The resultant of two vectors a and b of magnitudes 6 and 10 respective...
Problem:
The resultant of two vectors a and b of magnitudes 6 and 10 respectively is at right angles to the vector a. Find their resultant and also the angle between the given vectors.
Solution:
Given, magnitudes of vectors a and b are 6 and 10 respectively.
Let the angle between vectors a and b be θ.
Let's assume vector a is in the direction of x-axis.
The magnitude of the resultant R is given by the Pythagorean theorem as:
R² = a² + b²
R² = 6² + 10²
R² = 136
R = √136
R = 2√34
Let's assume vector b is at an angle of θ with the positive x-axis.
The dot product of vectors a and b is given by:
a . b = |a| |b| cosθ
Since a and b are at right angles, cosθ = 0
Therefore, a . b = 0
The dot product of vectors a and b can also be written as:
a . b = ax bx + ay by
where ax and ay are the components of vector a and bx and by are the components of vector b.
Since a is in the direction of x-axis, its components are ax = 6 and ay = 0.
Therefore, a . b = 6bx
Substituting a . b = 0, we get:
6bx = 0
bx = 0
This implies that vector b is in the direction of y-axis.
Therefore, the angle between vectors a and b is 90 degrees.
Answer:
The magnitude of the resultant of vectors a and b is 2√34 and the angle between the vectors is 90 degrees.
The resultant of two vectors a and b of magnitudes 6 and 10 respective...
Resultant of a and b, R=(a+b)
As per Q,
R•a=0 [dot product]
(a+b)•a =0
a•a + b•a =0
|a|² + b•a =0
6² + 60 cosx =0
or cosx = - ⅗
or x = (180 - 53) [ cos is -ve, it must lie in 2nd or 3rd quad]
or x = 127 deg
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