question_answer

Find a vector \[\vec{A}\]and its magnitude with initial point P(1, 2, -1) and terminal point Q(3, 2, 2).

Answer:

                We know that and           \[|\vec{A}+\vec{B}|=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }\] \[|\vec{A}-\vec{B}|=\sqrt{{{A}^{2}}+{{B}^{2}}-2AB\cos \theta }\]                            As per questions              \[\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }=\sqrt{{{A}^{2}}+{{B}^{2}}-2AB\cos \theta }\] Squaring both the sides, we get \[\text{4 AB cos}\theta =0\] or \[\text{cos}\theta =0\]or\[\theta =\text{9}0{}^\circ \]. It means, the vectors \[\vec{A}\] and \[\vec{B}\]are perpendicular to each other.