Question

The vector
(i*a.b)i +(j*a.b)j+(k*a.b)k is equal to
Here i,j,k are unit vectors in x ,y and z directions...and
a.b represent the scalar product of two vectors a and b...

Answer 1

* symbol stands for cross product

Answer 2

@priyar

Answer 3

Options are
a*b
b*a
a
b

Answer 4

what is the cross product of a scalar and a vector?

Answer 5

Exactly !..i also thought the same

Answer 6

ya..it can't be that.. * must denote simple multiplication...not cross prod.

Answer 7

R u sure??
I mean how can we do then

Answer 8

coz a.b = scalar ..for ex:
say 2 so i*a.b = 2i

Answer 9

N its multiplication with i again will give us??

Answer 10

It will give us scalar ryt?

Answer 11

I mean if we assume your example only then 2i multiplied with i gives 2

Answer 12

ya but that is dot product.. is (2i)i = 2i.i ?

Answer 13

seeing the options i can say that they have intended * to be cross product.. so forget what we did abv.

Answer 14

Actually it cannot possibly be that simple product as we proposed it earlier

Answer 15

i think the question is wrong coz cross product of a vector and a scalar doesn't make any sense..like i x 2 =?? doesn't have any physical meaning..

Answer 16

Well, I don't think so bcoz it's not even given as bonus in ans key

Answer 17

bonus?

Answer 18

If any question is wromg or it"z options doesn't match with the correct ans then ans key gives bonus remark on it

Answer 19

we can use the theory of determinants, so we can write:
\[\Large i \times a = \det \left( {\begin{array}{*{20}{c}}
i&j&k \\
1&0&0 \\
{{a_x}}&{{a_y}}&{{a_z}}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{0,}&{ - {a_z},}&{{a_y}}
\end{array}} \right)\]
therefore, we get:
\[\Large \left( {i \times a} \right) \cdot b = {a_y}{b_z} - {a_z}{b_y} = {\left( {a \times b} \right)_x}\]
Similarly for other components
Then the the resultant vector is:
\[\Large {\left( {a \times b} \right)_x}i + {\left( {a \times b} \right)_y}j + {\left( {a \times b} \right)_z}k = a \times b\]

Answer 20

Gr8 sir .....wow!!!

Answer 21

thanks!! :D

Answer 22

yeah! why didn't we think of that.. just wrong grouping of terms!