Let {a}={i}+{j}, {c}={i}-{j} and a vector {b} be such that {a} × {b}={c} and {a} · {b}=3 then |{b}|=

Correct Answer is : 112{11/2}211

Correct Answer is : 112{11/2}211

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MHTCET 16 Apr 2026 Shift 1 Paper

Question : 137 of 150

Marks: +1, -0

Let

\bar{a}=\hat{i}+\hat{j}, \bar{c}=\hat{i}-\hat{j}

\bar{b}

\bar{a} \times \bar{b}=\bar{c}

\bar{a} \cdot \bar{b}=3

|\bar{b}|=

Solution:

Concept:

Use the vector identity

|\bar{a} \times \bar{b}|^2 + (\bar{a} \cdot \bar{b})^2 = |\bar{a}|^2 |\bar{b}|^2

to determine

|\bar{b}|

Explanation:

First, calculate

|\bar{a}|^2

for

\bar{a} = \hat{i} + \hat{j}

|\bar{a}|^2 = 1^2 + 1^2 = 2.

Since

\bar{a} \times \bar{b} = \bar{c}

, we have:

|\bar{a} \times \bar{b}|^2 = |\bar{c}|^2 = 1^2 + (-1)^2 = 2.

The dot product is given as

\bar{a} \cdot \bar{b} = 3

, so

(\bar{a} \cdot \bar{b})^2 = 9

Substitute these values into the identity:

2 + 9 = 2|\bar{b}|^2.

This gives:

11 = 2|\bar{b}|^2.

Hence:

|\bar{b}|^2 = \frac{11}{2} \quad \Rightarrow \quad |\bar{b}| = \sqrt{\frac{11}{2}}.

Answer:

|\bar{b}| = \sqrt{\frac{11}{2}}

, which corresponds to Option C.

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