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If $\overrightarrow{\mathrm{p}} \neq \overrightarrow{0}$ and the conditions $\overrightarrow{\mathrm{p}} \overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{r}}$ and $\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{r}}$ hold
simultaneously, then which one of the following is correct?
Options:
simultaneously, then which one of the following is correct?
Solution:
1503 Upvotes
Verified Answer
The correct answer is:
$\overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{r}}$
Given that $\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{r}}$
$\Rightarrow \overrightarrow{\mathrm{p}} \cdot(\overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{r}})=0$
$\Rightarrow \overrightarrow{\mathrm{p}}$ is perpendicular to $\overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{r}}$
Also, $\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{r}}$ (given).
$\Rightarrow \overrightarrow{\mathrm{p}} \times(\overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{r}})=0$
$\Rightarrow \overrightarrow{\mathrm{p}}$ is parallel to $\overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{r}}$
Which is not possible simultaneously unless either $\mathrm{p}$ or $\overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{r}}$ is zero, since $\overrightarrow{\mathrm{p}} \neq 0, \Rightarrow \overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{r}}=0$
Thus, the given conditions hold simultaneously if $\overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{r}}$
$\Rightarrow \overrightarrow{\mathrm{p}} \cdot(\overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{r}})=0$
$\Rightarrow \overrightarrow{\mathrm{p}}$ is perpendicular to $\overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{r}}$
Also, $\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{r}}$ (given).
$\Rightarrow \overrightarrow{\mathrm{p}} \times(\overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{r}})=0$
$\Rightarrow \overrightarrow{\mathrm{p}}$ is parallel to $\overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{r}}$
Which is not possible simultaneously unless either $\mathrm{p}$ or $\overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{r}}$ is zero, since $\overrightarrow{\mathrm{p}} \neq 0, \Rightarrow \overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{r}}=0$
Thus, the given conditions hold simultaneously if $\overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{r}}$
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