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If the angle between the planes $\bar{r} \cdot(11 \hat{i}-2 \hat{j}+\alpha \hat{k})=7$ and $\bar{r} \cdot(2 \hat{i}+4 \hat{j}-2 \hat{k})=5$ is $\frac{\pi}{2}$, then $\alpha=$
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The correct answer is:
7
Given planes are $\bar{r} \cdot(11 \hat{i}-2 \hat{j}-\alpha \hat{k})=7$
and $\overline{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})=5$
$\Rightarrow \mathrm{n}_1 \cdot \mathrm{n}_2=0 \Rightarrow 22-8-2 \alpha=0$ or $\alpha=7$
and $\overline{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})=5$
$\Rightarrow \mathrm{n}_1 \cdot \mathrm{n}_2=0 \Rightarrow 22-8-2 \alpha=0$ or $\alpha=7$
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