Each of the displacement vectors vecA and vecB shown in...

Each of the displacement vectors vec{A} and vec{B} shown in Figure P3.11 has a magnitude of 3.00 m. Graphically find (a) vec{A} + vec{B}, (b) vec{A} - vec{B}, (c) vec{B} - vec{A}, and (d) vec{A} - 2vec{B}. Report all angles counterclockwise from the positive x axis.

Transcript

00:01 In this problem we have given vector a and vector b the magnitude of vector a is 3 meter and magnitude of vector b is also 3 meter first in first part we have to find summation of vector a plus summation of vector b so we know that we can write vector b will be equal to 3 j cap and we can also write vector a in the component form so the x component of vector a is a x and y component of vector a is a y so this will be a x i cap plus a y j cap so a x will be equal to three plus thirty i cap plus three sine 30 j cap so the vector will be equal to t cost 30 this is equal to 2 .60 iqq plus 1 point this is 3 into 530 jkab this is vector a and this is vector v so now vector a plus vector v will be equal to 2 plucks 60 i kf plus 1 .5 jk plus 3 jk so this will be equal to 2 .60 i cap plus 4 .50 j cap so this is summation vector a plus vector b right now this is vector a plus vector b right now this is vector a plus vector b we can draw this like this is vector a plus vector b this angle is supposed theta now so this theta will be equal to 10 inverse 4 .50 divided by 2 .60 so this will be equal to around 60 degree this is around 60 degree right now in the second part we have to find vector a minus vector b so this will be equal to 2 .60 i cap plus 1 .5 j cap minus vector b which is 3 jcap so it will be equal to 2 .60 i cap minus 1 .5 j cap this is vector a minus vector b right so now we can again draw this this is x component and this is y component this is positive x -axis this is positive y -axis so this is vector a minus vector b now this angle is suppose alpha so we can write alpha will be equal to 10 inverse 1 .50 2 .60 so this is equal to to around 30 degree the angle from positive x axis in the counterclockwise direction will be this theta which is equal to 360 minus 3 that is equal 3 30 degree right or we can say it is from positive x axis in the clockwise anti -clockwise direction now in part d in part c we have to find vector v minus vector a so vector v is 3 j gap and the vector a is 2 .60 i gap plus 1 .50 j gap so this is equal to minus 2 .60 i gap plus 1 .50 j gap this is vector b minus vector a now we can draw this this is 2 .60 towards the negative x -axis and this is is 1 .50...

Question

Each of the displacement vectors vec{A} and vec{B} shown in Figure P3.11 has a magnitude of 3.00 m. Graphically find (a) vec{A} + vec{B}, (b) vec{A} - vec{B}, (c) vec{B} - vec{A}, and (d) vec{A} - 2vec{B}. Report all angles counterclockwise from the positive x axis.

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