Question #74302

Given that vector c=a-b , vector c has a magnitude of 10 units and direction 270, vector a has a magnitude of 2 units and direction 150.
A) write the vector c in terms of unit vectors i,j,k.
B) write the vector a in terms of unit vectors i,j,k.
C) write the vector b in terms of unit vectors i,j,k.
D) find the magnitude and direction of the vector b.
1

Expert's answer

2018-03-06T09:29:07-0500

Question #74302, Physics / Other

Given that vector c=abc = a - b, vector cc has a magnitude of 10 units and direction 270, vector aa has a magnitude of 2 units and direction 150.

A) write the vector cc in terms of unit vectors i,j,ki,j,k.

B) write the vector aa in terms of unit vectors i,j,ki,j,k.

C) write the vector bb in terms of unit vectors i,j,ki,j,k.

D) find the magnitude and direction of the vector bb.

Solution

A) Vector components are determined as follows.


cx=10×cos270=0cy=10×sin270=10c=10j\begin{array}{l} c_x = 10 \times \cos 270{}^\circ = 0 \\ c_y = 10 \times \sin 270{}^\circ = -10 \\ c = -10j \\ \end{array}


B) ax=2×cos150=1.73a_x = 2 \times \cos 150{}^\circ = -1.73

ay=2×sin150=1a=1.73i+j\begin{array}{l} a_y = 2 \times \sin 150{}^\circ = 1 \\ a = -1.73i + j \\ \end{array}


C) b=ac=1.73i+j(10j)=1.73i+11jb = a - c = -1.73i + j - (-10j) = -1.73i + 11j

D) Vector magnitude is calculated as follows.


magnitude=(1.73)2+112=11.14direction=cos1(1.7311.14)=98.93\begin{array}{l} \text{magnitude} = \sqrt{\left(-1.73\right)^2 + 11^2} = 11.14 \\ \text{direction} = \cos^{-1}\left(\frac{-1.73}{11.14}\right) = 98.93{}^\circ \\ \end{array}


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