Interpreting Scalar Product (formula in equations), Please help!

Apr 2018
2
0
Wales
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 answers? Answers would be appreciated as I can work through the steps logically to see how it works. All responses are highly appreciated, Thank you everyone! P.S I have added a file attachment of the questions below which is a 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 = x∙y $\hspace{57px}$ (1)

and
x∙y = |x||y|cosθ (2)

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

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

a = a$_1$i + a$_2$j (3)
b = b$_1$i + b$_2$j (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.

a∙b = (a$_1$b$_1$) + (a$_2$b$_2$) (5)

|a| = √(a$_1^2$ + a$_2^2$) (6)

|b| = √(b$_1^2$ + b$_2^2$) (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 θ the subject of the formula in equation (2).
 

Attachments

Last edited by a moderator:

skipjack

Forum Staff
Dec 2006
21,481
2,470
Your attachment is a bit fuzzy.
 

mathman

Forum Staff
May 2007
6,932
774
To get M, use equation (5). To get $cos(\theta)$, use equations (5), (6), and (7) and insert answers into equation (2).
 

Country Boy

Math Team
Jan 2015
3,261
899
Alabama
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 answers? Answers would be appreciated as I can work through the steps logically to see how it works. All responses are highly appreciated, Thank you everyone! P.S I have added a file attachment of the questions below which is a 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 = x∙y $\hspace{57px}$ (1)

and
x∙y = |x||y|cosθ (2)

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

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

a = a$_1$i + a$_2$j (3)
b = b$_1$i + b$_2$j (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.

a∙b = (a$_1$b$_1$) + (a$_2$b$_2$) (5)

|a| = √(a$_1^2$ + a$_2^2$) (6)

|b| = √(b$_1^2$ + b$_2^2$) (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).
Using (1) m= (8i+ 6j).(8i-2j)= 8(8)+ (6)(-2)= 64- 12= 52.

Using (2), |8i+ 6j|= sqrt(8(8)+ 6(6))= sqrt{64+ 36}= sqrt{100}= 10, |8i- 2j|= sqrt(8(8)+ 2(2))= sqrt(64+ 4)= sqrt(68)= 2\sqrt(17)
so (8i+ 6j).(8i+ 2j)= 20sqrt(17)cos(θ) where θ is the angle between the two vectors.

Question b.
Solve for the angle between the directional paths ‘x’ and ‘y’ by making θ the subject of the formula in equation (2).[/QUOTE]
From the two different ways of finding the dot product, 20sqrt(17)cos(θ)= 52. θ= arcos(52/(20sqrt(17)). That's about 51 degrees.