If |vec"a"| = 3 and –1 $\leq$ k $\leq$ 2, then |"k"vec"a"| lies in the interval ______.

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If |vec"a"| = 3 and –1 $\leq$ k $\leq$ 2, then |"k"vec"a"| lies in the interval [0, 6]. Explanation: The smallest value of |"k"vec"a"| will exist at numerically smallest value of k i.e., at k = 0 Which gives |"k"vec"a"| = |"k"||vec"a"| = 0 $\times$ 3 = 0 The numerically greatest value of k is 2 at which |"k"vec"a"| = 6.

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