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