Question

If \(\overset{̄}{a}\cdot \overset{̄}{b} = β\) and \(\overset{̄}{a}\times \overset{̄}{b} = \overset{̄}{c}\) then \(\overset{̄}{a} =\)

• A. \(\frac{\overset{̄}{b}\times \overset{̄}{c}-β\overset{̄}{b}}{|\overset{̄}{b}|^2}\)
• B. \(\frac{\overset{̄}{b}\times \overset{̄}{c}-β\overset{̄}{c}}{|\overset{̄}{b}|^2}\)
• C. \(\frac{\overset{̄}{b}\times \overset{̄}{c}+β\overset{̄}{b}}{|\overset{̄}{b}|^2}\)
• D. \(\frac{\overset{̄}{b}\times \overset{̄}{c}+β\overset{̄}{c}}{|\overset{̄}{b}|^2}\)

Hint:

Compute b x c = b x (a x b) using the triple product expansion.

Answer 1

The Correct Option is C

Step 1: Decompose a along and across b:
Write $\bar a = \frac{\bar a\cdot\bar b}{|\bar b|^2}\bar b + \bar a_\perp$. The first piece is $\frac{\beta}{|\bar b|^2}\bar b$.

Step 2: Recover the perpendicular part from c:
$\bar c = \bar a\times\bar b = \bar a_\perp\times\bar b$. Cross with $\bar b$ on the left: $\bar b\times\bar c = \bar b\times(\bar a_\perp\times\bar b) = \bar a_\perp|\bar b|^2$ (since $\bar a_\perp\perp\bar b$). So $\bar a_\perp = \frac{\bar b\times\bar c}{|\bar b|^2}$.

Step 3: Add the two parts:
$\bar a = \frac{\bar b\times\bar c + \beta\bar b}{|\bar b|^2}$.

Final Answer:
Option (C). \[ \boxed{\frac{\bar b\times\bar c+\beta\bar b}{|\bar b|^2} \text{ (C)}} \]