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If $\mathbf{p} \times \mathbf{q}=\mathbf{p} \times \mathbf{r}$ and $\mathbf{p} \cdot \mathbf{q}=\mathbf{p} \cdot \mathbf{r}$, then $\ldots . . .$.
Options:
Solution:
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Verified Answer
The correct answer is:
$q=r$
$\mathbf{p} \times \mathbf{q}=\mathbf{p} \times \mathbf{r}$
$$
\begin{aligned}
& \Rightarrow \quad \mathbf{p} \times(\mathbf{q}-\mathbf{r})=0 \\
& \text { p.q }=\text { p. } r \\
& \Rightarrow \quad \text { p. }(\mathbf{q}-\mathbf{r})=0 \\
&
\end{aligned}
$$
From Eqs. (i) and (ii) we can say that $\mathbf{p}$ is neither parallel nor Perpendicular to $(\mathbf{q}-\mathbf{r})$
$$
\begin{aligned}
& \Rightarrow \quad \mathbf{q}-\mathbf{r}=0 \\
& {[\because \mathbf{p} \neq 0]} \\
& \Rightarrow \quad \mathbf{q}=\mathbf{r} \\
&
\end{aligned}
$$
Hence, option (2) is correct.
$$
\begin{aligned}
& \Rightarrow \quad \mathbf{p} \times(\mathbf{q}-\mathbf{r})=0 \\
& \text { p.q }=\text { p. } r \\
& \Rightarrow \quad \text { p. }(\mathbf{q}-\mathbf{r})=0 \\
&
\end{aligned}
$$
From Eqs. (i) and (ii) we can say that $\mathbf{p}$ is neither parallel nor Perpendicular to $(\mathbf{q}-\mathbf{r})$
$$
\begin{aligned}
& \Rightarrow \quad \mathbf{q}-\mathbf{r}=0 \\
& {[\because \mathbf{p} \neq 0]} \\
& \Rightarrow \quad \mathbf{q}=\mathbf{r} \\
&
\end{aligned}
$$
Hence, option (2) is correct.
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