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Fundamentals Of Physics
Found in: Page 91

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Short Answer

The position vector for a proton is initially r=5.0i^-6.0j^+2.0k^ and then later is role="math" localid="1657003791208" r=-2.0i^-6.0j^+2.0k^ all in meters. (a) What is the proton’s displacement vector, and (b) to what plane is that vector parallel?

(a) The proton’s displacement vector is-7i^+12j^m

(b) dis parallel to the X-Y plane

See the step by step solution

Step by Step Solution

Step 1: Given data

1) Initial position vector ri=5.0i^-6.0j^+2.0k^

2) Final position vector rf=-2.0i^-6.0j^+2.0k^

Step 2: Understanding the concept of vector addition and subtraction

Taking the difference between position vectors of proton, we can find the displacement vector. We have to use vector laws of addition for this.

Formula:

d=rf-ri

Step 3: (a) Determining the displacement vector

To determine the displacementd, we need to use the equation,

d=rf-ri =-2.0i^+6.0j^+2.0k^m-5.0i^-6.0j^+2.0k^m =-7.0i^+12j^m

Therefore, the displacement vector is -7.0i^+12j^m

Step 4: (b) Determining the direction of the plane

The displacement vector d does not have any component along Z-axis. It means that the displacement vector lies in XY plane.Therefore, the displacement vector is parallel to the XY plane.

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