Problem 10 - Calculus: Early Transcendentals

Question

Find \(|\boldsymbol{v}|, \boldsymbol{v}+\boldsymbol{w}, \boldsymbol{v}-\boldsymbol{w},|\boldsymbol{v}+\boldsymbol{w}|,|\boldsymbol{v}-\boldsymbol{w}|\) and \(-2 \boldsymbol{v}\) for \(\boldsymbol{v}=\langle 3,2,1\rangle\) and \(\boldsymbol{w}=\langle-1,-1,-1\rangle .\)

Step-by-Step Solution

Answer

|\boldsymbol{v}| = \sqrt{14}, \boldsymbol{v} + \boldsymbol{w} = \langle 2, 1, 0 \rangle, \boldsymbol{v} - \boldsymbol{w} = \langle 4, 3, 2 \rangle. |\boldsymbol{v} + \boldsymbol{w}| = \sqrt{5}, |\boldsymbol{v} - \boldsymbol{w}| = \sqrt{29}, -2\boldsymbol{v} = \langle -6, -4, -2 \rangle.

Step 1: Find Magnitude of \( \boldsymbol{v} \)

To find the magnitude of vector \( \boldsymbol{v} = \langle 3, 2, 1 \rangle \), use the formula for the magnitude of a vector: \[ |\boldsymbol{v}| = \sqrt{3^2 + 2^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14} \] Thus, the magnitude of \( \boldsymbol{v} \) is \( \sqrt{14} \).

Step 2: Add Vectors \( \boldsymbol{v} \) and \( \boldsymbol{w} \)

The sum of vectors \( \boldsymbol{v} \) and \( \boldsymbol{w} \) is found by adding their corresponding components: \[ \boldsymbol{v} + \boldsymbol{w} = \langle 3, 2, 1 \rangle + \langle -1, -1, -1 \rangle = \langle 3 + (-1), 2 + (-1), 1 + (-1) \rangle = \langle 2, 1, 0 \rangle \] So, \( \boldsymbol{v} + \boldsymbol{w} = \langle 2, 1, 0 \rangle \).

Step 3: Subtract Vector \( \boldsymbol{w} \) from \( \boldsymbol{v} \)

Subtract \( \boldsymbol{w} \) from \( \boldsymbol{v} \) by subtracting their components: \[ \boldsymbol{v} - \boldsymbol{w} = \langle 3, 2, 1 \rangle - \langle -1, -1, -1 \rangle = \langle 3 - (-1), 2 - (-1), 1 - (-1) \rangle = \langle 4, 3, 2 \rangle \] Therefore, \( \boldsymbol{v} - \boldsymbol{w} = \langle 4, 3, 2 \rangle \).

Step 4: Find Magnitude of \( \boldsymbol{v} + \boldsymbol{w} \)

Use the magnitude formula on the resultant vector from Step 2: \[ |\boldsymbol{v} + \boldsymbol{w}| = \sqrt{2^2 + 1^2 + 0^2} = \sqrt{4 + 1 + 0} = \sqrt{5} \] The magnitude of \( \boldsymbol{v} + \boldsymbol{w} \) is \( \sqrt{5} \).

Step 5: Find Magnitude of \( \boldsymbol{v} - \boldsymbol{w} \)

Use the magnitude formula on the vector from Step 3: \[ |\boldsymbol{v} - \boldsymbol{w}| = \sqrt{4^2 + 3^2 + 2^2} = \sqrt{16 + 9 + 4} = \sqrt{29} \] The magnitude of \( \boldsymbol{v} - \boldsymbol{w} \) is \( \sqrt{29} \).

Step 6: Multiply \( \boldsymbol{v} \) by \(-2\)

Multiply each component of \( \boldsymbol{v} = \langle 3, 2, 1 \rangle \) by \(-2\): \[ -2 \boldsymbol{v} = -2 \langle 3, 2, 1 \rangle = \langle -6, -4, -2 \rangle \] Thus, \(-2 \boldsymbol{v} = \langle -6, -4, -2 \rangle \).

Key Concepts

Vector Magnitude

Understanding the magnitude of a vector is crucial since it gives us the "length" or "size" of the vector. It is a non-negative scalar quantity that represents the distance from the origin to the point described by the vector. For a vector \( \boldsymbol{v} = \langle x, y, z \rangle \), the magnitude is calculated using the formula:

• \( |\boldsymbol{v}| = \sqrt{x^2 + y^2 + z^2} \)

To illustrate, let's calculate the magnitude of \( \boldsymbol{v} = \langle 3, 2, 1 \rangle \). We plug the components into the formula:
\[ |\boldsymbol{v}| = \sqrt{3^2 + 2^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14} \]Hence, the magnitude of \( \boldsymbol{v} \) is \( \sqrt{14} \). This length tells you how far the vector extends in 3-dimensional space, providing insight into its impact in vector operations.

Vector Addition

Vector addition is a straightforward concept where two vectors are added together component-wise. This operation results in a new vector. When adding two vectors \( \boldsymbol{v} = \langle 3, 2, 1 \rangle \) and \( \boldsymbol{w} = \langle -1, -1, -1 \rangle \), we perform the following calculations for each respective component:

• Add the first components: \( 3 + (-1) = 2 \)
• Add the second components: \( 2 + (-1) = 1 \)
• Add the third components: \( 1 + (-1) = 0 \)

After completing these additions, the resultant vector is \( \langle 2, 1, 0 \rangle \).
This operation can help you understand phenomena such as combined forces in physics or resultant velocity in dynamics. Essentially, vector addition combines two directional influences into one.

Vector Subtraction

Vector subtraction is similar to vector addition, except it involves subtracting the components of one vector from the corresponding components of another. Given vectors \( \boldsymbol{v} = \langle 3, 2, 1 \rangle \) and \( \boldsymbol{w} = \langle -1, -1, -1 \rangle \), subtraction works as follows:

• Subtract the first components: \( 3 - (-1) = 4 \)
• Subtract the second components: \( 2 - (-1) = 3 \)
• Subtract the third components: \( 1 - (-1) = 2 \)

The resulting vector from this operation is \( \langle 4, 3, 2 \rangle \).
Vector subtraction is useful in determining direction differences or displacement vectors. It easily allows for seeing the change in positions or states in physical systems.

Scalar Multiplication

Scalar multiplication involves multiplying each component of a vector by a given number, or scalar. It essentially scales the vector, changing its magnitude but not its direction (as long as the scalar is positive). For the vector \( \boldsymbol{v} = \langle 3, 2, 1 \rangle \) and scalar \(-2\), the operation proceeds as follows:

• Multiply the first component: \(-2 \times 3 = -6 \)
• Multiply the second component: \(-2 \times 2 = -4 \)
• Multiply the third component: \(-2 \times 1 = -2 \)

This gives a new vector: \( \langle -6, -4, -2 \rangle \).
Scalar multiplication can be used to inversely reflect a vector or to extend its reach. In graphics and physics, it modifies how vectors affect objects by scaling their influence up or down.