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Ch. 4 - Laws of Sines and Cosines; Vectors
Blitzer - Trigonometry 3rd Edition
Blitzer3rd EditionTrigonometryISBN: 9780137316601Not the one you use?Change textbook
All textbooksBlitzer 3rd EditionCh. 4 - Laws of Sines and Cosines; VectorsProblem 55
Chapter 4, Problem 55

In Exercises 53–56, let u = -2i + 3j, v = 6i - j, w = -3i. Find each specified vector or scalar. ||u + v||² - ||u - v||²

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First, recall the formula for the magnitude squared of a vector: for any vector \( \mathbf{a} \), \( ||\mathbf{a}||^2 = \mathbf{a} \cdot \mathbf{a} \), where \( \cdot \) denotes the dot product.
Express the vectors \( \mathbf{u} + \mathbf{v} \) and \( \mathbf{u} - \mathbf{v} \) explicitly by adding and subtracting their components: \( \mathbf{u} + \mathbf{v} = (-2 + 6)\mathbf{i} + (3 - 1)\mathbf{j} \) and \( \mathbf{u} - \mathbf{v} = (-2 - 6)\mathbf{i} + (3 + 1)\mathbf{j} \).
Calculate \( ||\mathbf{u} + \mathbf{v}||^2 = (\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) \) by taking the dot product of \( \mathbf{u} + \mathbf{v} \) with itself.
Similarly, calculate \( ||\mathbf{u} - \mathbf{v}||^2 = (\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v}) \) by taking the dot product of \( \mathbf{u} - \mathbf{v} \) with itself.
Finally, subtract the two results to find \( ||\mathbf{u} + \mathbf{v}||^2 - ||\mathbf{u} - \mathbf{v}||^2 \). You can also use the identity \( ||\mathbf{u} + \mathbf{v}||^2 - ||\mathbf{u} - \mathbf{v}||^2 = 4(\mathbf{u} \cdot \mathbf{v}) \) to simplify the calculation.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Vector Addition and Subtraction

Vector addition combines corresponding components of two vectors to form a new vector, while subtraction finds the difference between components. Understanding how to add and subtract vectors component-wise is essential for calculating expressions like u + v and u - v.
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Adding Vectors Geometrically

Vector Norm (Magnitude)

The norm or magnitude of a vector is the length of the vector in space, calculated as the square root of the sum of the squares of its components. For a vector a = xi + yj, ||a|| = √(x² + y²). Squared norms (||a||²) simplify calculations by avoiding the square root.
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Finding Magnitude of a Vector

Properties of Dot Product and Norms

The dot product relates to vector magnitudes and angles, and satisfies the identity ||u + v||² = ||u||² + 2(u·v) + ||v||². Using this, the expression ||u + v||² - ||u - v||² simplifies to 4(u·v), linking vector norms and dot products for efficient computation.
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Introduction to Dot Product
Related Practice
Textbook Question

Find the magnitude ||v||, to the nearest hundredth, and the direction angle θ, to the nearest tenth of a degree, for each given vector v.


v = 2i - 8j

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Textbook Question

In Exercises 61–64, find the magnitude ||v||, to the nearest hundredth, and the direction angle θ, to the nearest tenth of a degree, for each given vector v. v = -10i + 15j

Textbook Question

Write the vector v in terms of i and j whose magnitude ||v|| and direction angle θ are given.


||v|| = 10, θ = 330°

Textbook Question

In Exercises 47–52, write the vector v in terms of i and j whose magnitude ||v|| and direction angle θ are given. ||v|| = 1/2, θ = 113°

Textbook Question

In Exercises 61–64, find the magnitude ||v||, to the nearest hundredth, and the direction angle θ, to the nearest tenth of a degree, for each given vector v. v = (4i - 2j) - (4i - 8j)

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Textbook Question
In Exercises 53–56, letu = -2i + 3j, v = 6i - j, w = -3i.Find each specified vector or scalar.4u - (2v - w)