Question

Question ;-

=> Write all the unit vector in XY - plane .



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Answer 1

$\rm\underline\bold{ SolUtIon \red{\huge{\checkmark}}}$

Let the unit be å

We know that

å = xî + yỹ + zk

Since the vector is in X-Y plane, there is no Z- coordinate. Hence,

å = xî + yỹ + Ok

d = xî + yỹ

Taking a general vector à, making an angle with X- axis

Unit vector in direction of x axis is î & in y axis is ſ

Given that

ā makes an angle of 0 with x-axis

So, angle between å & î is e

Now, we know that,

ä.b = lä||b| cos 0,

Putting å = å , b =i&e=0

å î = Jä||î] cos 0

å î = 1x 1 x cos

ä .î = cos e

(xî + yf + Ok). î = cos e

(xi + y) + Ok). (1î + of + Ok) = cos e x.1 + y.0 + 0.0 = cos 0 X = cos e

Similarly,

å makes an angle of (90°-) with y-axis So, angle between a & j is (90° - )

Now, we know that,

ä.b = lä|b cos 0,

Putting a = a, b = j & = (90°-0)

å .j = lalli cos (90° – 0)

.j = 1x1x cos (90°-0)

a .j = cos (90°- 8)

ä .j = sin e

(xî + yỹ + Ok). ĵ = sin e

(xî + yỹ + 0k). (0î + 1j + Ok) = sin 0

x.0 + y.1 + 0.0 = sin 0

y = sin 0

So, x= cos 0 and y = sin 0

So, x = cos and y = sin 0

Thus,

d = xî + yỹ

= cos eî + sin 0 î

teachoo.com

: a = cos e î + sin 0 j

This value will be true in all quadrants

Quadrant 1: Os 0 < 90°

Quadrant 2: 90° x< 180°

Quadrant 3: 180° <0 < 270°

Quadrant 4: 270° se< 360°

90°

x(t)

So, 0° < 0s360°

i.e. 0 s 0s2M

So, d = cos Bî + sinej; for 0 ses 2n

Answer 2

Every vector in xy plane can be written as :

$\sf{\vec{u}=x\hat{i}+y\hat{j}}$

• But as $\sf{\vec{u}}$ is a unit vector : $\sf{x^2+y^2=1}$

So magnitude cannot change but we can change the direction of unit vector to get new vector, so :

$\large\boxed{\sf{\vec{u}=cos\theta\:\hat{i}+sin\theta\: \hat{j}}}$