Question

Q5. (a) If vec V =(2xy^ 2 ) hat i -(yz) hat j +(3xz^ 3 ) hat k find curl curl vec v at (1, 1, 1)​

Answer 1

$\huge\red{A}\pink{N}\orange{S} \green{W}\blue{E}\gray{R} =$

Let the given vectors are a = -25+ 3k and

b = 31-2f+R

lal = √1² + (-2)² + 3² = √√√1+4+9= √14

||5| = √√3² + (−2)² + 1² = /9+4+1=√14

Now, a b = (-2+ 3k)(3î − 2ĵ) + k)

Also, we know that a blabl cos 0

= 1.3 + (-2)(-2) + 3.1

= 3+4+3 = 10

.. 10 = √14√14 cos 0

$\cos0 = \frac{10}{14}$

$0 = \cos^{ - 1} ( \frac{5}{7} )$

Answer 2

Answer:

Let the given vectors are a = -25+ 3k and

b = 31-2f+R

lal = √1² + (-2)² + 3² = √√√1+4+9= √14

||5| = √√3² + (−2)² + 1² = /9+4+1=√14

Now, a b = (-2+ 3k)(3î − 2ĵ) + k)

Also, we know that a blabl cos 0

= 1.3 + (-2)(-2) + 3.1

= 3+4+3 = 10

.. 10 = √14√14 cos 0

cos 0 = 10/14.

0 = cos −1( 5/7 ).

$thanks$