Question

the angle between \(\underline{a}\) and \(\underline{b}\) is \((5 \pi / 6)\), projectile of a over \(\underline{b}\) is \((6 / \sqrt{3})\) then the value \(|\underline{\mathrm{a}}|=\underline{(\underline{\mathrm{a}}, \underline{\mathrm{b}}, \neq \underline{0})}\) (a) 12 (b) 6 (c) 4 (d) \((\sqrt{3} / 2)\)

Short answer

The magnitude of vector \(\underline{a}\) is 4 (c).

Step-by-step solution

Formula for projection of a vector

The projection of vector \(\underline{a}\) over \(\underline{b}\) can be found using the formula: \[proj_{\underline{b}}(\underline{a}) = \frac{\underline{a} \cdot \underline{b}}{|\underline{b}|}\]

Angle provided, use projection-appraoch

Since we are given the angle between the vectors, it is easier to use the approach: \[proj_{\underline{b}}(\underline{a}) = |\underline{a}|\cos(\theta)\] We know that the angle between \(\underline{a}\) and \(\underline{b}\) is \((5 \pi / 6)\) and the projection of \(\underline{a}\) over \(\underline{b}\) is \((6 / \sqrt{3})\). We can plug these values into the equation.

Solving for the magnitude of vector \(\underline{a}\)

With the given values of \(\theta\) and the projectile of \(\underline{a}\) over \(\underline{b}\), we can solve for the magnitude of vector \(\underline{a}\). \[ \frac{6}{\sqrt{3}} = |\underline{a}|\cos\left(\frac{5 \pi}{6}\right) \] Since \(\cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2}\), the equation becomes: \[ \frac{6}{\sqrt{3}} = |\underline{a}|\left(-\frac{\sqrt{3}}{2}\right) \]

Isolate |\(\underline{a}\)| and calculate its value

Now, we need to isolate \(|\underline{a}|\) and find its value. Divide both sides of the equation by \((-\frac{\sqrt{3}}{2})\): \[ |\underline{a}| = \frac{6}{\sqrt{3}}\times \frac{-2}{\sqrt{3}} \] \[ |\underline{a}| = -\frac{12}{3} \] Since the magnitude of a vector is always non-negative, we consider only the positive value of the expression: \[ |\underline{a}| = 4 \] So, the magnitude of vector \(\underline{a}\) is 4, which corresponds to the answer (c).

Key concepts

Angle Between Vectors

Understanding the angle between two vectors is key when working with vector projections and relationships. Two vectors, let's call them \( \underline{a} \) and \( \underline{b} \), can form an angle \( \theta \) between them. This angle can be calculated using the dot product of the vectors along with their magnitudes. The formula is:\[ \cos(\theta) = \frac{\underline{a} \cdot \underline{b}}{|\underline{a}| |\underline{b}|} \]By finding \( \cos(\theta) \), you can determine the angle \( \theta \) through trigonometric functions. In our problem, the angle is already given as \((5 \pi / 6)\), which directly influences the projection of one vector onto another. This angle determines the degree to which the vectors are aligned and is crucial for finding projections and solving for other vector properties. Note that a negative cosine indicates an obtuse angle, which affects other calculations, as observed in the given exercise.

Magnitude of a Vector

The magnitude of a vector, often denoted as \(|\underline{a}|\), represents its length or size. It is a scalar quantity that describes how long the vector is, without concern for its direction. To calculate the magnitude of a vector \(\underline{a} = (x, y, z) \) in a 3-dimensional space, use the formula:\[ |\underline{a}| = \sqrt{x^2 + y^2 + z^2} \]In scenarios involving projections or angles between vectors, the magnitude also helps us solve equations by providing insights into the vector projections over other vectors.For instance, the magnitude is used here to find the length of \(\underline{a}\) based on its projection length over another vector \(\underline{b}\). Through the calculation, the projection leads to solving an expression involving \( \cos(\theta) \) times the magnitude, and eventually discovering that the magnitude \(|\underline{a}| = 4\). This is crucial as it validates the solution when analysing vector interactions.

Dot Product

The dot product, also known as the scalar product, describes an algebraic operation between two vectors. It plays a significant role in determining projections, angles, and relationships between vectors. The formula for the dot product of two vectors \( \underline{a} = (a_1, a_2, a_3) \) and \( \underline{b} = (b_1, b_2, b_3) \) is:\[ \underline{a} \cdot \underline{b} = a_1b_1 + a_2b_2 + a_3b_3 \]The dot product yields a scalar that directly relates to the angle between two vectors through:\[ \cos(\theta) = \frac{\underline{a} \cdot \underline{b}}{ |\underline{a}| |\underline{b}|} \]In the problem, knowing \( \underline{a} \cdot \underline{b} \) is vital for understanding how these vectors interact, guiding us in calculating their alignment and projection. The sign and magnitude of the dot product signal whether vectors are pointing in the same direction (positive dot product) or opposite directions (negative dot product). Furthermore, a zero dot product indicates perpendicular vectors. These characteristics underscore the importance of the dot product in vector mathematics.