If a¯,b¯ are non collinear unit vectors, |a¯+b¯|=2 , then (2a¯+5b¯).(3a¯+b¯) =
|a+b| <= |a|+|b|
|a+b| = |a|+|b| only if a and b are collinear
so, bogus problem. they can't be unit vectors whose sum has magnitude 2 unless they are collinear.
1/2
To find the value of the expression (2a¯+5b¯).(3a¯+b¯), where a¯ and b¯ are given non-collinear unit vectors, we can use the dot product formula.
The dot product of two vectors a¯ and b¯ is given by: a¯ · b¯ = |a¯| |b¯| cos(θ), where |a¯| and |b¯| are the magnitudes of the vectors, and θ is the angle between them.
Given that a¯ and b¯ are non-collinear unit vectors, this means that |a¯| = |b¯| = 1.
We are also given that |a¯+b¯| = 2. This implies that the magnitude of the vector a¯+b¯ is 2.
Using the properties of the dot product, we can expand the expression:
(2a¯+5b¯).(3a¯+b¯) = (2a¯).(3a¯+b¯) + (5b¯).(3a¯+b¯)
Let's calculate the individual dot products:
(2a¯).(3a¯+b¯) = 2a¯ · 3a¯ + 2a¯ · b¯
= 6(a¯ · a¯) + 2(a¯ · b¯)
= 6|a¯|^2 + 2(a¯ · b¯)
= 6(1) + 2(a¯ · b¯)
= 6 + 2(a¯ · b¯)
Similarly,
(5b¯).(3a¯+b¯) = 5b¯ · 3a¯ + 5b¯ · b¯
= 15(b¯ · a¯) + 5(b¯ · b¯)
= 15(a¯ · b¯) + 5|b¯|^2
= 15(a¯ · b¯) + 5(1)
= 15(a¯ · b¯) + 5
Now we can substitute these values back into the original expression:
(2a¯+5b¯).(3a¯+b¯) = 6 + 2(a¯ · b¯) + 15(a¯ · b¯) + 5
Simplifying further:
(2a¯+5b¯).(3a¯+b¯) = 21 + 17(a¯ · b¯)
Therefore, the value of (2a¯+5b¯).(3a¯+b¯) is 21 + 17 times the dot product of vectors a¯ and b¯.