Problem 76 - Precalculus with Calculus Previews

Question

Find a scalar \(c\) so that the angle between the vectors \(\mathbf{u}=\mathbf{i}+c \mathbf{j}\) and \(\mathbf{v}=\mathbf{i}+\mathbf{j}\) is \(45^{\circ} .\)

Step-by-Step Solution

Answer

The scalar \(c\) is 0.

Step 1 – State the Formula for the Dot Product and Angle between Vectors

The angle \(\theta\) between two vectors \(\mathbf{u}\) and \(\mathbf{v}\) can be found using the formula for the dot product: \[\mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\| \|\mathbf{v}\| \cos(\theta).\] Here, \(\theta = 45^{\circ}\) hence \(\cos(45^{\circ}) = \frac{1}{\sqrt{2}}.\) Our goal is to find \(c\) satisfying this relationship.

Step 2 – Calculate the Dot Product

The dot product of \(\mathbf{u} = \mathbf{i} + c \mathbf{j}\) and \(\mathbf{v} = \mathbf{i} + \mathbf{j}\) is calculated as follows: \[\mathbf{u} \cdot \mathbf{v} = 1 \cdot 1 + c \cdot 1 = 1 + c.\]

Step 3 – Determine the Magnitudes of the Vectors

Calculate the magnitudes of the vectors \(\mathbf{u}\) and \(\mathbf{v}\). For \(\mathbf{u}\), the magnitude is \[\|\mathbf{u}\| = \sqrt{1^2 + c^2} = \sqrt{1 + c^2},\] and for \(\mathbf{v}\), the magnitude is \[\|\mathbf{v}\| = \sqrt{1^2 + 1^2} = \sqrt{2}.\]

Step 4 – Substitute into the Dot Product Formula

Insert the results from previous steps into the dot product equation: \[(1 + c) = \sqrt{1 + c^2} \cdot \sqrt{2} \cdot \frac{1}{\sqrt{2}}.\] Simplifying this, we get \[1 + c = \sqrt{1 + c^2}.\]

Step 5 – Solve the Equation for c

Square both sides to remove the square root: \[(1 + c)^2 = 1 + c^2.\] This expands to \[1 + 2c + c^2 = 1 + c^2.\] Subtract \(1 + c^2\) from both sides to get \[2c = 0.\] Solving, we find \[c = 0.\]

Key Concepts

Dot Product

The dot product, also known as the scalar product, is an operation that takes two vectors and returns a scalar. If you have two vectors \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} \) and \( \mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} \), the dot product is calculated as: \[\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2.\]This operation is fundamental in determining the relationship between two vectors. It helps in finding the angle between them, projecting one vector onto another, and even in mechanics and physics problems.
The dot product measures how much one vector extends in the direction of another. For instance, if two vectors are parallel, the dot product is maximal. If they're perpendicular, the dot product is zero. In our problem, the dot product formula \( \mathbf{u} \cdot \mathbf{v} = \| \mathbf{u} \| \| \mathbf{v} \| \cos(\theta) \) is used to find the angle of \( 45^{\circ} \) between vectors.

Angle Between Vectors

Finding the angle between vectors involves understanding how two vectors relate directionally. The angle \( \theta \) between two vectors \( \mathbf{u} \) and \( \mathbf{v} \) can be derived from their dot product as expressed:\[\cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{u} \| \| \mathbf{v} \|}.\]For our exercise, the angle provided is \( 45^{\circ} \), which simplifies calculations with a known cosine value \( \cos(45^{\circ}) = \frac{1}{\sqrt{2}} \).
This concept tells us about the geometrical relationship between vectors. If the angle is acute (less than 90 degrees), the vectors trend closer together. If it's obtuse (more than 90 degrees), they point away from each other. At \( 90^{\circ} \), they are orthogonal, meaning entirely separate directions spatially.

Scalar Multiplication

Scalar multiplication involves taking a vector and multiplying it by a scalar value (a real number). This operation stretches or shrinks the vector without changing its direction. If a vector \( \mathbf{u} = x \mathbf{i} + y \mathbf{j} \), then multiplying by a scalar \( c \) will transform it to:\[c \mathbf{u} = c x \mathbf{i} + c y \mathbf{j}.\]This concept is essential in involving vectors in real-life applications, where we often scale quantities up or down.
In our problem, the vector \( \mathbf{u} = \mathbf{i} + c \mathbf{j} \) means that \( c \) scales the unit vector along the \( y \)-axis. By solving for \( c \), we adjusted the vector to match the required conditions of the given exercise.

Magnitude of a Vector

The magnitude or length of a vector \( \mathbf{v} = v_1 \mathbf{i} + v_2 \mathbf{j} \) can be calculated using the Pythagorean theorem. Its magnitude \( \| \mathbf{v} \| \) is given by:\[\| \mathbf{v} \| = \sqrt{v_1^2 + v_2^2}.\]The magnitude tells us the "size" of a vector, which is always a positive scalar quantity.
For our vectors \( \mathbf{u} = \mathbf{i} + c \mathbf{j} \) and \( \mathbf{v} = \mathbf{i} + \mathbf{j} \), their magnitudes \( \| \mathbf{u} \| \) and \( \| \mathbf{v} \| \) were calculated to find the relation needed to determine \( c \). This concept helps us determine distances and scalar projections in various fields like physics, engineering, and computer graphics.