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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,$m^3+1 = 28 \implies m^3 = 27 \implies m = 3$.

Similar Questions

Observe the following statements:
$A$. Three vectors are coplanar if one of them is expressible as a linear combination of the other two.
$R$. Any three coplanar vectors are linearly dependent.
Then, which of the following is true?

If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors and $\lambda$ is a real number,then for what value of $\lambda$ does the equation $[\lambda(\vec{a} + \vec{b}), \lambda^2\vec{b}, \lambda\vec{c}] = [\vec{a}, \vec{b} + \vec{c}, \vec{b}]$ hold?

The sum of all real values of $\lambda$ for which the vectors $\vec{a} = \lambda \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{i} + \lambda \hat{j} + 2\hat{k}$, and $\vec{c} = 2\hat{i} + 3\hat{j} + \lambda \hat{k}$ are coplanar is:

If $a, b$ and $c$ are three non-coplanar vectors,then $(a + b + c) \cdot [(a + b) \times (a + c)]$ is equal to

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