Calculating Vector Quantity with Given Magnitude and Direction

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In summary, the conversation discusses how to determine the quantity |A + B|^2 − |A − B|^2, where Vector A has a magnitude of 4.0 units and is directed θA = 15◦ counterclockwise from the positive x-axis, and Vector B has a magnitude of 4.0 units and is directed θB = 85◦ counterclockwise from the positive x-axis. The steps involved in finding the answer include calculating the components for both vectors, creating a new vector C by adding A and B, finding the length of C, and using this length in the expression to find the answer. The conversation also mentions the importance of understanding the notation |A| and
  • #1
Physicsnoob90
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Homework Statement


Vector A has magnitude A = 4.0 units and is directed θA = 15◦ counterclockwise from the positive x-axis. Vector B has magnitude B = 4.0 units and is directed θB = 85◦ counterclockwise from the positive x-axis. Determine the following quantity: |A + B|^2 − |A − B|^2.

Homework Equations

The Attempt at a Solution


Steps) 1. i calculated their components: Ax= 4.0 cos 15º = 3.86, Ay= 4.0 sin 15º = 1.04 , A= 4.898979486
Bx= 4.0 cos 85º = 0.35 , By= 4.0 sin 85º = 3.98 , B= 4.333401763

2. plug into the format: |(4.898979486)^2 + (4.333401763)^2| - |(4.898979486)^2 - (4.333401763)^2| = 38
 
Last edited:
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  • #2
Look at the A+B and A-B terms in your expression are they vectors or scalars?
 
  • #3
they are vectors
 
  • #4
So what you wrote added the lengths of the components of A and B together which is not right.

when instead you should find the length of the new vector A+B and square it for your expression.
 
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  • #5
i thought it was asking to add the components of the vector A and B while also figuring out the quantity |A+B|^2 - |A-B|^2?
 
  • #6
The notation |A+B| means the length of the vector A+B just as |A| means the length of The vector A.
 
  • #7
So would creating a new vector like vector C = A+B?
 
  • #8
Yes and then find the length of C to use in your expression similarly for the |A-B| term which is the length of the vector A-B.

Have you drawn these four vectors on paper? The A and B represent the sides of a parallelogram and the A+B is one diagonal.

Do you know what the other vector is?
 
  • #9
isn't the other vector the opposite diagonal to A+B?
 
  • #10
Yes it is.

Did you figure out the answer now?
 

Related to Calculating Vector Quantity with Given Magnitude and Direction

What is a vector quantity?

A vector quantity is a physical quantity that has both magnitude and direction. This means that in addition to having a numerical value, it also has a specific direction in space. Examples of vector quantities include velocity, force, and displacement.

How is a vector represented?

A vector is typically represented graphically as an arrow, with its length representing the magnitude and its direction indicating the direction of the quantity. In mathematical notation, a vector is often denoted with a boldface letter or an arrow above the letter.

What is the difference between a vector and a scalar?

A scalar is a physical quantity that has only magnitude, while a vector has both magnitude and direction. For example, speed is a scalar quantity as it only tells us how fast an object is moving, while velocity is a vector quantity as it also includes the direction of motion.

How are vectors added and subtracted?

To add or subtract vectors, we use the parallelogram rule or the head-to-tail method. This involves placing the vectors end-to-end and then drawing a line from the tail of the first vector to the head of the second vector. The resulting vector, called the resultant, is the sum or difference of the original vectors.

What is the dot product of two vectors?

The dot product is a mathematical operation that takes two vectors and produces a scalar quantity. It is calculated by multiplying the magnitude of one vector by the magnitude of the component of the other vector in the same direction. The result is a scalar value that represents the amount of overlap or similarity between the two vectors.

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