If 'a' and 'b' are unit vectors that make an angle of 60 degrees with each other, calculate

l 3a - 5b l and l 8a + 3b l

*the 'a' and 'b' have a carat of top of them*

How do i answer this without using components?

From point O draw the two unit vectors a and b with a 60° angle between them.

Extend vector a to a point A, so that OA = 3 units
In the opposite direction of vector b, draw OB = 5 units
Complete the parallelogram AOBC.
You now have vector OC = 3a - 5b , where OB = 5, BC = 3, angle B = 60°

cosine law:
|OC|^2 = 5^2 + 3^2 - 2(3)(5)cos60°
= 19
|3a - 5b| = √19

I will leave it up to you to find the 2nd one in a similar way

check using components

let a = (1,0)
then b = (1/2, √3/2)
3a - 5b = (3,0) - (5/2 , 5√3/2)
|3a - 5b| = √((5/2 - 3)^2 + (5√3/2-0)^2 )
= √( 1/4 + 75/4) = √19

To calculate the magnitude of a vector without using components, you can use the dot product formula. The magnitude of vector A, denoted as ||A||, can be calculated using the dot product of A with itself:

||A|| = √(A · A)

Let's apply this formula to the given vectors.

1. For 3a - 5b:
Start by calculating the dot product of (3a - 5b) with itself:
(3a - 5b) · (3a - 5b) = (3a · 3a) + (3a · -5b) + (-5b · 3a) + (-5b · -5b)

Now, since 'a' and 'b' are unit vectors, their magnitudes are both 1. Additionally, the dot product of two orthogonal (perpendicular) unit vectors is 0.

Substituting these values, the equation simplifies to:
(3^2 · 1) + (3a · -5b) + (-5a · 3b) + (-5^2 · 1) = 9 + (-15) + (-15) + 25 = 4

Finally, take the square root of the result to find the magnitude:
||3a - 5b|| = √4 = 2

2. For 8a + 3b:
Similarly, calculate the dot product of (8a + 3b) with itself:
(8a + 3b) · (8a + 3b) = (8a · 8a) + (8a · 3b) + (3b · 8a) + (3b · 3b)

Applying the same logic as before, the equation simplifies to:
(8^2 · 1) + (8a · 3b) + (3a · 8b) + (3^2 · 1) = 64 + 24 + 24 + 9 = 121

Take the square root of the result to find the magnitude:
||8a + 3b|| = √121 = 11

Therefore, the magnitude of 3a - 5b is 2, and the magnitude of 8a + 3b is 11.

To answer this question without using components, we'll use the fact that the length or magnitude of a vector is given by the formula:

|v| = √(v · v)

where v · v represents the dot product of the vector with itself.

Given that 'a' and 'b' are unit vectors that make an angle of 60 degrees with each other, we can calculate the dot product of 'a' and 'b' using the relationship:

a · b = |a| |b| cos(θ)

where |a| and |b| are the magnitudes of 'a' and 'b' respectively, and θ is the angle between them.

In this case, since both 'a' and 'b' are unit vectors, their magnitudes |a| and |b| are equal to 1. Therefore, we have:

a · b = 1 * 1 * cos(60°)

Now, let's calculate the dot product 'a · b':

a · b = cos(60°)

Using the value of cos(60°) = 0.5, we can substitute it into the equation:

a · b = 0.5

Now, let's calculate the magnitudes of the given vectors:

|3a - 5b| = √((3a - 5b) · (3a - 5b))

Substituting the dot product formula:

|3a - 5b| = √((3a · 3a) - (3a · 5b) - (5b · 3a) + (5b · 5b))

Since 'a' and 'b' are unit vectors, their dot products with themselves are equal to 1:

|3a - 5b| = √((3^2) - (3 * 5 * a · b) - (5 * 3 * a · b) + (5^2))

Substituting the value of a · b = 0.5:

|3a - 5b| = √((9) - (3 * 5 * 0.5) - (5 * 3 * 0.5) + (25))

Simplifying the equation:

|3a - 5b| = √(9 - 7.5 - 7.5 + 25)

|3a - 5b| = √(19)

Therefore, the magnitude of the vector 3a - 5b is √(19).

Similarly, let's calculate the magnitude of the vector 8a + 3b:

|8a + 3b| = √((8a + 3b) · (8a + 3b))

Substituting the dot product formula:

|8a + 3b| = √((8a · 8a) + (8a · 3b) + (3b · 8a) + (3b · 3b))

Using the value of a · b = 0.5:

|8a + 3b| = √((8^2) + (8 * 3 * a · b) + (3 * 8 * a · b) + (3^2))

Simplifying the equation:

|8a + 3b| = √(64 + 24 + 24 + 9)

|8a + 3b| = √(121)

Therefore, the magnitude of the vector 8a + 3b is √(121) = 11.