Problem 35 - APEX Calculus

Question

In Exercises 35-38, find a unit vector orthogonal to both \(\vec{u}\) and \(\vec{v} .\) \(\vec{u}=\langle 1,1,1\rangle, \quad \vec{v}=\langle 2,0,1\rangle\)

Step-by-Step Solution

Answer

The unit vector orthogonal to both is \(\langle \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{-2}{\sqrt{6}} \rangle\).

Step 1: Understand the Problem

We need to find a unit vector that is orthogonal (perpendicular) to both given vectors \(\vec{u}\) and \(\vec{v}\). These vectors are in 3D space, and a common method to find a vector orthogonal to both is to use the cross product of the two vectors.

Step 2: Calculate the Cross Product

The cross product \(\vec{w} = \vec{u} \times \vec{v}\) of vectors \(\vec{u} = \langle 1, 1, 1 \rangle\) and \(\vec{v} = \langle 2, 0, 1 \rangle\) is calculated using the determinant formula:\[\vec{w} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \1 & 1 & 1 \2 & 0 & 1 \end{vmatrix}\]The calculated result is \(\vec{w} = \langle 1, 1, -2 \rangle\).

Step 3: Confirm Orthogonality

We confirm that \(\vec{w}\) is orthogonal to both \(\vec{u}\) and \(\vec{v}\) by checking that their dot products are zero:For \(\vec{u} \cdot \vec{w}\):\[1(1) + 1(1) + 1(-2) = 0\]For \(\vec{v} \cdot \vec{w}\):\[2(1) + 0(1) + 1(-2) = 0\]Since both dot products are zero, \(\vec{w}\) is orthogonal to both \(\vec{u}\) and \(\vec{v}\).

Step 4: Find the Unit Vector

The unit vector \(\vec{u_w}\) in the direction of \(\vec{w}\) is found by dividing \(\vec{w}\) by its magnitude. The magnitude \(||\vec{w}||\) is calculated as:\[||\vec{w}|| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{6}\]Thus, the unit vector \(\vec{u_w}\) is:\[\vec{u_w} = \left\langle \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{-2}{\sqrt{6}} \right\rangle\]

Key Concepts

Cross Product

The concept of the cross product is fundamental in vector calculus, especially in three dimensions. The cross product is a binary operation on two vectors that results in another vector which is orthogonal (perpendicular) to both of the original vectors. This is particularly useful for determining a vector that lies perpendicular to a plane defined by two other vectors.
To calculate the cross product of two vectors, \boldsymbol{a} and \boldsymbol{b}, you use the determinant of a 3x3 matrix, with unit vectors \hat{i}, \hat{j}, \hat{k} as the first row, the components of vector \boldsymbol{a} as the second, and those of vector \boldsymbol{b} as the third. For example, if \boldsymbol{a} = \langle a_1, a_2, a_3 \rangle and \boldsymbol{b} = \langle b_1, b_2, b_3 \rangle, the cross product \boldsymbol{a} \times \boldsymbol{b} results in a new vector:\

• \(\boldsymbol{a} \times \boldsymbol{b} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle\)

In the given exercise, the vectors \boldsymbol{u} and \boldsymbol{v} were \(\langle 1, 1, 1 \rangle\) and \(\langle 2, 0, 1 \rangle\) respectively, leading to the cross product \(\boldsymbol{w} = \langle 1, 1, -2 \rangle\). This demonstrates the classic orthogonal vector in vector space.

Orthogonal Vectors

Orthogonal vectors are a key concept when working with vectors, particularly in higher dimensions. Two vectors are said to be orthogonal if their dot product is zero. This implies that the angle between them is 90 degrees, making them perpendicular.
Here, orthogonality is essential for problems where we're interested in vectors that don't influence each other in the direction they span. By using the cross product, we obtain a vector that is orthogonal to the plane formed by two given vectors.
To check whether two vectors, say \(\boldsymbol{a} = \langle a_1, a_2, a_3 \rangle\) and \(\boldsymbol{b} = \langle b_1, b_2, b_3 \rangle\), are orthogonal, calculate their dot product:

• \(\boldsymbol{a} \cdot \boldsymbol{b} = a_1b_1 + a_2b_2 + a_3b_3 = 0\)

In the original exercise, the vector \(\boldsymbol{w} = \langle 1, 1, -2 \rangle\) is confirmed to be orthogonal to both \(\boldsymbol{u} = \langle 1, 1, 1 \rangle\) and \(\boldsymbol{v} = \langle 2, 0, 1 \rangle\) because \(\boldsymbol{u} \cdot \boldsymbol{w} = 0\) and \(\boldsymbol{v} \cdot \boldsymbol{w} = 0\). This makes \(\boldsymbol{w}\) a sought-after orthogonal vector to both \(\boldsymbol{u}\) and \(\boldsymbol{v}\).

Unit Vector

A unit vector is a vector with a magnitude of 1 that indicates a direction in space. It is a fundamental concept as it simplifies vector manipulation-related calculations, such as finding directions in physical space.
To convert a vector \(\boldsymbol{a} = \langle a_1, a_2, a_3 \rangle\) into a unit vector, you divide each component by the vector's magnitude. The magnitude \(\|\boldsymbol{a}\|\) is calculated as follows:

• \(\|\boldsymbol{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2}\)

The resulting unit vector \(\boldsymbol{a}_u\) is then:

• \(\boldsymbol{a}_u = \left\langle \frac{a_1}{\|\boldsymbol{a}\|}, \frac{a_2}{\|\boldsymbol{a}\|}, \frac{a_3}{\|\boldsymbol{a}\|} \right\rangle\)

In the given problem, once the orthogonal vector \(\boldsymbol{w} = \langle 1, 1, -2 \rangle\) is found, its magnitude is computed as \(\sqrt{6}\). Thus, the unit vector \(\boldsymbol{u}_w\) is determined as \(\left\langle \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{-2}{\sqrt{6}} \right\rangle\). This unit vector maintains the direction of \(\boldsymbol{w}\) but scales it to a magnitude of 1.