Question

For three non–coplanar vectors \(\vec a,\vec b,\vec c\), if \[ (\vec b+\vec c)\cdot\big((\vec c+\vec a)\times(\vec a+\vec b)\big) = \alpha \,[\vec a\,\vec b\,\vec c] \] and \[ (\vec a+\vec b)\cdot\big((\vec b+\vec c)\times(\vec a+\vec b+\vec c)\big) = \beta \,[\vec a\,\vec b\,\vec c], \] then \(\alpha+\beta\) is equal to:

• A. \(-3\)
• B. \(-1\)
• C. \(1\)
• D. \(3\)

Hint:

Use the multilinearity and cyclic properties of the scalar triple product. Most expanded terms vanish due to repetition of vectors.

Answer 1

The Correct Option is D

Step 1: Evaluate \(\alpha\) \[ (\vec b+\vec c)\cdot\big((\vec c+\vec a)\times(\vec a+\vec b)\big) =[\vec b+\vec c,\ \vec c+\vec a,\ \vec a+\vec b] \] Using multilinearity of scalar triple product: \[ =[\vec b,\vec c,\vec a]+[\vec c,\vec a,\vec b] \] Both are cyclic permutations of \([\vec a,\vec b,\vec c]\), hence: \[ \alpha=2 \] Step 2: Evaluate \(\beta\) \[ (\vec a+\vec b)\cdot\big((\vec b+\vec c)\times(\vec a+\vec b+\vec c)\big) =[\vec a+\vec b,\ \vec b+\vec c,\ \vec a+\vec b+\vec c] \] Expanding and retaining only non–zero terms: \[ =[\vec a,\vec b,\vec c]+[\vec a,\vec c,\vec b]+[\vec b,\vec c,\vec a] \] Now, \[ [\vec a,\vec c,\vec b]=- [\vec a,\vec b,\vec c],\qquad [\vec b,\vec c,\vec a]=[\vec a,\vec b,\vec c] \] Thus, \[ \beta=1 \] Step 3: Final result \[ \alpha+\beta=2+1=3 \] \[ \boxed{3} \]