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

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प्रश्न

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

विकल्प

  • [0, 6]

  • [– 3, 6]

  • [3, 6]

  • [1, 2]

MCQ
रिक्त स्थान भरें

उत्तर

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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अध्याय 10: Vector Algebra - Solved Examples [पृष्ठ २१४]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 10 Vector Algebra
Solved Examples | Q 21 | पृष्ठ २१४

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