\hat{w}} = \left(\frac{-19}{\sqrt{846}}\right)\mathbf{i} + \left(\frac{17}{\sqrt{846}}\right)\mathbf{j} + \left(\frac{14}{\sqrt{846}}\right)\mathbf{k}

This is the unit vector orthogonal to both u\mathbf{u} and v\mathbf{v}.

Would you like any further details or have any questions?

Here are 5 related questions to consider:

  1. How do you find the magnitude of a vector in three dimensions?
  2. What does it mean for two vectors to be orthogonal?
  3. How can you verify that a vector is orthogonal to two given vectors?
  4. What is the relationship between the cross product and vector areas?
  5. How do you find the direction of the cross product vector?

Tip: A vector is orthogonal to another if their dot product is zero. This property can be useful to verify your results.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Vectors
Cross Product
Unit Vectors

Formulas

Cross product formula
Unit vector formula

Theorems

-

Suitable Grade Level

High School