Finding Vector Magnitude Using Identity

The problem asks us to find the magnitude of vector \(\vec{a}\), denoted as \(|\vec{a}|\), given a relationship involving the magnitudes of the cross product and dot product of vectors \(\vec{a}\) and \(\vec{b}\), and the magnitude of vector \(\vec{b}\).

We are given the following information:

• The equation: \((\vec{a} \times \vec{b})^2 + (\vec{a} \cdot \vec{b})^2 = 144\)
• The magnitude of vector \(\vec{b}\): \(|\vec{b}| = 4\)

We need to find the value of \(|\vec{a}|\).

Applying the Vector Identity

There is a fundamental vector identity that relates the magnitudes of the cross product and dot product to the magnitudes of the individual vectors:

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

This identity is derived from the definitions of the cross product and dot product:

• \(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta\)
• \(|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta\)

where \(\theta\) is the angle between vectors \(\vec{a}\) and \(\vec{b}\).

Squaring these and adding them gives:

\((\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta\)

\(|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta\)

Summing them:

\(|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta + |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta\)

\(|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 (\cos^2 \theta + \sin^2 \theta)\)

Since \(\cos^2 \theta + \sin^2 \theta = 1\), we get:

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

Solving for \(|\vec{a}|\)

We are given \((\vec{a} \times \vec{b})^2 + (\vec{a} \cdot \vec{b})^2 = 144\). Using the identity, we can substitute 144 for the left side of the identity equation:

\(144 = |\vec{a}|^2 |\vec{b}|^2\)

We are also given that \(|\vec{b}| = 4\). Substitute this value into the equation:

\(144 = |\vec{a}|^2 (4)^2\)

\(144 = |\vec{a}|^2 \times 16\)

Now, we can solve for \(|\vec{a}|^2\):

\(|\vec{a}|^2 = \frac{144}{16}\)

\(|\vec{a}|^2 = 9\)

To find \(|\vec{a}|\), we take the square root of both sides. Since magnitude must be non-negative:

\(|\vec{a}| = \sqrt{9}\)

\(|\vec{a}| = 3\)

Therefore, the value of \(|\vec{a}|\) is 3.

Revision Table: Key Concepts in Vector Magnitude

Additional Information: Properties of Vector Operations

Understanding vector operations is crucial in physics and mathematics. Here are some key properties:

• Dot Product Properties:
  • Commutative: \(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}\)
  • Distributive: \(\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}\)
  • Scalar Multiplication: \((c\vec{a}) \cdot \vec{b} = c(\vec{a} \cdot \vec{b})\)
• Cross Product Properties:
  • Anti-commutative: \(\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})\)
  • Distributive: \(\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}\)
  • Scalar Multiplication: \((c\vec{a}) \times \vec{b} = c(\vec{a} \times \vec{b})\)
  • Cross product of parallel vectors is the zero vector: If \(\vec{a} \parallel \vec{b}\), then \(\vec{a} \times \vec{b} = \vec{0}\) (since \(\sin 0^\circ = \sin 180^\circ = 0\))
  • Cross product of a vector with itself is the zero vector: \(\vec{a} \times \vec{a} = \vec{0}\)
• Geometric Interpretations:
  • The magnitude of the dot product is related to the projection of one vector onto another.
  • The magnitude of the cross product represents the area of the parallelogram formed by the two vectors.