Problem 10 Let \(\mathbf{u}=\mathbf{i}+3 \m... [FREE SOLUTION]

Chapter 14: Problem 10

Let $\mathbf{u}=\mathbf{i}+3 \mathbf{j}-2 \mathbf{k}$ and $\mathbf{v}=2 \mathbf{i}+\mathbf{j} .$ Find a) $$|\mathbf{u}+\mathbf{v}|$$ b) $$|\mathbf{u}|+|\mathbf{v}|$$ c) $$|-3 u|+|3 v|$$ d) $$\frac{1}{|\mathbf{u}|} \mathbf{u}$$ e) $\left|\frac{1}{|\mathbf{u}|} \mathbf{u}\right|$

Short Answer

Expert verified

a) $ \sqrt{29} $, b) $ \sqrt{14} + \sqrt{5} $, c) $ 3\sqrt{14} + 3\sqrt{5} $, d) $ \frac{1}{\sqrt{14}}(\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}) $, e) 1

Step by step solution

Find $ \mathbf{u} + \mathbf{v} $

Add the corresponding components of vectors $ \mathbf{u} = \mathbf{i} + 3\mathbf{j} - 2\mathbf{k} $ and $ \mathbf{v} = 2\mathbf{i} + \mathbf{j} $: $ \mathbf{u} + \mathbf{v} = (1+2)\mathbf{i} + (3+1)\mathbf{j} + (-2+0)\mathbf{k} = 3\mathbf{i} + 4\mathbf{j} - 2\mathbf{k} $.

Compute $ |\mathbf{u} + \mathbf{v}| $

Use the magnitude formula: $ |\mathbf{u} + \mathbf{v}| = \sqrt{3^2 + 4^2 + (-2)^2} = \sqrt{9 + 16 + 4} = \sqrt{29} $.

Find $ |\mathbf{u}| $ and $ |\mathbf{v}| $

Compute each separately. For $ \mathbf{u} $, the magnitude is $ |\mathbf{u}| = \sqrt{1^2 + 3^2 + (-2)^2} = \sqrt{1 + 9 + 4} = \sqrt{14} $. For $ \mathbf{v} $, the magnitude is $ |\mathbf{v}| = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} $.

Compute $ |\mathbf{u}| + |\mathbf{v}| $

Add the magnitudes found in Step 3: $ |\mathbf{u}| + |\mathbf{v}| = \sqrt{14} + \sqrt{5} $.

Compute $ |-3\mathbf{u}| $ and $ |3\mathbf{v}| $

For $-3\mathbf{u}$, the magnitude is $ |-3|\cdot|\mathbf{u}| = 3\sqrt{14} $ since magnitude is always positive. Similarly, $ |3\mathbf{v}| = 3|\mathbf{v}| = 3\sqrt{5} $.

Find $ |-3\mathbf{u}| + |3\mathbf{v}| $

Add the results from Step 5: $ 3\sqrt{14} + 3\sqrt{5} $.

Compute $ \frac{1}{|\mathbf{u}|} \mathbf{u} $

This is the unit vector in the direction of $ \mathbf{u} $. Divide each component of $ \mathbf{u} $ by $ |\mathbf{u}| $: $ \frac{1}{\sqrt{14}}(\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}) $.

Compute $ \left|\frac{1}{|\mathbf{u}|}\mathbf{u}\right| $

The magnitude of a unit vector is always 1: $ \left|\frac{1}{|\mathbf{u}|}\mathbf{u}\right| = 1 $.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Addition

Vector addition involves combining two or more vectors to create a new vector. This is typically achieved by adding corresponding components of the vectors together. For example, if you have vectors $ \mathbf{u} = a_1 \mathbf{i} + b_1 \mathbf{j} + c_1 \mathbf{k} $ and $ \mathbf{v} = a_2 \mathbf{i} + b_2 \mathbf{j} + c_2 \mathbf{k} $, their sum, $ \mathbf{u} + \mathbf{v} $, would be

$(a_1 + a_2) \mathbf{i} + (b_1 + b_2) \mathbf{j} + (c_1 + c_2) \mathbf{k}$.

This step-wise addition allows us to handle multiple dimensions easily. In practice, always ensure that vectors are expressed in the same coordinate system before adding them.

Magnitude of a Vector

The magnitude of a vector, often referred to as the length or norm, is a measure of how long the vector is. For a three-dimensional vector $ \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} $, its magnitude is calculated using the formula:

$ |\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} $.

This formula comes from the application of the Pythagorean theorem in three dimensions. It is a crucial concept because it provides a scalar quantity that represents the vector's size irrespective of its direction. Magnitude is always a non-negative value.

Unit Vector

A unit vector is simply a vector that has a magnitude of 1. It is often used to indicate direction only since it has been normalized from any original length. To find a unit vector $ \mathbf{u} $ in the direction of a given vector $ \mathbf{a} $, divide every component of $ \mathbf{a} $ by its magnitude $ |\mathbf{a}| $. This can be expressed as

$ \mathbf{u} = \frac{1}{|\mathbf{a}|} \mathbf{a} = \frac{1}{|\mathbf{a}|}(a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k}) $.

Unit vectors are vital in physics and engineering because they simplify calculations involving vector direction, making them easier to interpret and apply. This concept ensures that any vector can be converted to a purely directional form.

Scalar Multiplication

Scalar multiplication involves scaling a vector by a scalar value $ c $. When a vector $ \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} $ is multiplied by a scalar $ c $, the result is

$ c\mathbf{a} = (ca_1)\mathbf{i} + (ca_2)\mathbf{j} + (ca_3)\mathbf{k} $.

This operation either stretches or shrinks the vector depending on whether $ c $ is greater than 1, less than 1, or negative. If $ c $ is negative, it will also reverse the direction of the vector. Scalar multiplication maintains the vector's direction when $ c $ is positive, illustrating the scalar's effect as a simple rescaling of the vector's magnitude without changing its original direction.

91影视

Let \(\mathbf{u}=\mathbf{i}+3 \mathbf{j}-2 \mathbf{k}\) and \(\mathbf{v}=2 \mathbf{i}+\mathbf{j} .\) Find a) $$|\mathbf{u}+\mathbf{v}|$$ b) $$|\mathbf{u}|+|\mathbf{v}|$$ c) $$|-3 u|+|3 v|$$ d) $$\frac{1}{|\mathbf{u}|} \mathbf{u}$$ e) \(\left|\frac{1}{|\mathbf{u}|} \mathbf{u}\right|\)

Short Answer

Step by step solution

Find \( \mathbf{u} + \mathbf{v} \)

Compute \( |\mathbf{u} + \mathbf{v}| \)

Find \( |\mathbf{u}| \) and \( |\mathbf{v}| \)

Compute \( |\mathbf{u}| + |\mathbf{v}| \)

Compute \( |-3\mathbf{u}| \) and \( |3\mathbf{v}| \)

Find \( |-3\mathbf{u}| + |3\mathbf{v}| \)

Compute \( \frac{1}{|\mathbf{u}|} \mathbf{u} \)

Compute \( \left|\frac{1}{|\mathbf{u}|}\mathbf{u}\right| \)

Key Concepts

Vector Addition

Magnitude of a Vector

Unit Vector

Scalar Multiplication

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