If|A| = 3 units, |B| = 4 units and the angle between vectors A and B is 90°, then the magnitude and direction of resultant , R with vector A are
(a) R = 25 units and tan 0 = 4/3
(b) R = 5 units and tai – 4/3
(c) R = 25 units and tan 0 = 3/4
(d) R = 5 units and tan 0 = 3/4
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Answer:
|A| = 3 units
|B| = 4 units
therefore, R = √ A² + B² +2ABcostheta
= √ 9 + 16 + 2.3.4.cos90°
= √ 25
= 5 units
tan alpha = B sin theta
A + B cos theta
= 4 × sin 90°
3 + 4 cos 90°
= 4/3
R = 5 units and tan alpha = 4/3
Answer:
The correct option is d) R = 5 units and tan θ = [tex]\frac{3}{4}[/tex]
Explanation:
Given
IAI = 3 units
IBI = 4 units
An angle between A and B vectors is 90°
A vector has both magnitude and direction as well.
The magnitude of Resultant vector R is,
R = [tex]\sqrt{A^2 + B^2 + 2AB cos\alpha }[/tex]
where α is an angle between the vector A and B
R = [tex]\sqrt{3^2 + 4^2+2*3*4*cos 90}[/tex]
cos 90° = 0
R = [tex]\sqrt{3^2 +4 ^2}[/tex]
R= [tex]\sqrt{9+16\\[/tex]
R= [tex]\sqrt{25}[/tex]
R = 5 units
Direction of Resultant vector R is
tan θ = [tex]\frac{B sin\alpha}{A+Bcos\alpha }[/tex]
sin 90° = 1
tan θ = [tex]\frac{3}{4}[/tex]
So, R = 5 units and tan θ = [tex]\frac{3}{4}[/tex]