Vector Algebra ...

Let $$\vec{a}=\hat{i}-\hat{j},\vec{b}=\hat{i}-\hat{j}=\vec{c}=\hat{i}-\hat{j}$$, if $$\vec{d}$$ is a unit vector such that $$\vec{a}.\vec{d}=0=|\vec{b}\vec{c}\vec{d}|$$ then $$\vec{d}$$ equals:

If $$\vec a = 4 i + 5 j - k , \vec { b } = i - 4 j + 5 k , \vec c = 3 i + j - k$$ such that $$\vec { p } \perp \vec { a } , \vec { p } \perp \overline { b }$$ and $$\vec { p . c } = 21 ,$$ then $$p =$$

Vector $$\vec { x }$$ satisfying the relation $$\vec { A } . \overline { x } = c$$ and $$\vec { A } \times \vec { x } = \vec { B }$$ is

For any vector $$\vec a$$, the value of $${ (\vec a\times \hat i) }^{ 2 }+{ (\vec a\times \hat j) }^{ 2 }+{ (\vec a\times \hat k) }^{ 2 }$$ is equal to

In a parallelogram ABD, $$|\overset { \_  }{ A\overset { \rightarrow  }{ B }  } |=a,|\overset { \_  }{ A\overset { \rightarrow  }{ D }  } |=b$$ and $$|\overset { \_ \\ \quad \quad \rightarrow  }{ AC } |=c,$$, $$\overset { \_ \rightarrow  }{ AB } .\overset { \_ \rightarrow  }{ DB } $$ has the value :

If the position vectors of the vertices $$A,B$$ and $$C$$ of a $$\Delta ABC$$ are respectively $$4\hat{i}+7\hat{j}+8\hat{k},2\hat{i}+3\hat{j}+4\hat{k}$$ and $$2\hat{i}+5\hat{j}+7\hat{k}$$, then the position vector of the point, where the bisector of $$\angle A$$ meets $$BC$$ is:

let $$\bar { a } $$ be a unit vector and $$\bar { b } $$ be a non-zero vector not parallel to $$\bar { a } $$ if two sides of a triangle are represented by the vectors $$\sqrt { 3 } \left( \bar { a } \times \bar { b }  \right) \quad and\quad \bar { b } - \left( \bar { a } .\bar { b }  \right) \bar { a } $$ then the angles of triangle are 

In a triangle ABC, if $$A=(0, 0), B=(3, 3\sqrt{3}), C=(-3\sqrt{3}, 3)$$ then the vector of magnitude $$2\sqrt{2}$$ units directed along $$\overline{AO}$$, where O is the circumcentre of triangle ABC is?

If $$a=i-j,b=i+j,c=i+3j+5k$$ and $$n$$ is a unit vector such that $$b,n=0,a,n=0$$ then the value of $$|c,n|$$ is equal to

$$If\quad |\overrightarrow { a } |$$ =2 and $$\quad |\overrightarrow { b } |$$=3 and $$\quad |\overrightarrow { a } |$$.$$\quad |\overrightarrow { b } |$$ =0. Then (a x (a x (a x (a x b)))) is equal to