Finding the Angle Between Vectors Summing to Zero

The problem provides three vectors, \({\rm{\vec a}},{\rm{\;\vec b}}\) and \({\rm{\vec c}}\), with specific magnitudes: \(\left| {{\rm{\vec a}}} \right| = 10\), \({\rm{\;}}\left| {{\rm{\vec b}}} \right| = 6\) , and \({\rm{\;}}\left| {{\rm{\vec c}}} \right| = 14\). We are also given the condition that their sum is the zero vector: \({\rm{\vec a}} + {\rm{\;\vec b}} + {\rm{\;\vec c}} = \vec 0\). We need to find the angle between vector \({\rm{\vec a}}\) and vector \({\rm{\vec b}}\).

Using Vector Properties to Find the Angle

The condition \({\rm{\vec a}} + {\rm{\;\vec b}} + {\rm{\;\vec c}} = \vec 0\) can be rearranged to isolate the vectors we are interested in:

Now, let's take the dot product of both sides of this equation with themselves. The dot product of a vector with itself gives the square of its magnitude:

The magnitude squared of the sum of two vectors \({\rm{\vec a}}\) and \({\rm{\vec b}}\) is related to their individual magnitudes and the angle between them. If \(\theta\) is the angle between \({\rm{\vec a}}\) and \({\rm{\vec b}}\), the formula is:

Substituting this into our equation \(\left| {\rm{\vec a}} + {\rm{\;\vec b}} \right|^2 = \left| {\rm{\vec c}} \right|^2\), we get:

Now, we substitute the given magnitudes:

• \(\left| {\rm{\vec a}} \right| = 10\)
• \(\left| {\rm{\vec b}} \right| = 6\)
• \(\left| {\rm{\vec c}} \right| = 14\)

Plugging these values into the equation:

Now, we solve for \(\cos \theta\):

To find the angle \(\theta\), we take the inverse cosine:

The angle whose cosine is \(\frac{1}{2}\) is \(60^\circ\).

Thus, the angle between vector \({\rm{\vec a}}\) and vector \({\rm{\vec b}}\) is \(60^\circ\).

Revision Table: Key Concepts

Here's a quick summary of the concepts used:

• Vector addition: \({\rm{\vec a}} + {\rm{\;\vec b}} + {\rm{\;\vec c}} = \vec 0\) implies that the vectors can form a closed triangle when placed head-to-tail.
• Magnitude of a vector sum: \(\left| {\rm{\vec u}} + {\rm{\;\vec v}} \right|^2 = \left| {\rm{\vec u}} \right|^2 + \left| {\rm{\vec v}} \right|^2 + 2 \left| {\rm{\vec u}} \right| \left| {\rm{\vec v}} \right| \cos \phi\), where \(\phi\) is the angle between \({\rm{\vec u}}\) and \({\rm{\vec v}}\).
• Dot product: \({\rm{\vec u}} \cdot {\rm{\;\vec v}} = \left| {\rm{\vec u}} \right| \left| {\rm{\vec v}} \right| \cos \phi\). Using \(\left| {\rm{\vec u}} + {\rm{\;\vec v}} \right|^2 = \left( {\rm{\vec u}} + {\rm{\;\vec v}} \right) \cdot \left( {\rm{\vec u}} + {\rm{\;\vec v}} \right) = \left| {\rm{\vec u}} \right|^2 + \left| {\rm{\vec v}} \right|^2 + 2 {\rm{\vec u}} \cdot {\rm{\;\vec v}}\) yields the same result.

Additional Information on Vector Sum and Triangles

When three vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) satisfy \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\), they form the sides of a triangle (provided none are zero and they are not all parallel). The lengths of the sides of this triangle are the magnitudes \(|\vec{a}|\), \(|\vec{b}|\), and \(|\vec{c}|\). The angle between the vectors \(\vec{a}\) and \(\vec{b}\) when placed tail-to-tail is \(\theta\). In the triangle formed by placing them head-to-tail (e.g., \(\vec{a}\) followed by \(\vec{b}\), with \(-\vec{c}\) closing the loop), the angle between the side corresponding to \(|\vec{a}|\) and the side corresponding to \(|\vec{b}|\) (which is the angle opposite the side \(|\vec{c}|\)) is \(180^\circ - \theta\). This is because the direction of \(\vec{b}\) relative to \(\vec{a}\) in the triangle involves the internal angle, while the vector definition uses the angle between their original directions. The method used above, based on the magnitude of the resultant \(|\vec{a}+\vec{b}|\), correctly uses the angle \(\theta\) between \(\vec{a}\) and \(\vec{b}\) when placed tail-to-tail.

The Law of Cosines can also be applied to the triangle formed by the vectors. If we use the sides \(a=|\vec{a}|\), \(b=|\vec{b}|\), \(c=|\vec{c}|\), and \(\gamma\) is the internal angle opposite side \(c\), the Law of Cosines states \(c^2 = a^2 + b^2 - 2ab \cos \gamma\). Plugging in the values, \(14^2 = 10^2 + 6^2 - 2(10)(6) \cos \gamma\), which gives \(196 = 100 + 36 - 120 \cos \gamma\), leading to \(60 = -120 \cos \gamma\), so \(\cos \gamma = -1/2\). This means \(\gamma = 120^\circ\). The angle \(\theta\) between vectors \(\vec{a}\) and \(\vec{b}\) when placed tail-to-tail is supplementary to the internal angle \(\gamma\) opposite \(\vec{c}\) in the triangle formed by \(\vec{a}, \vec{b}, \vec{c}\) (specifically, if you draw \(\vec{a}\), then \(\vec{b}\) from the head of \(\vec{a}\), the angle between \(\vec{a}\) and \(\vec{b}\) in the triangle is \(180^\circ - \theta\)). So, \(\theta = 180^\circ - \gamma = 180^\circ - 120^\circ = 60^\circ\). Both methods yield the same result for the angle between the vectors \(\vec{a}\) and \(\vec{b}\).