Let a→ and b→ be two vectors such that |b→|=1 and |b→×a→|=2 Then |(b→×a→)-b→|^2 is equal to

Correct answer: (B) 5. Given: b→=1, b→×a→=2 Now, solving b→×a→-b→^2, ⇒b→×a→-b→^2=b→×a→^2+b→^2-2b→×a→·b→ Now, we know that, b→×a→·b→=b→×a→·a→=0 ⇒b→×a→-b→^2=2^2+1^2-2×0 ⇒b→×a→-b→^2=5

Q57 / 90 · JEE Main 2024 (30 Jan Shift 2) · Mathematics

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $| \vec{b} | = 1$ and $| \vec{b} \times \vec{a} | = 2$ Then $| (\vec{b} \times \vec{a}) - \vec{b} |^{2}$ is equal to

A
$3$
Correct answer
B
$5$
C
$1$
D
$4$

Solution

Given: $|\vec{b}| = 1, |\vec{b} \times \vec{a}| = 2$
Now, solving ${|\vec{b} \times \vec{a} - \vec{b}|}^{2}$ ,
$\Rightarrow {|\vec{b} \times \vec{a} - \vec{b}|}^{2} = {|\vec{b} \times \vec{a}|}^{2} + {|\vec{b}|}^{2} - 2 (\vec{b} \times \vec{a}) \cdot \vec{b}$
Now, we know that, $(\vec{b} \times \vec{a}) \cdot \vec{b} = (\vec{b} \times \vec{a}) \cdot \vec{a} = 0$
$\Rightarrow {|\vec{b} \times \vec{a} - \vec{b}|}^{2} = 2^{2} + 1^{2} - 2 \times 0$
$\Rightarrow {|\vec{b} \times \vec{a} - \vec{b}|}^{2} = 5$

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

Let a→ and b→ be two vectors such that |b→|=1 and |b→×a→|=2 Then |(b→×a→)-b→|2 is equal to

Solution

Relevant Questions

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $| \vec{b} | = 1$ and $| \vec{b} \times \vec{a} | = 2$ Then $| (\vec{b} \times \vec{a}) - \vec{b} |^{2}$ is equal to