IIT JEE , AIEEE Forum - Mathematics , Vectors

rphy

if p satisfies p X a = b and p.a=0 for non zero vectors a and b, then x is equal to :

karthik_abiram

x???

i dont understand ur question...........but i think this is wat u want.........

if p satisfies p X a = b and p.a=0 for non zero vectors a and b, then x is equal to :

#p.a=0

For non -zero vectors ,from this condition implies that "p" and "a" are perpendicular...................................eq1

#p X a = b

For non-zero vectors ,by definition of cross product,"b" is perpendicular both to "p" and "a"......................................eq2

From eq1 and eq2 we can infer that vectors "p","a" and "b" are mutually perpendicular.

I think this is what u want.............if u want any other plz specify.

rphy

oh sorry.... the question is: if p satisfies p X a = b and p.a=0 for non zero vectors a and b, then p is equal to :

karthik_abiram

Oh itz ok.........

SOLUTION:-

I think u know this..............

(p x a) ² + ( p . a)² = | p |²+ | a |²

--------eq1---------if not ,try to prove it...if u can't get it i will post the proof for this relation also.

substituting the known values in eq1 we get.........

( b )² + ( 0 )² = | p |² + | a |²

| b | ²

-------- = | p |²

| a |²

Taking square root,we get

| p | = +_ | b |

-----------

| a |

Note :-

+_ is plus r minus

rphy

thanks karthik, but we have to find the vector p and not just the magnitude..

also please post the proof of (p x a) ² + ( p . a)² = | p |²* | a |²
if possible

karthik_abiram

PROOF:-

( p x a )² + ( p . a )²

={ | p | | a | sin

( n cap ) }² + { | p | | a | cos

}²

=| p |² | a |²sin²

( 1 ) + | p |² | a |² cos²

=| p |² | a |² {sin²

+ cos²

}

=| p |² | a |² {1 }

=| p |² | a |²

------------------------------

p vector = +_ | b |( b x a )

--------- -------

| a | |b x a |

since , b x a is in the direction of p vector.

Aatish

hey Karthik plssssssss tell me that how u got

p vector = +_ | b |( b x a )
--------- -------
| a | |b x a |

since , b x a is in the direction of p vector.?????????????

Thanks.......

karthik_abiram

u got that modulus of p vector....dont u?

p , b , a are mutually perpendicular.

by definition of cross product, in ( b x a) the resultant will be perpendicular both to b and a.............the direction of p vector is also perpendicular both to b and a.

hence, p vector and ( b x a) are in the same direction.

magnitude of p vector is already found.

to show direction a unit vector in the direction of ( b x a) is taken.........i.e

( b x a )

---------------

| b x a |

karthik_abiram

The above solution that i gav is correct........there is also another solution.....

here it is

p . a = o-----------------------eq1

This implies that "p" and "a" are perpendicular.

Also given

p x a = b

crossing both sides with a vector we get

( p x a ) x a = b x a

( a . p ) a - ( a . a ) p = b x a

( 0 ) a - ( a )² p = b x a { from eq1 }

- ( a )² p = - a x b

( a )² p = a x b

p = a x b

----------

a²

phanindraramesh

If the magnintude of vector p is to be founded

thye given information is pxa=b

and pxa=0

so the angle between p,a is 90

So magnitude of pxa= /p/./a/=/b/

So /p/ =/b/ divided by/a/

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