`hati "and" hatj` are unit vectors along *x*- and *y*-axis respectively. What is the magnitude and direction of the vectors `hati+hatj` and `hati-hatj` ? What are the components of a vector `A = 2hati + 3hatj` along the directions of `hati + hatj` and `hati - hatj` ? [You may use graphical method]

#### Solution 1

Consider a vector `vecP` , given as:

`vecP = hati+hatj`

`P_x hati + p_y hatj = hati+hatj`

On comparing the components on both sides, we get:

`P_x = P_y = 1`

`|vecP| = sqrt(p_x^2+P_y^2) = sqrt(1^2 + 1^2) = sqrt2` ...(i)

Hence, the magnitude of the vector `hati+hatj is sqrt2`

Let Θ be the angle made by the vector` vecP`, with the *x*-axis, as shown in the following figure.

`:.tantheta = (P_y/P_x)`

`theta = tan^(-1)(1/1) = 45^@` ....(ii)

Hence the vector `hati + hatj` makesan angle of `45^@` with the x-axis.

Let `vecQ = hati-hatj`

`Q_x hati - Q_y hatj = hati -hatj`

`Q_x=Q_y = 1`

`|vecQ| = sqrt(Q_x^2+Q_y^2) =sqrt2` ... (iii)

Hence, the magnitude of the vector `hati -hatj` is `sqrt2`.

Let Θ be the angle made by the vector `vecQ`, with the *x*- axis, as shown in the following figure

:.tantheta = `(Q_y/Q_x)`

`theta = -tan^(-1)(-1/1) = -45` ...(iv)

Hence the vector `hati-hatj` makes an angle of `-45^@` with the x-axis.

It is given that

`vecA = 2hati + 3hatj`

`A_x hati+A_y hatj = 2hati + 3hatj`

On comparing the coefficients of `hati "and" hatj` , we have:

`A_x = 2 and A_y = 3`

`|vecA| =sqrt(2^2+3^3) = sqrt13`

Let `vecA_x` make an angle Θ with the *x*-axis, as shown in the following figure.

`:. tan theta = (A_y/A_x)`

`theta = tan^(-1)(3/2)`

`=tan^(-1)(1.5) = 56.31^@`

Angle between the vectors` (2hati + 3hatj)`

and `(hati+hatj)` , `theta = 56.31 - 45 = 11.31^@` Component of vector `vecA` , along the direction of `vecP`, making an angle Θ

=`(Acostheta)p = (Acos11.31) (hati+hatj)/sqrt2`

=`sqrt13xx 0.9806/sqrt2 (hati+hatj)`

=`2.5(hati+hatj)`

=`25/10 xx sqrt2`

=`5/sqrt2` ...(v)

Let Θ be the angle between the vectors `(2hati+3hatj)` and `(hati-hatj)`.

`theta^" = 45 + 56.31 = 101.31^@`

Component of vector `vecA`, along the direction of `vecQ`, making an angle `theta^"`

=`(Acostheta^")vecQ = (Acostheta^")(hati-hatj)/sqrt2`

=`sqrt13cos(901.31^@)(hati-hatj)/sqrt2`

`=sqrt(13/2) sin 11.30^@ (hati-hatj)`

`=-2.550xx 0.1961(hati-hatj)`

=`-0.5(hati-hatj)`

`=-5/10xxsqrt2`

`=-1/sqrt2` ... (vi)

#### Solution 2

1) `hati+hatj = sqrt((1)^2 + (1)^2+2xx1xx1xxcos 90^@) = sqrt2` = 1.414 unit

`tan theta = 1/1 = 1 :. theta = 45^@`

So the vector `hati+hatj` makes an angle `45^@` with x-axis

2) `|hati-hatj| = sqrt((1)^2 +(2)^2 - 2 xx 1xx1xxcos 90^@`

`= sqrt2` = 1.414 units

The vector `hati -hatj` makess an angle `-45^@` with x-axis

3)Let us now determined the component of `vecA = 2hati+3hatj` in the direction of `hati+hatj`

Let `vecB = hati+hatj`

`vecA.vecB = AB cos theta = (Acostheta)B`

So the component of `vecA` in the direction of `vecB` = `(vecA.vecB)/B`

`=((2hati+3hatj).(hati+hatj))/sqrt((1)^2+(1)^2) = (2hati.hati+2hati.hatj+3hatj.hati+3hatj.hatj)/sqrt2 = 5/sqrt2 units`

4) Component of `vecA` in the drection of `hati-hatj = ((2hati+3hatj).(hati-hatj))/sqrt2 = -1/sqrt2` units