Question

\(\mathrm{A}^{-}+\mathrm{B}^{-}\) is perpendicular to \(\mathrm{A}^{-}\) and \(\left|\mathrm{B}^{-}\right|=2\left|\mathrm{~A}^{-}+\mathrm{B}^{-}\right|\) What is the angle between \(\mathrm{A}^{-}\) and \(\mathrm{B}^{\rightarrow}\) \((\mathrm{A})(\pi / 6)\) (B) \((5 \pi / 6)\) (C) \((2 \pi / 3)\) (D) \((\pi / 3)\)

Short answer

The angle between \(\mathrm{A}^{-}\) and \(\mathrm{B}^{\rightarrow}\) is \(\boxed{(D) \frac{\pi}{3}}\).

Step-by-step solution

Find the dot product

Since \(\mathrm{A}^{-}+\mathrm{B}^{-}\) is perpendicular to \(\mathrm{A}^{-}\), their dot product is equal to 0. Thus, we can write: \((\mathrm{A}^{-}+\mathrm{B}^{-})\cdot \mathrm{A}^{-}=0\)

Expand the dot product

We can expand the dot product as: \(\mathrm{A}^{-}\cdot\mathrm{A}^{-}+\mathrm{B}^{-}\cdot\mathrm{A}^{-}=0\)

Use the magnitude relationship

Now, we use the relationship between the magnitudes of \(\mathrm{B}^{-}\) and \(\mathrm{A}^{-}+\mathrm{B}^{-}\) which is \(\left|\mathrm{B}^{-}\right|=2\left|\mathrm{A}^{-}+\mathrm{B}^{-}\right|\). Square both sides: \begin{align*} \left|\mathrm{B}^{-}\right|^2 &= 4\left|\mathrm{A}^{-}+\mathrm{B}^{-}\right|^2 \\ \left(\mathrm{B}^{-}\cdot\mathrm{B}^{-}\right) &= 4\left(\mathrm{A}^{-}+\mathrm{B}^{-}\right)\cdot\left(\mathrm{A}^{-}+\mathrm{B}^{-}\right) \end{align*}

Expand and simplify

Expand and simplify the equation above: \begin{align*} \mathrm{B}^{-}\cdot\mathrm{B}^{-}&=4\left(\mathrm{A}^{-}\cdot\mathrm{A}^{-}+2\mathrm{A}^{-}\cdot\mathrm{B}^{-}+\mathrm{B}^{-}\cdot\mathrm{B}^{-}\right) \\ \mathrm{B}^{-}\cdot\mathrm{B}^{-}&=4\mathrm{A}^{-}\cdot\mathrm{A}^{-}+8\mathrm{A}^{-}\cdot\mathrm{B}^{-}+4\mathrm{B}^{-}\cdot\mathrm{B}^{-} \\ 3\mathrm{B}^{-}\cdot\mathrm{B}^{-}&=4\mathrm{A}^{-}\cdot\mathrm{A}^{-}+8\mathrm{A}^{-}\cdot\mathrm{B}^{-} \end{align*}

Dot product of B with A

We can express the dot product of \(\mathrm{B}^{-}\) with \(\mathrm{A}^{-}\) in terms of the previous result: \(\mathrm{B}^{-}\cdot\mathrm{A}^{-}=-\frac{1}{2}\mathrm{A}^{-}\cdot\mathrm{A}^{-}\)

Cosine of the angle between A and B

Now, let \(\theta\) be the angle between \(\mathrm{A}^{-}\) and \(\mathrm{B}^{\rightarrow}\) (not \(\mathrm{B}^{-}\)). Recall that the dot product can be expressed in terms of the angle between two vectors and their magnitudes. So: \begin{align*} \mathrm{A}^{-}\cdot\mathrm{B}^{-}&=\left|\mathrm{A}^{-}\right|\left|\mathrm{B}^{\rightarrow}\right|\cos(\pi-\theta) \\ -\frac{1}{2}\mathrm{A}^{-}\cdot\mathrm{A}^{-}&=\left|\mathrm{A}^{-}\right|\left|\mathrm{B}^{-}\right|\cos(\pi-\theta) \end{align*}

Solve for the angle

Finally, we solve for the angle \(\theta\) by dividing both sides by \(\left|\mathrm{A}^{-}\right|\left|\mathrm{B}^{-}\right|\). Recall that \(\cos(\pi-\theta)=-\cos(\theta)\). So: \begin{align*} \cos(\theta)&=\frac{1}{2} \\ \theta&=\frac{\pi}{3} \end{align*} Thus, the angle between \(\mathrm{A}^{-}\) and \(\mathrm{B}^{\rightarrow}\) is \(\boxed{(D) \frac{\pi}{3}}\).

Key concepts

Dot Product

The dot product, also known as the scalar product, is an essential concept when dealing with vectors. It involves two vectors and results in a scalar quantity. You calculate the dot product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) using the formula: \[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos(\theta) \] where \( |\mathbf{A}| \) and \( |\mathbf{B}| \) are the magnitudes of vectors \( \mathbf{A} \) and \( \mathbf{B} \), and \( \theta \) is the angle between them.
For example, when two vectors are said to be perpendicular, this means their dot product is zero. This suggests that the angle between them is \( 90^\circ \) or \( \frac{\pi}{2} \) radians, a situation which played a crucial role in our original exercise.

Thus, the step indicating "\( (\mathrm{A}+\mathbf{B}) \cdot \mathrm{A} = 0 \)" corresponded to verifying the perpendicularity condition, as "plus \( \mathrm{B} \)" is orthogonal to \( \mathrm{A} \). Understanding dot products thus helps decipher how vectors interact in space.

Magnitude of Vectors

The magnitude of a vector gives an idea about the length or size of the vector. You can think of it as the "hypotenuse" in a right triangle formed by the vector plots on a coordinate plane.
Mathematically, if \( \mathbf{A} \) is a vector expressed as \( \mathbf{A} = (x, y, z) \), its magnitude is calculated using the formula: \[ |\mathbf{A}| = \sqrt{x^2 + y^2 + z^2} \]
In the context of the exercise, we had a unique condition: \( |\mathrm{B}| = 2|\mathrm{A} + \mathrm{B}| \). Squaring both sides and using the properties of dot products helped simplify the relationships among vector magnitudes, supporting further steps in solving for the angle \( \theta \).
Magnitude plays a crucial role because it helps compute operations like the dot product, ultimately linking back to understanding how vectors relate spatially.

Cosine Rule

The cosine rule is a vital tool that links the sides of a triangle to its angles, making it quite useful in vector analysis. It states: For a triangle with sides \( a \), \( b \), and \( c \) opposite respective angles \( A \), \( B \), and \( C \), we have: \[ c^2 = a^2 + b^2 - 2ab \cos(C) \]
However, this relationship is directly applied in vector calculations. When dealing with vectors, if you consider \( \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos(\theta) \), this echoes the periodic use of the cosine rule, wherein the angle \( \theta \) directly correlates with the dot product output.
In the exercise, finding \( \theta \) involved recognizing the transformation \( \cos(\pi - \theta) = -\cos(\theta) \). This knowledge allowed us to convert our symbolic findings into a precise angle measurement, linking directly to the given vector conditions.

Perpendicular Vectors

Perpendicular vectors are central to understanding dimensions of space and calculating angles. When two vectors are perpendicular to each other, their dot product equals zero. This concept is fundamental, much like understanding that lines in a plane form right angles when perpendicular.
In our specific problem, \( \mathrm{A} + \mathrm{B} \) was perpendicular to \( \mathrm{A} \), hence \( (\mathrm{A} + \mathrm{B}) \cdot \mathrm{A} = 0 \). This provided a foundational relationship that guided through finding other equations needed for the solution.
The perpendicularity of vectors not only helps in affirming conditions in complex problems but also guides using these insights about geometrical and spatial representations to solve for unknowns like angles. This fact shows why perpendicular vectors are pivotal in vector algebra and beyond.