The position vector for a proton is initially r=5.0i^-6.0j^+2.0k^and then later isrole="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?

Short Answer

Expert verified

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

(b) dis parallel to the X-Y plane

Step by step solution

01

Given data

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

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

02

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

03

(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

04

(b) Determining the direction of the plane

The displacement vector ddoes 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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