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Cross product of vectors or vector product of vectors

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Question

$\begin{array}{l} If the vectors, \vec{p} = (a + 1) \hat{i} + \hat{a j} + a \hat{k}, \vec{q} = a \hat{i} + (a + 1) \hat{j} + a \hat{k} and \vec{r} = a \hat{i} + \hat{j} + (a + 1) \hat{k} (a \in R) are \\ coplanar and 3 (\vec{p} \cdot \vec{q})^{2} - λ | \vec{r} \times \bar{q} |^{2} = 0, then the value of λ is \end{array}$

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Solution

$3 vectors \bar{p}, \bar{q}, \bar{r} are coplanar \Rightarrow |\begin{matrix} a + 1 & a & a \\ a & a + 1 & a \\ a & a & a + 1 \end{matrix}| = 0$
Expand the determinant,
$\begin{matrix} (a + 1) (a^{2} + 2 a + 1 - a^{2}) - a (a^{2} + a - a^{2}) + a (a^{2} - a^{2} - a) = 0 \\ (a + 1) (2 a + 1) - a^{2} - a^{2} = 0 \\ 3 a + 1 = 0 \\ a = - \frac{1}{3} \end{matrix}$
$\begin{array}{l} So, \bar{p} = \frac{2}{3} \hat{i} - \frac{1}{3} \hat{j} - \frac{1}{3} \hat{k}, \bar{q} = \frac{1}{3} (- \hat{i} + 2 \hat{j} - \hat{k}), \bar{r} = \frac{1}{3} (- \hat{i} - \hat{j} + 2 \hat{k}) \\ \bar{p} \cdot \bar{q} = \frac{1}{9} (- 2 - 2 + 1) = - \frac{1}{3} \\ \bar{r} \times \bar{q} = \frac{1}{9} |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 1 & - 1 & 2 \\ - 1 & 2 & - 1 \end{matrix}| = \frac{1}{9} (\hat{i} (1 - 4) - \hat{j} (1 + 2) + \hat{k} (- 2 - 1)) \\ = \frac{- 1}{9} (3 \hat{i} + 3 \hat{j} + 3 \hat{k}) = \frac{- (\hat{i} + \hat{j} + \hat{k)}}{3} \\ λ = \frac{3 (\bar{p} \cdot \bar{q})^{2}}{| \bar{r} \times \bar{q} |^{2}} = 1 \end{array}$

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