Understanding the Vector Relationship

We are given three vectors, a, b, and c, with the condition that their sum is the zero vector:

\[ a + b + c = 0 \]

We are also provided with information about the magnitudes of these vectors:

• Vector a is a unit vector, so its magnitude is \( |a| = 1 \).
• Vector b is a unit vector, so its magnitude is \( |b| = 1 \).
• The magnitude of vector c is given as \( |c| = 2 \).

Our goal is to find the angle between vectors b and c.

Manipulating the Vector Equation

From the given vector equation \( a + b + c = 0 \), we can rearrange it to isolate one of the vectors. Let's move vector a to the other side:

\[ a = -b - c \]

This equation can also be written as:

\[ a = -(b + c) \]

Using Magnitudes and Dot Product

Now, let's take the magnitude of both sides of the equation \( a = -(b + c) \). We know that the magnitude of a negative vector is the same as the magnitude of the positive vector, i.e., \( |-v| = |v| \).

\[ |a| = |-(b + c)| = |b + c| \]

We are given that \( |a| = 1 \), so:

\[ 1 = |b + c| \]

To get rid of the magnitude and involve the dot product (which relates to the angle between vectors), we can square both sides of the equation:

\[ 1^2 = |b + c|^2 \] \[ 1 = |b + c|^2 \]

Recall the property that for any vectors x and y, \( |x + y|^2 = |x|^2 + |y|^2 + 2(x \cdot y) \). Applying this to \( |b + c|^2 \):

\[ |b + c|^2 = |b|^2 + |c|^2 + 2(b \cdot c) \]

Substitute the known magnitudes \( |b| = 1 \) and \( |c| = 2 \) into this equation:

\[ |b + c|^2 = (1)^2 + (2)^2 + 2(b \cdot c) \] \[ |b + c|^2 = 1 + 4 + 2(b \cdot c) \] \[ |b + c|^2 = 5 + 2(b \cdot c) \]

Now, substitute the result from squaring the magnitude \( |b + c| \), which was 1:

\[ 1 = 5 + 2(b \cdot c) \]

Calculating the Dot Product

Solve the equation for the dot product \( b \cdot c \):

\[ 1 - 5 = 2(b \cdot c) \] \[ -4 = 2(b \cdot c) \] \[ b \cdot c = \frac{-4}{2} \] \[ b \cdot c = -2 \]

Finding the Angle Between Vectors b and c

The dot product of two vectors b and c is also defined as \( b \cdot c = |b| |c| \cos \theta \), where \(\theta\) is the angle between vectors b and c. We know \( b \cdot c = -2 \), \( |b| = 1 \), and \( |c| = 2 \). Substitute these values into the dot product formula:

\[ -2 = (1)(2) \cos \theta \] \[ -2 = 2 \cos \theta \]

Solve for \( \cos \theta \):

\[ \cos \theta = \frac{-2}{2} \] \[ \cos \theta = -1 \]

We need to find the angle \( \theta \) whose cosine is -1. This angle is \( 180^\circ \).

\[ \theta = 180^\circ \]

Thus, the angle between vectors b and c is \( 180^\circ \).

Revision Table: Vector Sum and Magnitudes

Additional Information: Unit Vectors and Vector Algebra

A unit vector is a vector with a magnitude of 1. Unit vectors are often used to indicate direction. For example, in a 3D Cartesian system, the standard unit vectors along the x, y, and z axes are denoted by \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\).

The equation \(a + b + c = 0\) means that the three vectors form a closed triangle when placed head-to-tail, or that they are in equilibrium if representing forces. In this specific case, since \(a = -(b+c)\) and \(|a| = |b+c|\), vectors \(a\) and \((b+c)\) have the same magnitude but opposite directions. The fact that \(b \cdot c = -2\) and \(|b|=1, |c|=2\) led to \(\cos \theta = -1\), which means vectors b and c are in exactly opposite directions. If two vectors are in opposite directions, their sum \(b+c\) would have a magnitude of \(||c| - |b|| = |2-1| = 1\). This aligns with \(|b+c|=1\), which is consistent with the problem statement \(|a|=1\). Since \(a = -(b+c)\) and b and c are opposite, \((b+c)\) points in the direction of the longer vector (c), and \(a\) points in the opposite direction of c. Therefore, a, b, and c are collinear vectors, with a and b opposing c, and a also opposing b.