Question

vector a is due west and vector b is due north. which of the following represents the subtraction of vector a - vector b?

Explanation:

To solve the problem of vector subtraction \(\vec{A} - \vec{B}\), we can use the concept of vector components and the rules of vector subtraction. Here's a step - by - step breakdown:

Step 1: Represent Vectors in Component Form

• Let's assume the positive \(x\) - axis is towards the East and the positive \(y\) - axis is towards the North.
• Vector \(\vec{A}\) is due West. If the magnitude of \(\vec{A}\) is \(A\), then in component form, \(\vec{A}=(-A,0)\) (because West is the negative \(x\) - direction).
• Vector \(\vec{B}\) is due North. If the magnitude of \(\vec{B}\) is \(B\), then in component form, \(\vec{B}=(0,B)\) (because North is the positive \(y\) - direction).

Step 2: Apply the Vector Subtraction Formula

The formula for vector subtraction \(\vec{C}=\vec{A}-\vec{B}\) in component form is given by:
If \(\vec{A}=(A_x,A_y)\) and \(\vec{B}=(B_x,B_y)\), then \(\vec{C}=\vec{A}-\vec{B}=(A_x - B_x,A_y - B_y)\)
Substituting the components of \(\vec{A}\) and \(\vec{B}\) into the formula:
\(\vec{A}-\vec{B}=(-A - 0,0 - B)=(-A,-B)\)

Step 3: Interpret the Resulting Vector

• The \(x\) - component of \(\vec{A}-\vec{B}\) is \(-A\) (which means it has a component towards the West) and the \(y\) - component is \(-B\) (which means it has a component towards the South).
• Geometrically, vector subtraction \(\vec{A}-\vec{B}\) is equivalent to \(\vec{A}+(-\vec{B})\). The vector \(-\vec{B}\) has the same magnitude as \(\vec{B}\) but is in the opposite direction (i.e., due South). So, when we add \(\vec{A}\) (due West) and \(-\vec{B}\) (due South), the resultant vector \(\vec{A}-\vec{B}\) will be in the direction of South - West. The magnitude of the resultant vector \(|\vec{A}-\vec{B}|=\sqrt{(-A)^2+(-B)^2}=\sqrt{A^{2}+B^{2}}\) (by the Pythagorean theorem, since the two components are perpendicular to each other).

If we were to represent this graphically:

1. Draw vector \(\vec{A}\) pointing to the West.
2. Draw vector \(-\vec{B}\) (the negative of vector \(\vec{B}\)) pointing to the South.
3. The resultant vector \(\vec{A}-\vec{B}\) is the vector from the tail of \(\vec{A}\) to the head of \(-\vec{B}\) when we place the tail of \(-\vec{B}\) at the head of \(\vec{A}\) (or using the parallelogram law of vector addition for \(\vec{A}\) and \(-\vec{B}\)).

So, the vector \(\vec{A}-\vec{B}\) has a magnitude of \(\sqrt{A^{2}+B^{2}}\) and is directed towards the South - West direction. If we are given options (even though they are not shown here), we can compare the direction (South - West) and magnitude (\(\sqrt{A^{2}+B^{2}}\)) with the given options to find the correct representation of \(\vec{A}-\vec{B}\).

Answer:

To solve the problem of vector subtraction \(\vec{A} - \vec{B}\), we can use the concept of vector components and the rules of vector subtraction. Here's a step - by - step breakdown:

Step 1: Represent Vectors in Component Form

• Let's assume the positive \(x\) - axis is towards the East and the positive \(y\) - axis is towards the North.
• Vector \(\vec{A}\) is due West. If the magnitude of \(\vec{A}\) is \(A\), then in component form, \(\vec{A}=(-A,0)\) (because West is the negative \(x\) - direction).
• Vector \(\vec{B}\) is due North. If the magnitude of \(\vec{B}\) is \(B\), then in component form, \(\vec{B}=(0,B)\) (because North is the positive \(y\) - direction).

Step 2: Apply the Vector Subtraction Formula

The formula for vector subtraction \(\vec{C}=\vec{A}-\vec{B}\) in component form is given by:
If \(\vec{A}=(A_x,A_y)\) and \(\vec{B}=(B_x,B_y)\), then \(\vec{C}=\vec{A}-\vec{B}=(A_x - B_x,A_y - B_y)\)
Substituting the components of \(\vec{A}\) and \(\vec{B}\) into the formula:
\(\vec{A}-\vec{B}=(-A - 0,0 - B)=(-A,-B)\)

Step 3: Interpret the Resulting Vector

• The \(x\) - component of \(\vec{A}-\vec{B}\) is \(-A\) (which means it has a component towards the West) and the \(y\) - component is \(-B\) (which means it has a component towards the South).
• Geometrically, vector subtraction \(\vec{A}-\vec{B}\) is equivalent to \(\vec{A}+(-\vec{B})\). The vector \(-\vec{B}\) has the same magnitude as \(\vec{B}\) but is in the opposite direction (i.e., due South). So, when we add \(\vec{A}\) (due West) and \(-\vec{B}\) (due South), the resultant vector \(\vec{A}-\vec{B}\) will be in the direction of South - West. The magnitude of the resultant vector \(|\vec{A}-\vec{B}|=\sqrt{(-A)^2+(-B)^2}=\sqrt{A^{2}+B^{2}}\) (by the Pythagorean theorem, since the two components are perpendicular to each other).

If we were to represent this graphically:

1. Draw vector \(\vec{A}\) pointing to the West.
2. Draw vector \(-\vec{B}\) (the negative of vector \(\vec{B}\)) pointing to the South.
3. The resultant vector \(\vec{A}-\vec{B}\) is the vector from the tail of \(\vec{A}\) to the head of \(-\vec{B}\) when we place the tail of \(-\vec{B}\) at the head of \(\vec{A}\) (or using the parallelogram law of vector addition for \(\vec{A}\) and \(-\vec{B}\)).

So, the vector \(\vec{A}-\vec{B}\) has a magnitude of \(\sqrt{A^{2}+B^{2}}\) and is directed towards the South - West direction. If we are given options (even though they are not shown here), we can compare the direction (South - West) and magnitude (\(\sqrt{A^{2}+B^{2}}\)) with the given options to find the correct representation of \(\vec{A}-\vec{B}\).