Solving Vector Sum and Dot Product Problems

The question provides us with three vectors, \({\rm{\vec a}},{\rm{\;\vec b}}\) and \({\rm{\vec c}}\), that satisfy a specific condition: their vector sum is the zero vector. This means \({\rm{\vec a}} + {\rm{\vec b}} + {\rm{\vec c}} = \vec 0\). We are also given the magnitudes of these vectors: \({\rm{\left| {{\rm{\vec a}}} \right| = 10}}\), \({\rm{\left| {{\rm{\vec b}}} \right| = 6}}\), and \({\rm{\left| {{\rm{\vec c}}} \right| = 14}}\). We need to find the value of the expression \({\rm{\vec a}} \cdot {\rm{\vec b}} + {\rm{\vec b}} \cdot {\rm{\vec c}} + {\rm{\vec c}} \cdot {\rm{\vec a}}\), which is the sum of the dot products of the vectors taken in pairs.

Using the Vector Sum Property

The key to solving this problem lies in the given vector sum equation \({\rm{\vec a}} + {\rm{\vec b}} + {\rm{\vec c}} = \vec 0\). A common technique when dealing with vector sums and magnitudes is to square both sides of the equation using the dot product. Squaring a vector equation involves taking the dot product of each side with itself.

Let's start with the given equation:

\({\rm{\vec a}} + {\rm{\vec b}} + {\rm{\vec c}} = \vec 0\)

Taking the dot product of both sides with \(({\rm{\vec a}} + {\rm{\vec b}} + {\rm{\vec c}})\):

\(({\rm{\vec a}} + {\rm{\vec b}} + {\rm{\vec c}}) \cdot ({\rm{\vec a}} + {\rm{\vec b}} + {\rm{\vec c}}) = \vec 0 \cdot \vec 0\)

The dot product of the zero vector with itself is 0, so the right side is 0.

\(({\rm{\vec a}} + {\rm{\vec b}} + {\rm{\vec c}}) \cdot ({\rm{\vec a}} + {\rm{\vec b}} + {\rm{\vec c}}) = 0\)

Expanding the Dot Product

Now, we expand the left side using the distributive property of the dot product. Remember that \(\vec v \cdot \vec v = |\vec v|^2\) and \(\vec u \cdot \vec v = \vec v \cdot \vec u\).

Expansion of \(({\rm{\vec a}} + {\rm{\vec b}} + {\rm{\vec c}}) \cdot ({\rm{\vec a}} + {\rm{\vec b}} + {\rm{\vec c}})\):

\({\rm{\vec a}} \cdot ({\rm{\vec a}} + {\rm{\vec b}} + {\rm{\vec c}}) + {\rm{\vec b}} \cdot ({\rm{\vec a}} + {\rm{\vec b}} + {\rm{\vec c}}) + {\rm{\vec c}} \cdot ({\rm{\vec a}} + {\rm{\vec b}} + {\rm{\vec c}})\)

\(= {\rm{\vec a}} \cdot {\rm{\vec a}} + {\rm{\vec a}} \cdot {\rm{\vec b}} + {\rm{\vec a}} \cdot {\rm{\vec c}} + {\rm{\vec b}} \cdot {\rm{\vec a}} + {\rm{\vec b}} \cdot {\rm{\vec b}} + {\rm{\vec b}} \cdot {\rm{\vec c}} + {\rm{\vec c}} \cdot {\rm{\vec a}} + {\rm{\vec c}} \cdot {\rm{\vec b}} + {\rm{\vec c}} \cdot {\rm{\vec c}}\)

Grouping the terms:

\(({\rm{\vec a}} \cdot {\rm{\vec a}} + {\rm{\vec b}} \cdot {\rm{\vec b}} + {\rm{\vec c}} \cdot {\rm{\vec c}}) + ({\rm{\vec a}} \cdot {\rm{\vec b}} + {\rm{\vec b}} \cdot {\rm{\vec a}}) + ({\rm{\vec b}} \cdot {\rm{\vec c}} + {\rm{\vec c}} \cdot {\rm{\vec b}}) + ({\rm{\vec c}} \cdot {\rm{\vec a}} + {\rm{\vec a}} \cdot {\rm{\vec c}})\)

Using \(\vec v \cdot \vec v = |\vec v|^2\) and \(\vec u \cdot \vec v = \vec v \cdot \vec u\):

\({\rm{\left| {{\rm{\vec a}}} \right|}}^2 + {\rm{\left| {{\rm{\vec b}}} \right|}}^2 + {\rm{\left| {{\rm{\vec c}}} \right|}}^2 + 2({\rm{\vec a}} \cdot {\rm{\vec b}}) + 2({\rm{\vec b}} \cdot {\rm{\vec c}}) + 2({\rm{\vec c}} \cdot {\rm{\vec a}})\)

So, the equation becomes:

\({\rm{\left| {{\rm{\vec a}}} \right|}}^2 + {\rm{\left| {{\rm{\vec b}}} \right|}}^2 + {\rm{\left| {{\rm{\vec c}}} \right|}}^2 + 2({\rm{\vec a}} \cdot {\rm{\vec b}} + {\rm{\vec b}} \cdot {\rm{\vec c}} + {\rm{\vec c}} \cdot {\rm{\vec a}}) = 0\)

Substituting Magnitudes and Solving

We are given the magnitudes: \({\rm{\left| {{\rm{\vec a}}} \right| = 10}}\), \({\rm{\left| {{\rm{\vec b}}} \right| = 6}}\), and \({\rm{\left| {{\rm{\vec c}}} \right| = 14}}\). Substitute these values into the equation:

\(10^2 + 6^2 + 14^2 + 2({\rm{\vec a}} \cdot {\rm{\vec b}} + {\rm{\vec b}} \cdot {\rm{\vec c}} + {\rm{\vec c}} \cdot {\rm{\vec a}}) = 0\)

Calculate the squares:

\(100 + 36 + 196 + 2({\rm{\vec a}} \cdot {\rm{\vec b}} + {\rm{\vec b}} \cdot {\rm{\vec c}} + {\rm{\vec c}} \cdot {\rm{\vec a}}) = 0\)

Sum the numbers:

\(332 + 2({\rm{\vec a}} \cdot {\rm{\vec b}} + {\rm{\vec b}} \cdot {\rm{\vec c}} + {\rm{\vec c}} \cdot {\rm{\vec a}}) = 0\)

Now, isolate the term we want to find:

\(2({\rm{\vec a}} \cdot {\rm{\vec b}} + {\rm{\vec b}} \cdot {\rm{\vec c}} + {\rm{\vec c}} \cdot {\rm{\vec a}}) = -332\)

Divide by 2:

\({\rm{\vec a}} \cdot {\rm{\vec b}} + {\rm{\vec b}} \cdot {\rm{\vec c}} + {\rm{\vec c}} \cdot {\rm{\vec a}} = -{\frac{332}{2}}\)

\({\rm{\vec a}} \cdot {\rm{\vec b}} + {\rm{\vec b}} \cdot {\rm{\vec c}} + {\rm{\vec c}} \cdot {\rm{\vec a}} = -166\)

Thus, the value of \({\rm{\vec a}} \cdot {\rm{\vec b}} + {\rm{\vec b}} \cdot {\rm{\vec c}} + {\rm{\vec c}} \cdot {\rm{\vec a}}\) is -166.

Revision Table: Key Concepts and Steps

Additional Information: Geometric Interpretation of Vector Sum

When three vectors \({\rm{\vec a}},{\rm{\;\vec b}}\) and \({\rm{\vec c}}\) satisfy \({\rm{\vec a}} + {\rm{\vec b}} + {\rm{\vec c}} = \vec 0\), it implies that if you place these vectors head-to-tail, they form a closed triangle. The magnitudes of the vectors then correspond to the lengths of the sides of this triangle. In our case, we have a triangle with side lengths 10, 6, and 14. The sum of the dot products \(\vec a \cdot \vec b + \vec b \cdot \vec c + \vec c \cdot \vec a\) is related to the angles of this triangle.

Recall that the dot product \({\rm{\vec u}} \cdot {\rm{\vec v}} = {\rm{\left| {{\rm{\vec u}}} \right|}}{\rm{\left| {{\rm{\vec v}}} \right|}}\cos\theta\), where \(\theta\) is the angle between the vectors. In the context of \({\rm{\vec a}} + {\rm{\vec b}} + {\rm{\vec c}} = \vec 0\), the angles between the vectors placed head-to-tail are the external angles of the triangle formed by them. The internal angles are supplementary to these external angles.

For example, consider \(\vec a + \vec b = -\vec c\). Squaring this gives \(|\vec a + \vec b|^2 = |-\vec c|^2\), which is \(|\vec a|^2 + |\vec b|^2 + 2\vec a \cdot \vec b = |\vec c|^2\). This is related to the Law of Cosines in trigonometry. Specifically, \(|\vec c|^2 = |\vec a|^2 + |\vec b|^2 - 2|\vec a||\vec b|\cos C\), where C is the angle opposite side c (formed by vectors a and b when placed tail-to-tail). Comparing this with \(|\vec a|^2 + |\vec b|^2 + 2\vec a \cdot \vec b = |\vec c|^2\), we see that \(2\vec a \cdot \vec b = -2|\vec a||\vec b|\cos C\), or \(\vec a \cdot \vec b = -|\vec a||\vec b|\cos C\). The method used in the solution is a more direct algebraic approach for the sum of all three dot products.