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The sum of all possible real values of ‘a’ for which vectors ${\vec{r}}_{1} = \ln | a | \hat{i} + a \hat{j} - \hat{k}$ and ${\vec{r}}_{2} = (1 + a^{2}) \hat{i} - \hat{j} - (a - 1) \hat{k}$ are orthogonal is

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-2

answer is B.

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Detailed Solution

${\vec{r}}_{1} \cdot {\vec{r}}_{2} = 0$

$\ln | a | (1 + a^{2}) - a + a - 1 = 0$

$\ln | a | = \frac{1}{1 + a^{2}}$

From the graph it is clear that number of values of ‘a’ are two i.e.,

$x_{1}, x_{2} and x_{1} + x_{2} = 0$

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