Let a→ and b→ are two vectors such that a→=2, b→=3 and a→⋅b→=4. If c→=3a→×b→-4b→, then c→ is equal to

Correct answer: 18. c→^2=3a→×b→-4b→^2=9a→×b→^2+16b→^2-24a→×b→⋅b→ ∵a→×b→^2=a→^2b→^2-a→⋅b→^2=4×9-16=20 and a→×b→⋅b→=0 Hence, c→^2=9×20+16×9=324⇒c→=18

Q33 / 100 · NTA Abhyas (JEE Main) 2020 · Mathematics · Vector Algebra

Let $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}| = 2, |\vec{b}| = 3$ and $\vec{a} \cdot \vec{b} = 4 .$ If $\vec{c} = (3 \vec{a} \times \vec{b}) - 4 \vec{b},$ then $|\vec{c}|$ is equal to

Correct answer: 18

{|\vec{c}|}^{2} = {|(3 \vec{a} \times \vec{b}) - 4 \vec{b}|}^{2} = 9 {|\vec{a} \times \vec{b}|}^{2} + 16 {|\vec{b}|}^{2} - 24 (\vec{a} \times \vec{b}) \cdot \vec{b}

∵ {|\vec{a} \times \vec{b}|}^{2} = {|\vec{a}|}^{2} {|\vec{b}|}^{2} - {(\vec{a} \cdot \vec{b})}^{2} = 4 \times 9 - 16 = 20

and

(\vec{a} \times \vec{b}) \cdot \vec{b} = 0

Hence,

{|\vec{c}|}^{2} = 9 \times 20 + 16 \times 9 = 324 \Rightarrow |\vec{c}| = 18

Hence, the correct answer is 18.