the answer is (b) true
Let v be a vector in Rn in the form v=(v1,v2,v3,.....,vn)v=(v_{1}, v_{2}, v_{3}, . . . . ., v_{n})v=(v1,v2,v3,.....,vn) ,
then
∣∣v1∣∣2=(v1)2+(v2)2+(v3)2+.....+(vn)2|| v_{1} ||^{2}= (v_{1})^{2}+ (v_{2})^{2}+(v_{3})^{2}+ . . . . .+(v_{n})^{2}∣∣v1∣∣2=(v1)2+(v2)2+(v3)2+.....+(vn)2
since v1,v2,v3,.....,vnv_{1}, v_{2}, v_{3}, . . . . ., v_{n}v1,v2,v3,.....,vn are real numbers , then
(v1)2+(v2)2+(v3)2+.....+(vn)2(v_{1})^{2}+ (v_{2})^{2}+(v_{3})^{2}+ . . . . .+(v_{n})^{2}(v1)2+(v2)2+(v3)2+.....+(vn)2 is a positive real number ( scalar )
Thus ∣∣v1∣∣=v12+v22+v32+...+vn2|| v_1 ||= \sqrt{v_1^2+ v_{2}^2+v_{3}^2+ ...+v_n^2}∣∣v1∣∣=v12+v22+v32+...+vn2 is a positive real number,
i.e. ∣∣v1∣∣|| v_{1} ||∣∣v1∣∣ is scalar.
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