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The position vectors of the vertices `A ,Ba n dC` of a triangle are three unit vectors ` vec a , vec b ,a n d vec c ,` respectively. A vector ` vec d` is such that ` vec ddot vec a= vec ddot vec ba n d vec d=lambda( vec b+ vec c)dot` Then triangle `A B C` is a. acute angled b. obtuse angled c. right angled d. none of these

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The position vectors of the vertices A ,Ba n dC of a triangle are three unit vectors vec a , vec b ,a n d vec c , respectively. A vector vec d is such that vecd dot vec a= vecd dot vec b= vec d dot vec ca n d vec d=lambda( vec b+ vec c)dot Then triangle A B C is a. acute angled b. obtuse angled c. right angled d. none of these

If vec a , vec b , vec c are three non coplanar vectors such that vec adot vec a= vec d vec b= vec ddot vec c=0 , then show that vec d is the null vector.

If vec a ,\ vec b ,\ vec c are three vectors such that vec adot vec b= vec adot vec c then show that vec a=0\ or ,\ vec b=c\ or\ vec a_|_( vec b- vec c)dot

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Find the angle between two vectors vec a\ a n d\ vec b , if | vec axx vec b|= vec adot vec bdot

Let vec a , vec b , vec c be unit vectors such that vec adot vec b= vec adot vec c=0 and the angle between vec ba n d vec c is pi/6,t h a t vec a=+-2( vec bxx vec c)dot

Prove that the points A ,Ba n d C with position vectors vec a , vec ba n d vec c respectively are collinear if and only if vec axx vec b+ vec bxx vec c+ vec cxx vec a= vec0dot

If vec a\ a n d\ vec b are two vectors such that | vec a|=| vec axx vec b|, write the angle between vec a\ a n d\ vec bdot

If vec a , vec b , vec ca n d vec d are distinct vectors such that vec axx vec c= vec bxx vec da n d vec axx vec b= vec cxx vec d , prove that ( vec a- vec d). (vec b- vec c)!=0,

If vec a\ a n d\ vec b are two vectors such that | vec axx vec b|=3\ a n d\ vec adot vec b=1, find the angle between vec a\ a n d\ vec b .

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