If a|a→| = 3 and –1 ≤ k ≤ 2, then ka|ka→| lies in the interval ______. - Mathematics

Question

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

Options

[0, 6]
[– 3, 6]
[3, 6]
[1, 2]

MCQ

Fill in the Blanks

Solution

If $| \vec{a} |$ = 3 and –1 ≤ k ≤ 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 × 3

= 0

The numerically greatest value of k is 2 at which $| k \vec{a} |$ = 6.

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If a|a→| = 3 and –1 ≤ k ≤ 2, then ka|ka→| lies in the interval ______. - Mathematics

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