If {a} · {b}=β and {a} × {b}={c}^{ {then }} {a}=

Correct Answer is : bˉ×cˉ+βbˉ∣bˉ∣2{{b} × {c}+β {b}}{|{b}|²}∣bˉ∣2bˉ×cˉ+βbˉ

Correct Answer is : bˉ×cˉ+βbˉ∣bˉ∣2{{b} × {c}+β {b}}{|{b}|²}∣bˉ∣2bˉ×cˉ+βbˉ

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MHTCET 18 Apr 2026 Shift 2 Paper

Question : 143 of 150

Marks: +1, -0

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

\bar{a} \times \bar{b}=\bar{c}^{\text {then }} \bar{a}=

Solution:

Concept:

Use the vector triple product identity to express

\bar{a}

in terms of

\bar{b}

and

\bar{c}

Explanation:

Given

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

and

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

Take the cross product of

\bar{b}

with

\bar{c}

on the left:

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

Using the identity

\bar{x} \times (\bar{y} \times \bar{z}) = \bar{y}(\bar{x} \cdot \bar{z}) - \bar{z}(\bar{x} \cdot \bar{y})

set

\bar{x} = \bar{b}

\bar{y} = \bar{a}

, and

\bar{z} = \bar{b}

Then,

\bar{b} \times (\bar{a} \times \bar{b}) = \bar{a}(\bar{b} \cdot \bar{b}) - \bar{b}(\bar{b} \cdot \bar{a})

Since

\bar{b} \cdot \bar{b} = |\bar{b}|^2

and

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

we get:

\bar{b} \times \bar{c} = |\bar{b}|^2 \bar{a} - \beta \bar{b}

Rearranging,

|\bar{b}|^2 \bar{a} = \bar{b} \times \bar{c} + \beta \bar{b}

Dividing both sides by

|\bar{b}|^2

\bar{a} = \frac{\bar{b} \times \bar{c} + \beta \bar{b}}{|\bar{b}|^2}

This matches Option C.

Answer:

\bar{a} = \frac{\bar{b} \times \bar{c} + \beta \bar{b}}{|\bar{b}|^2}

Correct option: C.

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