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(B) Given the scalar triple product equation: $[m\overline{a}+\overline{b} \quad m\overline{b}+\overline{c} \quad m\overline{c}+\overline{a}] = 28[\ov

Vedclass · Accepted Answer

$3$

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(B) Given the scalar triple product equation: $[m\overline{a}+\overline{b} \quad m\overline{b}+\overline{c} \quad m\overline{c}+\overline{a}] = 28[\overline{a} \quad \overline{b} \quad \overline{c}]$. Expanding the scalar triple product using the property $[\overline{u}+\overline{v} \quad \overline{w} \quad \overline{x}] = [\overline{u} \quad \overline{w} \quad \overline{x}] + [\overline{v} \quad \overline{w} \quad \overline{x}]$,we get: $[m\overline{a}+\overline{b} \quad m\overline{b}+\overline{c} \quad m\overline{c}+\overline{a}] = m^3[\overline{a} \quad \overline{b} \quad \overline{c}] + [\overline{b} \quad \overline{c} \quad \overline{a}] = m^3[\overline{a} \quad \overline{b} \quad \overline{c}] + [\overline{a} \quad \overline{b} \quad \overline{c}]$. Since $[\overline{b} \quad \overline{c} \quad \overline{a}] = [\overline{a} \quad \overline{b} \quad \overline{c}]$ by cyclic permutation. Thus,$(m^3+1)[\overline{a} \quad \overline{b} \quad \overline{c}] = 28[\overline{a} \quad \overline{b} \quad \overline{c}]$. Since the vectors are non-coplanar,$[\overline{a} \quad \overline{b} \quad \overline{c}] 
eq 0$. Therefore,$m^3+1 = 28 \implies m^3 = 27 \implies m = 3$.

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