Answer:
Yes, because \[\vec{a}+\vec{b}\] is represented by the diagonal of the parallelogram drawn with \[\vec{a}\] and \[\vec{b}\] as adjacent sides. The diagonal passes through the common tail of \[\vec{a}\] and\[\vec{b}\]. However, \[\vec{a}-\vec{b}\] is represented by the other diagonal of the same parallelogram not passing through the common tail of \[\vec{a}\] and \[\vec{b}\] Thus both \[\vec{a}+\vec{b}\] and \[\vec{a}-\vec{b}\] lie in plane of the same parallelogram.
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