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Q. If the area of the parallelogram with $\vec{a}$ and $\vec{b}$ as two adjacent sides is $15$ sq. units then the area of the parallelogram having $3 \vec{a}+2 \vec{b}$ and $\vec{a}+3 \vec{b}$ as two adjacent sides in sq. units is

KCETKCET 2021Vector Algebra

Solution:

$|\vec{ a } \times \vec{ b }|=15$
$|(3 \vec{ a }+2 \vec{ b }) \times(\vec{ a }+3 \vec{ b })|$
$=|9(\vec{ a }+\vec{ b }) \times 2(\vec{ b }+\vec{ a })|$
$=|7(\vec{ a } \times \vec{ b })|=7 \times 15=105$