Understanding the Vector Dot Product Problem

The question asks for the condition under which the equation \((\vec a + \vec b) \cdot (\vec a + \vec b) = {\left| {\vec a} \right|^2} + {\left| {\vec b} \right|^2}\) holds, given that \(|\vec a| \ne 0\) and \(|\vec b| \ne 0\). This involves understanding the properties of the vector dot product.

The dot product of a vector with itself gives the square of its magnitude. That is, \(\vec v \cdot \vec v = |\vec v|^2\). The dot product is also commutative, meaning \(\vec a \cdot \vec b = \vec b \cdot \vec a\), and distributive, meaning \(\vec u \cdot (\vec v + \vec w) = \vec u \cdot \vec v + \vec u \cdot \vec w\).

Solving the Vector Equation

Let's expand the left side of the given equation using the properties of the dot product:

The left side is \((\vec a + \vec b) \cdot (\vec a + \vec b)\).

Using the distributive property:

\((\vec a + \vec b) \cdot (\vec a + \vec b) = \vec a \cdot (\vec a + \vec b) + \vec b \cdot (\vec a + \vec b)\)

\(= \vec a \cdot \vec a + \vec a \cdot \vec b + \vec b \cdot \vec a + \vec b \cdot \vec b\)

Using the property \(\vec v \cdot \vec v = |\vec v|^2\) and the commutative property \(\vec a \cdot \vec b = \vec b \cdot \vec a\):

\(= |\vec a|^2 + \vec a \cdot \vec b + \vec a \cdot \vec b + |\vec b|^2\)

\(= |\vec a|^2 + 2(\vec a \cdot \vec b) + |\vec b|^2\)

Now, we set this equal to the right side of the given equation:

\(|\vec a|^2 + 2(\vec a \cdot \vec b) + |\vec b|^2 = |\vec a|^2 + |\vec b|^2\)

To find the condition, we subtract \(|\vec a|^2 + |\vec b|^2\) from both sides of the equation:

\(|\vec a|^2 + 2(\vec a \cdot \vec b) + |\vec b|^2 - (|\vec a|^2 + |\vec b|^2) = (|\vec a|^2 + |\vec b|^2) - (|\vec a|^2 + |\vec b|^2)\)

\(2(\vec a \cdot \vec b) = 0\)

Dividing by 2 (since 2 is a non-zero scalar):

\(\vec a \cdot \vec b = 0\)

Interpreting the Condition \(\vec a \cdot \vec b = 0\)

The dot product of two non-zero vectors \(\vec a\) and \(\vec b\) is defined as \(\vec a \cdot \vec b = |\vec a| |\vec b| \cos \theta\), where \(\theta\) is the angle between the two vectors.

We found that the condition for the equation to hold is \(\vec a \cdot \vec b = 0\). Substituting the definition:

\(|\vec a| |\vec b| \cos \theta = 0\)

The question states that \(|\vec a| \ne 0\) and \(|\vec b| \ne 0\). Therefore, for the product \(|\vec a| |\vec b| \cos \theta\) to be zero, the only possibility is that \(\cos \theta = 0\).

The angle \(\theta\) between two vectors such that \(\cos \theta = 0\) is \(90^\circ\) (or \(\frac{\pi}{2}\) radians). This means the vectors \(\vec a\) and \(\vec b\) are perpendicular to each other.

Conclusion

The equation \((\vec a + \vec b) \cdot (\vec a + \vec b) = {\left| {\vec a} \right|^2} + {\left| {\vec b} \right|^2}\) holds if and only if the dot product \(\vec a \cdot \vec b\) is zero. Given that both vectors are non-zero, a zero dot product implies that the angle between them is \(90^\circ\), meaning they are perpendicular.

Therefore, the condition is that \(\vec a\) and \(\vec b\) are perpendicular.

Revision Table: Vector Dot Product

Additional Information: Properties of Dot Product

The dot product is a fundamental operation in vector algebra with several important properties:

• Commutative Property: \(\vec a \cdot \vec b = \vec b \cdot \vec a\). The order of the vectors in the dot product does not matter.
• Distributive Property: \(\vec a \cdot (\vec b + \vec c) = \vec a \cdot \vec b + \vec a \cdot \vec c\). The dot product distributes over vector addition.
• Scalar Multiplication Property: \((k \vec a) \cdot \vec b = k (\vec a \cdot \vec b) = \vec a \cdot (k \vec b)\), where \(k\) is a scalar. A scalar factor can be moved outside the dot product.
• Dot Product with Zero Vector: \(\vec a \cdot \vec 0 = 0\). The dot product of any vector with the zero vector is zero.
• Dot Product in Component Form: If \(\vec a = a_x \hat i + a_y \hat j + a_z \hat k\) and \(\vec b = b_x \hat i + b_y \hat j + b_z \hat k\), then \(\vec a \cdot \vec b = a_x b_x + a_y b_y + a_z b_z\).

These properties are crucial for manipulating and solving vector equations like the one in this problem.