In this page, you will learn about the alternate representation of vectors other than `ai+bj+ck`?

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What is 'cosine ' of an angle?

- Cosine of an angle does not exist.
- the ratio of adjacent side to the hypotenuse in a right angled triangle
- the ratio of adjacent side to the hypotenuse in a right angled triangle

Answer is 'the ratio of adjacent side to the hypotenuse in a right angled triangle'.

While referring to vector quantities, we used to switch between two representations. One with components along the axes. What is the other?

- magnitude
- angles with the axes
- magnitude along with angles with the axes
- magnitude along with angles with the axes
- none of the above

Answer is 'magnitude along with angles with the axes'. An alternate representation to component form is to specify magnitude and angles made by the vector on the axes.

What information is complete to describe the vector `ai+bj` shown in the figure?

- `a`
- `r`
- `theta`
- `r` and `theta`
- `r` and `theta`

answer is '`r` and `theta`'. If these two parameters are available we can derive the vector notation `ai+bj`.

What is the value of `a` and `b` in the figure?

- `a=8cos60` and `b=8sin60`
- `a=8cos60` and `b=8sin60`
- `a=8` and `b=8`
- `a=cos60` and `b=sin60`
- `a=sqrt(8^2cos60)` and `b=sqrt(8^2sin60)`

answer is '`a=8 cos 60` and `b=8 sin 60`'.

What information is complete to describe the vector `ai+bj+ck` shown in figure?

- `a`
- only `r`
- `alpha, beta, gamma`
- `r` and `alpha, beta`
- `r` and `alpha, beta`

answer is '`r` and `alpha, beta`'. If these three parameters are available we can derive the vector notation `ai+bj+ck`.

Note that the third angle can be derived from the other two angles.

What are the values of `a`, `b`, and `c` in the figure?

- `a=3cos60`; `b=3cos45`; `c=3cos75`
- `a=3cos60`; `b=3cos45`; `c=3cos75`
- `a=3`; `b=3`; `c=3`
- `a=cos60`; `b=cos45`; `c=cos75`
- `a=60`; `b=45`; `c=75`

answer is '`a=3cos60`; `b=3cos45`; `c=3cos75`'

When a vector with magnitude `r` and angles `alpha,beta,gamma` is given, the coordinate form of the vector is

`r(cos alpha i+cos beta j+cos gamma k)`

`cos alpha`, `cos beta`, `cos gamma` are called the ** directional cosines** of the vector.

**Directional cosines** are the ratio of projections on to an axes to the magnitude.

**Directional cosines: ** Given that a vector `vec p = ai+bj+ck` makes angles `alpha,beta,gamma` with `x,y,z`-axes respectively, then the directional cosines of the vector are

`cos alpha= a/(|p|)`

`cos beta= b/(|p|)`

`cos gamma= c/(|p|)`

The mathematical representation of vectors is the component form `vec p = ai+bj+ck`

Directional cosines along with magnitude provide an alternate representation of a vector.

`vec p = r (cos alpha i + cos beta j + cos gamma k)`

What is the magnitude of the directional cosines vector : `cos alpha i+cos beta j+cos gamma k` ?

Note that

`cos alpha =a/r`

`cos beta =b/r`

`cos gamma =c/r`

`r^2=a^2+b^2+c^2`

- `1`
- `1`
- `0`
- `r`
- `sqrt(r)`

Answer is '`1`'. The magnitude = `sqrt(cos^2 alpha+cos^2 beta+cos^2 gamma)`, which evaluates to 1.

** Directional cosines make Unit Vector: ** For a given vector `ai+bj+ck` the directional cosine vector `li+mj+nk` is the unit vector in the direction of the given vector. Note that

`l = cos alpha =a/r`

`m = cos beta =b/r`

`n = cos gamma =c/r`

This also implies that

`l^2+m^2+n^2=1`

*Solved Exercise Problem: *

Find the direction cosines of a line which makes equal angles with the coordinate axes.

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Answer is

`(1/sqrt(3),1/sqrt(3),1/sqrt(3))` OR `(-1/sqrt(3),-1/sqrt(3),-1/sqrt(3))`

From the question, we know that `alpha=beta=gamma` and We know the property of directional cosines `cos^2 alpha+cos^2 beta+cos^2 gamma = 1 `

*Solved Exercise Problem: *

If a line makes angles `45,135, 90` with the x, y and z-axes respectively, find its direction cosines.

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Answer is '`cos45, cos135, cos90`'.

*Solved Exercise Problem: *

If a line makes angles `30,60,90` with the x, y and z-axes respectively, find its direction cosines.

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Answer is '`cos30, cos60, cos90`'

*Solved Exercise Problem: *

Find the directional cosine of the vector `2i-3j+sqrt(3)k`.

- tap for the answer

magnitude of the vector = 4

So directional cosines are `2/4 , -3/4, sqrt(3)/4`

*slide-show version coming soon*