Q23 / 100 · NTA Abhyas (JEE Main) 2020 · Mathematics · Vector Algebra

Consider a parallelogram constructed on the vectors A→=5p→+2q→ and B→=p→-3q→. If p→=2, q→=5, the angle between p→ and q→ is π3 and the length of the smallest diagonal of the parallelogram is k units, then the value of k2 is equal to

Correct answer: 109

Solution

Diagonals of the parallelogram are A→+B→, A→-B→
=6p→-q→,4p→+5q→
A→+B→=36p→2+q→2-12p→⋅q→
=144+25-12×2×5×cosπ3
=109
A→-B→=16p→2+25q→2+40p→⋅q→
=16×4+25×25+40×2×5×cosπ3
=889
So, length of the smaller diagonal is 109 units

Hence, the correct answer is 109.

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