Question

\(\underline{\mathrm{a}}=\underline{\mathrm{u}}-\underline{\mathrm{v}}, \underline{\mathrm{b}}=\underline{\mathrm{u}}+\underline{\mathrm{v}},|\underline{\mathrm{u}}|=|\underline{\mathrm{u}}|^{2}\) and \(|\underline{\mathrm{u}}|=|\underline{\mathrm{v}}|=2\) find \(|\underline{a} \times \underline{b}|=\) (b) \(\left.\sqrt{[} 4-(\underline{\mathrm{u}} \cdot \underline{\mathrm{v}})^{2}\right]\) (c) \(\sqrt{\left[16-(\underline{\mathrm{u}} \cdot \underline{\mathrm{v}})^{2}\right]}\) (d) \(\sqrt{\left[4-(\underline{\mathrm{u}} \cdot \underline{\mathrm{v}})^{2}\right]}\)

Short answer

The short answer is: \(|\underline{a}\times\underline{b}| = \sqrt{\left[16-(\underline{\mathrm{u}} \cdot \underline{\mathrm{v}})^{2}\right]}\) (option c).

Step-by-step solution

Expressing \(\underline{a}\) and \(\underline{b}\) in terms of \(\underline{u}\) and \(\underline{v}\)

We have \(\underline{\mathrm{a}}=\underline{\mathrm{u}}-\underline{\mathrm{v}}\) and \(\underline{\mathrm{b}}=\underline{\mathrm{u}}+\underline{\mathrm{v}}\).

Computing the cross product \(\underline{a} \times \underline{b}\)

We have \(\underline{a}\times\underline{b} = (\underline{u}-\underline{v})\times(\underline{u}+\underline{v})\). Expanding this expression using properties of cross product, we get \[ \underline{a}\times\underline{b} = \underline{u}\times\underline{u} + \underline{u}\times\underline{v} - \underline{v}\times\underline{u} - \underline{v}\times\underline{v}. \] Remember that the cross product of a vector with itself is zero. Also, the cross product is anti-commutative, which means \(\underline{v}\times\underline{u}=-\underline{u}\times\underline{v}\). So, we can simplify the expression as: \[ \underline{a}\times\underline{b} = 2(\underline{u}\times\underline{v}). \]

Finding the magnitude of \(\underline{a}\times\underline{b}\)

Now, we need to find the magnitude of the cross product, which will be \[ |\underline{a}\times\underline{b}| = 2|\underline{u}\times\underline{v}|. \] Recall that \(|\underline{u}\times\underline{v}|=|\underline{u}||\underline{v}|\sin\theta\). We are given that \(|\underline{\mathrm{u}}|=|\underline{\mathrm{v}}|=2\), so we get \[ |\underline{a}\times\underline{b}| = 2(2)(2)\sin\theta = 8\sin\theta. \]

Comparing with the given options

Now let's compare our result with the given options (b), (c), and (d). Notice that each option contains a term of the form \((\underline{u}\cdot\underline{v})^2\). We can use the fact that \(|\underline{u}\times\underline{v}|^2=|\underline{u}|^2|\underline{v}|^2-\left(\underline{u}\cdot\underline{v}\right)^2\) to eliminate this term and find the correct answer. First, we compute \(|\underline{u}\times\underline{v}|^2\): \[ |\underline{u}\times\underline{v}|^2 = \left(8\sin\theta\right)^2=64\sin^2\theta. \] Plugging back the values of \(|\underline{u}|\) and \(|\underline{v}|\), we get \[ 64\sin^2\theta=4^2(2^2)-(\underline{u}\cdot\underline{v})^2. \] Comparing this expression with the given options, we see that it matches option (c): \[ |\underline{a}\times\underline{b}| = \sqrt{\left[16-(\underline{\mathrm{u}} \cdot \underline{\mathrm{v}})^{2}\right]}. \]

Key concepts

Vector Magnitude

The magnitude of a vector is a measure of its length. For a vector \( \underline{v} = (x, y, z) \), its magnitude is given by the formula: \[ |\underline{v}| = \sqrt{x^2 + y^2 + z^2} \]. This formula is derived from the Pythagorean theorem. In the context of the exercise, we dealt with vectors \( \underline{u} \) and \( \underline{v} \), both with magnitudes of 2. This means both vectors have a length of 2 units in space.
When calculating cross products, knowing the magnitudes is important as they help determine the magnitude of the result through the sine of the angle between two vectors.

Dot Product

The dot product is an algebraic operation that takes two equal-length sequences of numbers, usually coordinate vectors, and returns a single number. For vectors \( \underline{a} = (a_1, a_2, a_3) \) and \( \underline{b} = (b_1, b_2, b_3) \), the dot product is: \[ \underline{a} \cdot \underline{b} = a_1b_1 + a_2b_2 + a_3b_3 \]. This operation is very useful in projecting one vector onto another, checking perpendicularity, and finding angles.
In our exercise, the dot product \( \underline{u} \cdot \underline{v} \) appears in the answer options. It is involved in calculating the squared magnitude of the cross product due to the identity: \[ |\underline{u} \times \underline{v}|^2 = |\underline{u}|^2|\underline{v}|^2 - (\underline{u} \cdot \underline{v})^2 \]. This relation is used in simplifying the calculation of the cross product's magnitude.

Trigonometric Identities

Trigonometric identities are mathematical equations that relate the angles and sides of a triangle to trigonometric functions. In vector mathematics, they often come into play when dealing with angles between vectors. For example:

• \( \sin(\theta) \) appears in calculations of vector norms, particularly when finding the magnitude of the cross product, according to \[ |\underline{u} \times \underline{v}| = |\underline{u}||\underline{v}|\sin\theta \].
• \( \cos(\theta) \) can be derived using the dot product: \( \cos(\theta) = \frac{\underline{u} \cdot \underline{v}}{|\underline{u}||\underline{v}|} \).

Understanding these relationships is essential in simplifying complex expressions and solving vector-related problems. In the context of the provided exercise, identifying \( \sin(\theta) \) allowed us to directly compute the magnitude of the cross product.

Orthogonal Vectors

Two vectors are orthogonal if they are perpendicular to each other. This means the angle between them is \( 90^\circ \) or \( \pi/2 \) radians. A critical property of orthogonal vectors is that their dot product equals zero: \[ \underline{a} \cdot \underline{b} = 0 \]. This property is often used to prove perpendicularity without visual analysis.
In a cross product context, if the two vectors were orthogonal, then the magnitude of their cross product would be maximized since \( \sin(\pi/2) = 1 \). Although our exercise didn't specifically state orthogonality, understanding this concept helps to recognize when vectors produce maximum cross product magnitude. It's a handy concept when seeking maximal vector interactions where angles can impact outcomes.