Let a→ and b→ are unit vectors such that a→+b→=3/2, then the value of 2a→+7b→⋅4a→+3b→+2020a→×b→ is equal to

Correct answer: (A) 133/4. a→+b→^2=9/4⇒1+1+2a→⋅b→=9/4⇒a→⋅b→=1/8 2a→+7b→⋅4a→+3b→+2020a→×b→ =8a→^2+6a→⋅b→+0+28a→⋅b→+21b→^2+0 =8+6×1/8+28×1/8+21=29+17/4=133/4

Q47 / 100 · NTA Abhyas (JEE Main) 2020 · Mathematics · Vector Algebra

Let $\vec{a}$ and $\vec{b}$ are unit vectors such that $|\vec{a} + \vec{b}| = \frac{3}{2}$ , then the value of $(2 \vec{a} + 7 \vec{b}) \cdot (4 \vec{a} + 3 \vec{b} + 2020 \vec{a} \times \vec{b})$ is equal to

Correct answer
A
$\frac{133}{4}$
B
$133$
C
$30$
D
$120$

Solution

{|\vec{a} + \vec{b}|}^{2} = \frac{9}{4} \Rightarrow 1 + 1 + 2 \vec{a} \cdot \vec{b} = \frac{9}{4} \Rightarrow \vec{a} \cdot \vec{b} = \frac{1}{8}

(2 \vec{a} + 7 \vec{b}) \cdot (4 \vec{a} + 3 \vec{b} + 2020 (\vec{a} \times \vec{b}))

= 8 {|\vec{a}|}^{2} + 6 \vec{a} \cdot \vec{b} + 0 + 28 \vec{a} \cdot \vec{b} + 21 {|\vec{b}|}^{2} + 0

= 8 + 6 \times \frac{1}{8} + 28 \times \frac{1}{8} + 21 = 29 + \frac{17}{4} = \frac{133}{4}

Hence, the correct answer is option: 'A'.

Let a→ and b→ are unit vectors such that a→+b→=32, then the value of 2a→+7b→⋅4a→+3b→+2020a→×b→ is equal to

Solution

More Vector Algebra Questions

Let $\vec{a}$ and $\vec{b}$ are unit vectors such that $|\vec{a} + \vec{b}| = \frac{3}{2}$ , then the value of $(2 \vec{a} + 7 \vec{b}) \cdot (4 \vec{a} + 3 \vec{b} + 2020 \vec{a} \times \vec{b})$ is equal to