A)
then angle between a and b is 
.
B)
is a vector in the direction of angle bisector of vectors 
.
C)
D)
Vector perpendicular to 
and coplanar with 
and
.
Correct Answer: C
Solution :
[a] \[|\vec{a}|\,\,=\,\,|\vec{b}|\,\,=\,\left| \vec{a}-\vec{b} \right|=1\] \[{{\operatorname{a}}^{2}}+{{b}^{2}}-2\vec{a}.\vec{b}=1\] \[\therefore \,Angle between \vec{a}\And \vec{b} \,is\, \frac{\pi }{3}\] [b] A vector in the direction of angle bisector of vectors \[\vec{a} and \vec{b} \,is\,\,\vec{a}+\vec{b}\] \[\therefore \] The given statement is not correct. [c] \[{{(\vec{a}.\hat{i})}^{2}}+{{(\vec{a}.\hat{j})}^{2}}+{{(\vec{a}.\hat{k})}^{2}}={{\left| \,\vec{a}\, \right|}^{2}}={{a}^{2}}\] [d] Any vector in the plane \[\hat{i}\,\,+\,\,\hat{j}\,\,+\hat{k}\] and \[-\hat{i}\,\,+\,\,\hat{j}\,\,+\hat{k}\] is of the form \[\alpha \,(\hat{i}\,+\hat{j}+\hat{k})+\,\beta (-\hat{i}+\hat{j}+\hat{k})\] \[=\,\,\,(\alpha -\beta )\hat{i}\,\,+\,\,(\alpha +\beta )\hat{j}\,\,+(\alpha +\beta )\,\hat{k}\] This will be perpendicular with \[-\hat{i} +\hat{j}+\hat{k}\] if \[-\left( \alpha -\beta \right)\,\hat{i}\,+\left( \alpha +\beta \right)\,\hat{j}\,+\left( \alpha +\beta \right)=0 \Rightarrow \,\,\alpha \,=-\,3\,\beta \] Hence the required vector is of the form \[\left( -\,4i\,-2j\,-2k \right)\] \[\therefore \,\,statement is false.\]
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