the answer is (b) true
Let v be a vector in Rn in the form "v=(v_{1}, v_{2}, v_{3}, . . . . ., v_{n})" ,
then
"|| v_{1} ||^{2}= (v_{1})^{2}+ (v_{2})^{2}+(v_{3})^{2}+ . . . . .+(v_{n})^{2}"
since "v_{1}, v_{2}, v_{3}, . . . . ., v_{n}" are real numbers , then
"(v_{1})^{2}+ (v_{2})^{2}+(v_{3})^{2}+ . . . . .+(v_{n})^{2}" is a positive real number ( scalar )
Thus "|| v_1 ||= \\sqrt{v_1^2+ v_{2}^2+v_{3}^2+ ...+v_n^2}" is a positive real number,
i.e. "|| v_{1} ||" is scalar.
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