Problem 7 - 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 \(\boldsymbol{v}=\langle 1,2,3\rangle\) and \(\boldsymbol{w}=\langle-1,2,-3\rangle .\)

Step-by-Step Solution

Answer

\(|\boldsymbol{v}| = \sqrt{14}\), \(\boldsymbol{v} + \boldsymbol{w} = \langle 0, 4, 0 \rangle\), \(\boldsymbol{v} \cdot \boldsymbol{w} = -6\), \(|\boldsymbol{v} + \boldsymbol{w}| = 4\), \(|\boldsymbol{v} - \boldsymbol{w}| = 2\sqrt{10}\), \(-2\boldsymbol{v} = \langle -2, -4, -6 \rangle\).

Step 1: Find the magnitude of vector \(\boldsymbol{v}\)

To find the magnitude of \(\boldsymbol{v} = \langle 1, 2, 3 \rangle\), use the formula \(|\boldsymbol{v}| = \sqrt{x^2 + y^2 + z^2}\). In this case, replace \(x = 1\), \(y = 2\), and \(z = 3\), so: \(|\boldsymbol{v}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}\).

Step 2: Find the sum of vectors \(\boldsymbol{v}\) and \(\boldsymbol{w}\)

The sum of vectors \(\boldsymbol{v}\) and \(\boldsymbol{w}\) is given by \(\boldsymbol{v} + \boldsymbol{w} = \langle v_1 + w_1, v_2 + w_2, v_3 + w_3 \rangle \). For \(\boldsymbol{v} = \langle 1, 2, 3 \rangle\) and \(\boldsymbol{w} = \langle -1, 2, -3 \rangle\), the result is: \(\boldsymbol{v} + \boldsymbol{w} = \langle 1 + (-1), 2 + 2, 3 + (-3) \rangle = \langle 0, 4, 0 \rangle\).

Step 3: Find the dot product of vectors \(\boldsymbol{v}\) and \(\boldsymbol{w}\)

The dot product \(\boldsymbol{v} \cdot \boldsymbol{w}\) is calculated using the formula: \(v_1w_1 + v_2w_2 + v_3w_3\). Substituting the values, we get: \(1(-1) + 2(2) + 3(-3) = -1 + 4 - 9 = -6\).

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

We already calculated \(\boldsymbol{v} + \boldsymbol{w} = \langle 0, 4, 0 \rangle\) in Step 2. Now, find its magnitude: \(|\boldsymbol{v} + \boldsymbol{w}| = \sqrt{0^2 + 4^2 + 0^2} = \sqrt{16} = 4\).

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

First calculate \(\boldsymbol{v} - \boldsymbol{w} = \langle 1 - (-1), 2 - 2, 3 - (-3) \rangle = \langle 2, 0, 6 \rangle\). Then, find the magnitude: \(|\boldsymbol{v} - \boldsymbol{w}| = \sqrt{2^2 + 0^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}\).

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

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

Key Concepts

Vector Magnitude

The magnitude of a vector is like its length, representing how much space it occupies in its direction. Calculating vector magnitude is straightforward with its formula. Given a vector \( \boldsymbol{v} = \langle x, y, z \rangle \), the magnitude \(|\boldsymbol{v}|\) is calculated by:

• Squaring each component: \(x^2, y^2, z^2\).
• Adding these squared values together.
• Taking the square root of the sum.

For example, the magnitude of \(\boldsymbol{v} = \langle 1, 2, 3 \rangle\) is:

• \(|\boldsymbol{v}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}\).

Understanding vector magnitude helps determine how long a vector is in space.

Vector Addition

Vector addition is the process of "adding" two vectors together. This does not refer to simple numeric addition but to adding corresponding components of each vector. For vectors \(\boldsymbol{v} = \langle v_1, v_2, v_3 \rangle \) and \(\boldsymbol{w} = \langle w_1, w_2, w_3 \rangle\), the sum \(\boldsymbol{v} + \boldsymbol{w}\) is:

• \(\langle v_1 + w_1, v_2 + w_2, v_3 + w_3 \rangle\).

Taking the vectors \(\boldsymbol{v} = \langle 1, 2, 3 \rangle\) and \(\boldsymbol{w} = \langle -1, 2, -3 \rangle\):

• The result is \(\langle 0, 4, 0 \rangle\).

This operation represents the point at which both vectors meet in a coordinate system, highlighting how their "forces" combine.

Dot Product

The dot product of two vectors is a scalar value showing the extent to which two vectors point in the same direction. It's calculated by multiplying corresponding components and then adding them up. For vectors \(\boldsymbol{v} = \langle v_1, v_2, v_3 \rangle\) and \(\boldsymbol{w} = \langle w_1, w_2, w_3 \rangle\), the dot product \(\boldsymbol{v} \cdot \boldsymbol{w}\) is:

• \(v_1w_1 + v_2w_2 + v_3w_3\).

Using the vectors \(\boldsymbol{v} = \langle 1, 2, 3 \rangle\) and \(\boldsymbol{w} = \langle -1, 2, -3 \rangle\):

• The dot product is \(-6\).

The negative result here indicates the vectors point in partly opposite directions.

Vector Subtraction

Vector subtraction involves "subtracting" the components of one vector from another, similar to vector addition but using subtraction. For vectors \(\boldsymbol{v} = \langle v_1, v_2, v_3 \rangle \) and \(\boldsymbol{w} = \langle w_1, w_2, w_3 \rangle\), the difference \(\boldsymbol{v} - \boldsymbol{w}\) is:

• \(\langle v_1 - w_1, v_2 - w_2, v_3 - w_3 \rangle\).

For example, subtracting \(\boldsymbol{w} = \langle -1, 2, -3 \rangle\) from \(\boldsymbol{v} = \langle 1, 2, 3 \rangle\) gives:

• \(\langle 2, 0, 6 \rangle\).

This represents the vector from the tip of \(\boldsymbol{w}\) to the tip of \(\boldsymbol{v}\) in the coordinate system.

Scalar Multiplication

Scalar multiplication involves multiplying every component of a vector by the same scalar value. It changes the magnitude of the vector but not its direction. Given a vector \(\boldsymbol{v} = \langle v_1, v_2, v_3 \rangle\) and a scalar \(c\), the product is:

• \(\langle c v_1, c v_2, c v_3 \rangle\).

Considering \(-2\) as the scalar for \(\boldsymbol{v} = \langle 1, 2, 3 \rangle\):

• \(-2\boldsymbol{v} = \langle -2, -4, -6 \rangle\).

This operation reverses the direction due to the negative scalar, affecting the magnitude by doubling it.