Question

find the vector in \(\mathrm{R}^{2}\) such as perpendicular with \(\underline{\mathrm{x}}=(3,4)\) as well as acute angle with Y-axes. (a) \([(4 / 5),(3 / 5)]\) (b) \([-(4 / 5),(3 / 5)]\) (c) \([-(4 / 5),-(3 / 5)]\) (d) \([(3 / 5),-(4 / 5)]\)

Short answer

Option (b) $[-(\frac{4}{5}),(\frac{3}{5})]$ is the correct answer, as it has the correct form and positive $b$ component, which creates an acute angle with the Y-axis.

Step-by-step solution

Find the perpendicular vector

To find the vector that is perpendicular to \(\underline{\mathrm{x}}=(3,4)\), we can use the fact that the dot product of two perpendicular vectors is equal to zero. Let \(\underline{\mathrm{y}}=(a, b)\) be the perpendicular vector, we have: \((3,4) \cdot (a,b) = 3a + 4b = 0\)

Express one component in terms of other

From the equation in step 1, we have: \(3a + 4b = 0\) Rearranging it for one variable, we can get for example the expression of \(a\) in terms of \(b\): \(a = -\frac{4}{3}b\)

Look for options with correct vector and acute angle

Now the vector looks like the form \((-\frac{4}{3}b, b)\), so let's see which of the given options have this form as well as create an acute angle with the Y-axis. An acute angle with the Y-axis means that the \(b\) component must be positive. (a) \([(\frac{4}{5}), (\frac{3}{5})]\) (b) \([-(\frac{4}{5}), (\frac{3}{5})]\) (c) \([-(\frac{4}{5}), -(\frac{3}{5})]\) (d) \([(\frac{3}{5}), -(\frac{4}{5})]\)

Identify the correct option

Only option (b) has the correct form and positive \(b\) component which create an acute angle with the Y-axis. So the correct answer is: \(\underline{\mathrm{y}} = [-(\frac{4}{5}), (\frac{3}{5})]\)

Key concepts

Dot Product

The dot product, also known as the scalar product, is a fundamental operation in vector calculus. It takes two vectors and returns a single number (a scalar), and this value is a direct measure of the vectors' magnitude and the cosine of the angle between them.

For two-dimensional vectors \( \textbf{a} = (a_1, a_2) \) and \( \textbf{b} = (b_1, b_2) \), the dot product is calculated as \( \textbf{a} \cdot \textbf{b} = a_1b_1 + a_2b_2 \). If this value is zero, the vectors are said to be perpendicular or orthogonal to each other, meaning they form a right angle.

In the context of the exercise, finding the vector in \(\mathbb{R}^{2}\) that is perpendicular to \( \textbf{x} = (3,4) \) means we're looking for a second vector whose dot product with \( \textbf{x} \) is zero. This leads to the equation \( 3a + 4b = 0 \) which must hold true for the vectors to be perpendicular.

Vector Components

Each vector in a space is described by its components, which are projections onto the space's axes. In two-dimensional space, \(\mathbb{R}^{2}\), a vector \( \textbf{v} \) is commonly written as \( \textbf{v} = (v_x, v_y) \), where \( v_x \) and \( v_y \) are the vector components along the x and y axis, respectively.

Understanding vector components is crucial when dealing with geometric problems, like finding perpendicular vectors. Components give us a sense of direction and magnitude. In our exercise, we solved for one variable in terms of the other (\( a = -\frac{4}{3}b \) ), converting the perpendicular condition into an equation using components. This interdependence of components in the example reflects the intrinsic geometric properties of vectors and how altering one component impacts the vector's direction.

Acute Angle with Axis

When we mention an acute angle with an axis, we refer to the angle that a vector makes with that axis being less than 90 degrees. It's important when dealing with the orientation of vectors in two-dimensional space.

Specifically, for a vector to form an acute angle with the y-axis, its y-component must be positive. This results from the way angles are measured from the positive direction of the axes. The significance of this in practice, as shown in the exercise, is determining the direction of the vector. If we have a vector with components \( (a, b) \) and \( b > 0 \), the vector will make an acute angle with the y-axis.

Hence, the vector with a positive second component in choice (b) \( [-(\frac{4}{5}), (\frac{3}{5})] \) not only satisfies the perpendicularity condition but also ensures that the vector makes an acute angle with the y-axis. This illustrates a practical application of vector components in defining the orientation of vectors in the plane.