I and J Are Unit Vectors Along X- and Y-axis Respectively. What is the Magnitude and Direction of the Vectors I+J and I−J What Are the Components of a Vector A=2i+3ja=2i+3j Along the Directions of I+Ji+J and I−Ji-j ? [You May Use Graphical Method] - Physics

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

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APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 4 Motion in a Plane
Q 22 | Page 87