Problem 66 - Fundamentals of Physics

Question

Consider \(\vec{a}\) in the positive direction of \(x, \vec{b}\) in the positive direction of \(y\), and a scalar \(d\). What is the direction of \(\vec{b} / d\) if \(d\) is (a) positive and (b) negative? What is the magnitude of (c) \(\vec{a} \cdot \vec{b}\) and (d) \(\vec{a} \cdot \vec{b} / d ?\) What is the direction of the vector resulting from (e) \(\vec{a} \times \vec{b}\) and (f) \(\vec{b} \times \vec{a}\) ? (g) What is the magnitude of the vector product in (e)? (h) What is the magnitude of the vector product in (f)? What are (i) the magnitude and (j) the direction of \(\vec{a} \times \vec{b} / d\) if \(d\) is positive?

Step-by-Step Solution

Answer

(a) Positive \(y\), (b) Negative \(y\). (c) 0, (d) 0. (e) Positive \(z\), (f) Negative \(z\). (g), (h) \(|\vec{a}||\vec{b}|\). (i) \(|\vec{a}||\vec{b}| / d\), (j) Positive \(z\).

Step 1: Understanding the Vector Division for Scalar d

For part (a), if the scalar \(d\) is positive, then \(\vec{b}/d\) maintains the same direction as \(\vec{b}\). For part (b), if \(d\) is negative, \(\vec{b}/d\) points in the opposite direction of \(\vec{b}\), so it will be in the negative direction of the \(y\)-axis.

Step 2: Calculating the Magnitude of Dot Product

For part (c), the dot product \(\vec{a} \cdot \vec{b}\) is zero because \(\vec{a}\) and \(\vec{b}\) are perpendicular vectors (one along \(x\) and one along \(y\) direction). Hence, their magnitude is \(0\). For part (d), \(\vec{a} \cdot \vec{b} / d\) remains \(0\) since dividing zero by any non-zero scalar \(d\) gives zero.

Step 3: Finding the Direction of Cross Product

For part (e), the cross product \(\vec{a} \times \vec{b}\) results in a vector along the positive \(z\)-axis using the right-hand rule. For part (f), \(\vec{b} \times \vec{a}\) results in a vector along the negative \(z\)-axis, again using the right-hand rule.

Step 4: Calculating the Magnitude of Vector Product

For part (g), the magnitude of \(\vec{a} \times \vec{b}\) can be found using \(|\vec{a}||\vec{b}| \sin(\theta)\), where \(\theta = 90^\circ\). The magnitude is \(|\vec{a}||\vec{b}|\). Part (h) yields the same magnitude for \(\vec{b} \times \vec{a}\).

Step 5: Determining the Magnitude and Direction for Divided Cross Product

For (i), the magnitude of the vector \(\vec{a} \times \vec{b} / d\) when \(d\) is positive, is \(|\vec{a}||\vec{b}| / d\). For (j), the direction remains along the positive \(z\)-axis as only the magnitude is affected by \(d\), not the direction.

Key Concepts

Dot Product

The dot product is a fundamental concept in vector algebra that allows us to quantify the interaction between two vectors in a specific manner. When calculating the dot product of two vectors, we're essentially measuring how much one vector projects onto another. For two vectors, \( \vec{a} = \begin{bmatrix} a_1 \ a_2 \ a_3 \end{bmatrix} \) and \( \vec{b} = \begin{bmatrix} b_1 \ b_2 \ b_3 \end{bmatrix} \), the dot product is calculated as:

• \( \vec{a} \cdot \vec{b} = a_1 \cdot b_1 + a_2 \cdot b_2 + a_3 \cdot b_3 \)

This is a scalar product, meaning it results in a single number, not another vector.

A notable property of the dot product is that it equals zero when the two vectors are perpendicular to each other, which means they form a 90-degree angle. In the exercise, vectors \(\vec{a}\) and \(\vec{b}\) are along the x-axis and y-axis respectively, showing they are perpendicular. Therefore, \( \vec{a} \cdot \vec{b} = 0 \), an important point that will always hold for perpendicular vectors. When such a product is divided by a scalar \(d\), the result remains zero because zero divided by any non-zero number is still zero.

Cross Product

The cross product is another crucial operation in vector algebra which gives us a vector that is perpendicular to the two initial vectors. Given two vectors \(\vec{a}\) and \(\vec{b}\), their cross product, \(\vec{a} \times \vec{b}\), is calculated as:

• \( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} \)

This involves the determinant of a matrix built from the components of \(\vec{a}\) and \(\vec{b}\), leading to a new vector.

The resultant vector from the cross product has a magnitude calculated as \(|\vec{a}||\vec{b}|\sin(\theta)\), where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\). If \(\vec{a}\) and \(\vec{b}\) are perpendicular, as in our exercise, \(\sin(90^\circ) = 1\), meaning the magnitude is simply \(|\vec{a}||\vec{b}|\).

An interesting attribute to note is changing the order of the vectors in the cross product results in a vector pointing in the opposite direction. This is observable in \(\vec{a} \times \vec{b}\) compared to \(\vec{b} \times \vec{a}\). The direction in the cross product is always perpendicular to the plane formed by the two vectors.

Vector Magnitude

The magnitude of a vector, often perceived as its length, is a measure of how much space it covers. For a vector \( \vec{v} = \begin{bmatrix} x \ y \ z \end{bmatrix} \), the magnitude \(|\vec{v}|\) is calculated using the formula:

• \(|\vec{v}| = \sqrt{x^2 + y^2 + z^2}\)

This is derived from the Pythagorean theorem, extended into three dimensions. It gives us a scalar value representing how 'long' the vector is.

In the context of cross products, as seen in our example, the magnitude helps quantify the strength or size of the resulting vector. Particularly for \(\vec{a} \times \vec{b}\), the magnitude was determined purely by the magnitudes of \(\vec{a}\) and \(\vec{b}\) and their perpendicular relationship. When you divide this value by a scalar \(d\), the magnitude of the new resultant vector remains \(|\vec{a} \times \vec{b}| / d\). This operation affects only the size, not the direction.

Right-Hand Rule

The right-hand rule is an easy and quick way to determine the direction of a vector resulting from a cross product. This intuitive rule lets us visualize the three-dimensional orientation of vectors. By positioning your right hand:

• Point your right thumb in the direction of the first vector (\(\vec{a}\)).
• Extend your fingers in the direction of the second vector (\(\vec{b}\)).
• The direction your palm faces is the direction of the cross product \(\vec{a} \times \vec{b}\).

This rule is crucial in understanding the orientation of new vectors created through cross products.

In exercises involving vectors on the x and y axes, such as in our exercise, employing the right-hand rule demonstrates that \(\vec{a} \times \vec{b}\) points along the positive z-axis. Conversely, \(\vec{b} \times \vec{a}\) flips its direction to the negative z-axis, showcasing the right-hand rule's reversibility based on the order of vectors.