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

Given: | b| = 1 and | b a| = 2. Find: |( b a) - b|^2. Since b a is perpendicular to b, we use |( b a) - b|^2 = | b a|^2 + | b|^2 - 2( b a) b Now, ( b a) b = 0 because a cross product is perpendicular to each of the vectors involved. Therefore, |( b a) - b|^2 = 2^2 + 1^2 = 4 + 1 = 5 Hence, the value is 5. So the correct option is B. The solution appears unrelated to this question, so the working above is obtained directly from the given question data.

Let a and b be two vectors such that | b| = 1 and | b a| = 2. The… — Solution | JEE Mathematics

MCQEasyJEE 2024Cross Product

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
$3$
B
$5$
C
$1$
D
$4$

Answer

Correct answer:B

Step-by-step solution

Standard Method

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

Find: $|(\vec{b} \times \vec{a}) - \vec{b}|^2$ .

Since $\vec{b} \times \vec{a}$ is perpendicular to $\vec{b}$ , we use

|(\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,

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

because a cross product is perpendicular to each of the vectors involved.

Therefore,

|(\vec{b} \times \vec{a}) - \vec{b}|^2 = 2^2 + 1^2 = 4 + 1 = 5

Hence, the value is $5$ . So the correct option is B.

The solution appears unrelated to this question, so the working above is obtained directly from the given question data.

Common mistakes

Assuming $\vec{b} \times \vec{a}$ is parallel to $\vec{b}$ . This is wrong because a cross product is always perpendicular to both vectors. Use $(\vec{b} \times \vec{a}) \cdot \vec{b} = 0$ .
Using $|\vec{x} - \vec{y}|^2 = |\vec{x}|^2 - |\vec{y}|^2$ . This identity is incorrect. The correct expansion is $|\vec{x} - \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2\vec{x} \cdot \vec{y}$ .
Ignoring the square in the required expression and finding $|(\vec{b} \times \vec{a}) - \vec{b}|$ instead of $|(\vec{b} \times \vec{a}) - \vec{b}|^2$ . Read the target expression carefully before simplifying.

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Answer

Step-by-step solution

Common mistakes

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