Let a and b be two vectors such that | b | = 1 and | b × a | = 2 . Then | ( b × a ) – b | 2 is equal to [2024]

Question

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 [2024]

Smart Achievers · Accepted Answer

(1)

Given, $| \vec{b} |$ $| \vec{b} |$ and $| \vec{b} \times \vec{a} |$ $= 2$

Now, $| (\vec{b} \times \vec{a}) - \vec{b} |^{2}$ $=$ $| \vec{b} \times \vec{a} |^{2} - 2 (\vec{b} \times \vec{a}) \cdot \vec{b} + | \vec{b} |^{2}$

$= {(2)}^{2} + 1 - 2 (0) = 5$ .

Solution

Q.
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 [2024]

1 5

2 1

3 4

4 3

Ans.
(1)

Given, $| \vec{b} |$ $= 1$ and $| \vec{b} \times \vec{a} |$ $= 2$

Now, $| (\vec{b} \times \vec{a}) - \vec{b} |^{2}$ $=$ $| \vec{b} \times \vec{a} |^{2} - 2 (\vec{b} \times \vec{a}) \cdot \vec{b} + | \vec{b} |^{2}$

$= {(2)}^{2} + 1 - 2 (0) = 5$ .

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Solution

Q. Let a→ and b→ be two vectors such that |b→|=1 and |b→×a→|=2. Then |(b→×a→)–b→|2 is equal to [2024]

1 5

2 1

3 4

4 3

Ans. (1) Given, |b→|=1 and |b→×a→|=2 Now, |(b→×a→)–b→|2=|b→×a→|2–2(b→×a→)·b→+|b→|2 =(2)2+1–2(0)=5.

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Q.
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 [2024]

Ans.
(1)

Given, $| \vec{b} |$ $= 1$ and $| \vec{b} \times \vec{a} |$ $= 2$

Now, $| (\vec{b} \times \vec{a}) - \vec{b} |^{2}$ $=$ $| \vec{b} \times \vec{a} |^{2} - 2 (\vec{b} \times \vec{a}) \cdot \vec{b} + | \vec{b} |^{2}$

$= {(2)}^{2} + 1 - 2 (0) = 5$ .