A) 0
B) 1
C) ? 1
D) Arbitrary scalar
Correct Answer: D
Solution :
Since \[|\mathbf{a}\times \mathbf{r}{{|}^{2}}+|\mathbf{a}\,.\,\mathbf{r}{{|}^{2}}=|\mathbf{a}{{|}^{2}}|\mathbf{r}{{|}^{2}}\] \[\Rightarrow \,|\mathbf{j}{{|}^{2}}+{{(\mathbf{a}\,.\,\mathbf{r})}^{2}}=\,|\mathbf{a}{{|}^{2}}|\mathbf{r}{{|}^{2}}\]\[\Rightarrow \,(\mathbf{a}\,.\,\mathbf{r})=\pm \sqrt{|\mathbf{a}{{|}^{2}}|\mathbf{r}{{|}^{2}}-1}\] This shows that \[\mathbf{a}\,.\,\mathbf{r}\] depends on \[|\mathbf{r}|\] for given \[\mathbf{a}.\] Hence \[\mathbf{a}\,.\,\mathbf{r}\] is arbitrary scalar.
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