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In scalar multiplication we are considering only logical values and doing multiplication but here we are taking care of the direction of the force magnitude of the force applied to the body for the perfect accurate value which helps in making of the real world application.

here scalar values will not work for making any real thing, So in your question cos (a,b) suggests the resultant direction of the resultant force and |a||b| are the magnitude.

For more details on vector multiplication :: http://mathworld.wolfram.com/VectorMultiplication.html

Posted on Sep 20, 2008

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Posted on Jan 02, 2017

The arccosine function is defined for arguments in the range from -1 to +1. 50/30 is greater than 1 and thus is out of the domain.

Feb 06, 2014 | Texas Instruments TI 30XIIS Scientific...

variables not defined, 3/2 not a valid argument. How did this pop up under my bissell proheat 2 search?

Oct 22, 2013 | Western Digital My Book Essential Edition...

Make sure the argument is between negative one and positive one, inclusive. The inverse cosine is defined only for arguments in that domain, anything else will give you a domain error.

Oct 26, 2011 | Texas Instruments TI-84 Plus Calculator

Let your system of linear equations be [A][X]=[B] where [A] is a 3x3 matrix, [X] is column vector ( a 3x1 matrix) , and [B] is also a column vector or (3x1) matrix.

Assuming you are able to invert the [A] matrix to find [A^-1].

Multiplying on the keft by [A^-1] the system of equations you get

[A^-1].[A][X]=[A^-1][B].

But since the product [A^-1][A] is the (3x3) Identity matrix, the left side of the equation is jut [X], while the right side is [A^-1][B]. As you can see it is not [B][A^-1] .

To solve your problem, you should define your [A] matrix, define the [B] column vector (3x1) matrix, then perform the operation [A^-1][B] using the [X to -1] power key to calculate the inverse [A^-1].

Assuming you are able to invert the [A] matrix to find [A^-1].

Multiplying on the keft by [A^-1] the system of equations you get

[A^-1].[A][X]=[A^-1][B].

But since the product [A^-1][A] is the (3x3) Identity matrix, the left side of the equation is jut [X], while the right side is [A^-1][B]. As you can see it is not [B][A^-1] .

To solve your problem, you should define your [A] matrix, define the [B] column vector (3x1) matrix, then perform the operation [A^-1][B] using the [X to -1] power key to calculate the inverse [A^-1].

Jan 10, 2011 | Casio FX-115ES Scientific Calculator

One not way would be to define a 4x4 matrix Mat A to hold the coefficients of the linear system. Then define a 4x1 column vector Mat V to hold the constants on the right.

Define a third 4x4 matrix Mat B you may leave filled with 0.

On command line, in Run Mat screen enter (Mat A) ^(-1) and store it in the zero-filled matrix Mat B. this is the inverse of Mat A.

If the inverse of Mat A exists, and it does in this case, the solution of the system is obtained as the column vector, resulting from the multiplication of Mat B by column vector Mat V

You can even shorten the procedure by just calculating ((Mat A)^-1)X (Mat V) [EXE]

To summarize

Multiplication operator is the regular [times] key.

Define a third 4x4 matrix Mat B you may leave filled with 0.

On command line, in Run Mat screen enter (Mat A) ^(-1) and store it in the zero-filled matrix Mat B. this is the inverse of Mat A.

If the inverse of Mat A exists, and it does in this case, the solution of the system is obtained as the column vector, resulting from the multiplication of Mat B by column vector Mat V

You can even shorten the procedure by just calculating ((Mat A)^-1)X (Mat V) [EXE]

To summarize

- Create 4x4 Mat A and type in the coefficients of the linear system.
- Create a 4x1 column vector Mat V for the right-hand sides
- Obtain you solution vector as ((Mat A)^-1)X (Mat V) [EXE]

- use catalog or
- in RunMat screen, press [OPTN] followed by [F2:Mat], then [F1:Mat].
- At this point the identifier is on command line, and you have to press [ALPHA] [X,Theta, T] to enter letter A.
- You use a similar key sequence to enter Mat V

Multiplication operator is the regular [times] key.

Nov 16, 2010 | Casio FX-9750GPlus Calculator

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Nov 07, 2010 | Vector 20 Million Power Series Spotlight

There is not much you can do with the information you provide. One has to assume the application point to be (0,0) and the angle the vector makes with the positive X-axis is 60 degrees

The X-Component is X_A=10*cos(60 deg)

The Y-component is Y_A=10*sin(60 deg)

To calculate the values of the X, and Y components you must make sure the angle unit is degree.

To get X_a type in 10 [COS] 60 [ENTER]

The X-Component is X_A=10*cos(60 deg)

The Y-component is Y_A=10*sin(60 deg)

To calculate the values of the X, and Y components you must make sure the angle unit is degree.

- Press [MODE]
- Move cursor to the line that says RADIAN DEGREE
- Highlight DEGREE and press [ENTER].
- Press [2nd][QUIT] to exit the configuration screen.

To get X_a type in 10 [COS] 60 [ENTER]

Oct 08, 2010 | Texas Instruments TI-84 Plus Calculator

You are indeed committing an error. The sequence [2nd][COS] is activating the function arcosine or arccos or cos^-1, the inverse of the cosine function. If you remember the properties of the cosine functions, you know that cos(x) is defined over the real line ]- infinity to infinity[, but its range spans the interval [-1,1].

Since the arcosine function is the inverse of the cosine, its domain of definition is the range of the cosine, namely the closed interval [-1,1].

Thus if you enter [2nd][COS][3180.04] the calculator flags this as a domain error, because 3180.04 is outside the interval [-1,1]

Restrict the argument of cos^-1 to any value inside the closed interval [-1,1].

When manipulating the trigonometric functions and their inverses you must keep in mind that the results you get are dependent on the angle unit your calculator is configured for (deg, rad).

Since the arcosine function is the inverse of the cosine, its domain of definition is the range of the cosine, namely the closed interval [-1,1].

Thus if you enter [2nd][COS][3180.04] the calculator flags this as a domain error, because 3180.04 is outside the interval [-1,1]

Restrict the argument of cos^-1 to any value inside the closed interval [-1,1].

When manipulating the trigonometric functions and their inverses you must keep in mind that the results you get are dependent on the angle unit your calculator is configured for (deg, rad).

Jun 29, 2010 | Texas Instruments TI-84 Plus Calculator

Resultant of addition of two vectors A and B is calculated as the square root of (square of A + square of B + 2AB cosx) where x is the angle between the two vectors.

If the angle between the vectors increases, we need to keep in mind the behaviour of cos for different values of x. The resultant be maximum for x=0 and continue to decrease till x=180 but will again increase to a max value for x=360; the same as for x=0.

If the angle between the vectors increases, we need to keep in mind the behaviour of cos for different values of x. The resultant be maximum for x=0 and continue to decrease till x=180 but will again increase to a max value for x=360; the same as for x=0.

Feb 03, 2009 | Computers & Internet

Hello,

**The e is the same, it is the exponential**. According to Euler's relation

**e^(i theta) = cos(theta) + i sin(theta),** where** i** is the imaginary unit.

When represented on the complex plane (x,iy) the point (cos(theta), sin(theta)) is at the extremity of a vector of length 1 and making an angle theta with the real axis.

In (plane) polar coordinates, a point is defined by the radius r, and the angle, theta, it makes with the x axis, measured in the trigonometric (counterclockwise) direction. It is structurally equaivalent to representing it in the complex plane as r*e^(i*theta). Since r is the measure ot is radius, and the theta is it argument (angle). The complex notation is used for its convenience when adding vectors (as is AC circuits)

That is the theory.

I am inserting a clipping from the book to show you how to convert between polar and rectangular coordinates.

When represented on the complex plane (x,iy) the point (cos(theta), sin(theta)) is at the extremity of a vector of length 1 and making an angle theta with the real axis.

In (plane) polar coordinates, a point is defined by the radius r, and the angle, theta, it makes with the x axis, measured in the trigonometric (counterclockwise) direction. It is structurally equaivalent to representing it in the complex plane as r*e^(i*theta). Since r is the measure ot is radius, and the theta is it argument (angle). The complex notation is used for its convenience when adding vectors (as is AC circuits)

That is the theory.

I am inserting a clipping from the book to show you how to convert between polar and rectangular coordinates.

Oct 10, 2008 | Casio FX1.0 Plus Calculator

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