Question

Given vectors \(\mathbf{u}=x \mathbf{i}+2 \mathbf{j}\) and \(\mathbf{v}=7 \mathbf{i}-3 \mathbf{j}\). find \(x\) so that the angle between the vectors is \(30^{\circ} .\)

Short answer

The value of \(x\) can be found by solving \(7x - 6 = \frac{\sqrt{3\cdot 58(x^{2} + 4)}}{2}\).

Step-by-step solution

- Dot Product Formula

The dot product formula of two vectors \(\mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\| \|\mathbf{v}\| \cos(\theta)\) where \(\theta = 30^{\circ}\) can be used to find \(x\).

- Compute the Dot Product

The dot product of vectors \(\mathbf{u} = x \mathbf{i} + 2 \mathbf{j}\) and \(\mathbf{v}=7 \mathbf{i} - 3 \mathbf{j}\) is calculated as follows:\(\mathbf{u} \cdot \mathbf{v} = x \cdot 7 + 2 \cdot (-3) = 7x - 6\).

- Magnitude of \(\mathbf{u}\)

The magnitude of vector \(\mathbf{u}\) is given by:\(\|\mathbf{u}\| = \sqrt{x^{2} + 2^{2}} = \sqrt{x^{2} + 4}\).

- Magnitude of \(\mathbf{v}\)

The magnitude of vector \(\mathbf{v}\) is given by:\(\|\mathbf{v}\| = \sqrt{7^{2} + (-3)^{2}} = \sqrt{49 + 9} = \sqrt{58}\).

- Apply Cosine Angle Relationship

Using the angle \(\theta = 30^{\circ}\) in the dot product formula, we get:\(7x - 6 = \sqrt{x^{2} + 4} \cdot \sqrt{58} \cdot \cos(30^{\circ})\). \(\cos(30^{\circ}) = \frac{\sqrt{3}}{2}\), thus:\(7x - 6 = \sqrt{58(x^{2} + 4)} \cdot \frac{\sqrt{3}}{2}\).

- Solve for \(x\)

Rearrange and solve the equation: \(7x - 6 = \frac{\sqrt{3\cdot 58(x^{2} + 4)}}{2}\). Square both sides to eliminate the square root, solve the resulting quadratic equation for \(x\).

Key concepts

Dot Product

The dot product is a fundamental operation in vector mathematics. It combines two vectors and results in a scalar. For two vectors \(\textbf{a}\) and \(\textbf{b}\), the dot product is calculated as: \[ \textbf{a} \bullet \textbf{b} = a_1b_1 + a_2b_2 \] Here, each component of the first vector is multiplied by the corresponding component of the second vector, then summed up. In our example, the dot product of vectors \(\textbf{u} = x \textbf{i} + 2 \textbf{j}\) and \(\textbf{v} = 7 \textbf{i} - 3 \textbf{j}\) is: \[ \textbf{u} \bullet \textbf{v} = x \bullet 7 + 2 \bullet (-3) = 7x - 6 \] This scalar value describes in some way the projection of one vector onto another.

Vectors

A vector represents a quantity with both direction and magnitude. It is often depicted as an arrow. The direction of the arrow shows the direction of the vector, and the length represents the magnitude. Commonly, vectors are written in terms of their components, like \(\textbf{u} = x \textbf{i} + 2 \textbf{j}\), where \(\textbf{i}\) and \(\textbf{j}\) are the unit vectors in the horizontal and vertical directions, respectively. Here, \(\textbf{u}\) has a horizontal component of \(\textbf{x}\) and a vertical component of \(\textbf{2}\). Remember:

• Vectors are not defined by their location but by their direction and magnitude.
• They are useful in many fields including physics, engineering, and computer science.

Magnitude of a Vector

The magnitude (or length) of a vector can be thought of as the distance from its tail to its head. For a vector \(\textbf{u} = x \textbf{i} + y \textbf{j}\), its magnitude is found using the Pythagorean theorem: \[ \|\textbf{u}\| = \sqrt{x^2 + y^2} \] For our vector \(\textbf{u} = x \textbf{i} + 2 \textbf{j}\), the magnitude is: \[ \|\textbf{u}\| = \sqrt{x^2 + 2^2} = \sqrt{x^2 + 4} \] Similarly, for \(\textbf{v} = 7 \textbf{i} - 3 \textbf{j}\), the magnitude is: \[ \|\textbf{v}\| = \sqrt{7^2 + (-3)^2} = \sqrt{49 + 9} = \sqrt{58} \] The magnitude gives an idea of how 'long' or 'strong' the vector is.

Cosine of an Angle

The cosine of an angle in a right triangle is the ratio of the adjacent side to the hypotenuse. In our context, \(\cos\theta\) is used to relate the angle between two vectors to their dot product and magnitudes: \[ \textbf{u} \bullet \textbf{v} = \|\textbf{u}\| \textbf{v}\| \cos(\theta) \] For \(\theta = 30^{\circ}\), the cosine value is \(\cos(30^{\circ}) = \frac{\sqrt{3}}{2}\). Including this in our dot product equation: \[ 7x - 6 = \sqrt{x^2 + 4} \bullet \sqrt{58} \bullet \frac{\sqrt{3}}{2} \] This will help us find the value of \(\textbf{x}\).

Solving Quadratic Equations

To solve for \(\textbf{x}\), we need to rearrange and solve a quadratic equation. Starting with: \[ 7x - 6 = \frac{\sqrt{3 \bullet 58(x^2 + 4)}}{2} \] Let's simplify this step by step:

• First, clear the fraction by multiplying both sides by 2.
• Square both sides to eliminate the square root.
• Combine like terms and move everything to one side of the equation.
• Use the quadratic formula if needed: \[ x = \frac{-b \pm \textbackslashsqrt{b^2 - 4ac}}{2a} \]

Finally, you will get the possible values for \(\textbf{x}\). This method is crucial for dealing with quadratic problems in various areas of mathematics and science.