Hello all,
I have recently been advancing my knowledge of mathematics by working through worksheets online. However, I am stumped at these particular questions, and have no clue where to begin and answer! Any chance of any explanations and answers? All responses are highly appreciated, Thank you everyone! P.S I have added a file attachment of the questions below which is little clearer than what I have typed out

The dot scalar product (M) of two directional paths 'x' and 'y' is mathematically defined as follows:

M = xy (1)

and
xy = |x||y|cos0 (2)

where |x| is the magnitude of directional path 'x' and |y| is the magnitude of directional path 'y' and 0 is the angle between paths 'x' and 'y'

Generally, for two directional paths 'a' and 'b' defined as follows:

a = a1i + a2j (3)
b = b
1i + b2j (4)

The following formulas are given for the dot or scalar product of ‘a’ and ‘b’ and their respective magnitudes. Remember the notations ‘i’ and ‘j’ represent the spatial direction of the paths.

ab = (a1b1) + (a2b2) (5)

|a| = √ ̅ (a12+ a22) (6)

|b| =
√ ̅ (b12+ b22 ) (7)

If the directional paths ‘x’ and ‘y’ are defined as follows:

x = 3i + 6j (8)
y = 8i - 2j (9)

Question a.
Solve for M by interpreting all the given formulas in equations (1) to (9).

Question b.
Solve for the angle between the directional paths ‘x’ and ‘y’ by making  the subject of the formula in equation (2).

What have you been able to do so far? Knowing that will enable us to help you better.

-Dan

Hi Dan, unfortunately I have not attempted it as of yet as I am unsure on how to answer these particular questions, I would appreciate any help if possible, thank you!

One way to do the dot product of two vectors is defined as: $\displaystyle x = a i + b j$, $\displaystyle y = c i + d j$, $\displaystyle x \cdot y = (a c) + (b d)$ (Line 5)
a) So what is the dot product of $\displaystyle x = 3 i - 6 j$ and $\displaystyle y = 8 i - 2 j$? (From lines 8 and 9.)
b) Now that you have $\displaystyle x \cdot y$ can you use equation 2 to find $\displaystyle \theta$?