Question

More Vectors! Let A = 1.2 cmi + 5.4cmj. B-1.3 emi + 2.4 cm3- : 4.4 cmi Find: a. The magnitude and direction of A,B, and C. Express the direction as an angle, in degrees, assuming 0 is the positive axis and counterclockwise angles are positive. A: Magnitude B: Magnitude = C: Magnitude cm Direction cm Direction = i cm Direction b. The magnitude and direction of the vector R-A 1.5B+3C. Magnitude R i cm Direction

Answer 1

$\vec{A}=(1.2i+5.4j)cm$

$A=\sqrt{A_{x}^{2}+A_{y}^{2}}$

$A=\sqrt{(1.2)^{2}+(5.4)^{2}}$

ANSWER: ${\color{Red} A=5.532cm}$

Direction $\theta =tan^{-1}(A_{y}/A_{x})$

$\theta =tan^{-1}(5.4cm/1.2cm)$

ANSWER: ${\color{Red} \theta =77.47^{\circ}}$

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$\vec{B}=(1.3i+2.4j)cm$

$B=\sqrt{B_{x}^{2}+B_{y}^{2}}$

$B=\sqrt{(1.3)^{2}+(2.4)^{2}}$

ANSWER: ${\color{Red} B=2.729cm}$

Direction $\theta =tan^{-1}(B_{y}/B_{x})$

$\theta =tan^{-1}(2.4cm/1.3cm)$

ANSWER: ${\color{Red} \theta =61.56^{\circ}}$

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$\vec{C}=(4.4i)cm$

$C=\sqrt{C_{x}^{2}+C_{y}^{2}}$

$C=\sqrt{(4.4)^{2}+(0)^{2}}$

ANSWER: ${\color{Red} C=4.4cm}$

Direction $\theta =tan^{-1}(C_{y}/C_{x})$

$\theta =tan^{-1}(0/4.4cm)$

ANSWER: ${\color{Red} \theta =0^{\circ}}$

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$\vec{R}=\vec{A}-1.5\vec{B}+3\vec{C}$

$\vec{R}=(1.2i+5.4j)-1.5(1.3i+2.4j)+3(4.4i)$

$\vec{R}=(1.2i+5.4j)-(1.95i+3.6j)+(13.2i)$

$\vec{R}=1.2i+5.4j-1.95i-3.6j+13.2i$

$\vec{R}=(1.2-1.95+13.2)i+(5.4-3.6)j$

${\color{Red} \vec{R}=12.45i+1.8j}$

$R=\sqrt{R_{x}^{2}+R_{y}^{2}}$

$R=\sqrt{(12.45)^{2}+(1.8)^{2}}$

ANSWER: ${\color{Red} R=12.579cm}$

Direction $\theta =tan^{-1}(R_{y}/R_{x})$

$\theta =tan^{-1}(1.8cm/12.45cm)$

ANSWER: ${\color{Red} \theta =8.227^{\circ}}$

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