A) \[\mathbf{r}.(4\mathbf{i}-12\mathbf{j}-3\mathbf{k})=0\]
B) \[\mathbf{r}.(4\mathbf{i}-12\mathbf{j}-3\mathbf{k})=32\]
C) \[\mathbf{r}.(4\mathbf{i}-12\mathbf{j}-3\mathbf{k})=12\]
D) None of these
Correct Answer: B
Solution :
The equation of a plane parallel to the plane \[\mathbf{r}.(4\mathbf{i}-12\mathbf{j}-3\mathbf{k})-7=0\]is \[\mathbf{r}.(4\mathbf{i}-12\mathbf{j}-3\mathbf{k})+\lambda =0\]. This passes through\[2\mathbf{i}-\mathbf{j}-4\mathbf{k}\]. Therefore,\[(2\mathbf{i}-\mathbf{j}-4\mathbf{k}).(4\mathbf{i}-12\mathbf{j}-3\mathbf{k})+\lambda =0\] Þ \[8+12+12+\lambda =0\Rightarrow \lambda =-32\] So, the required plane is \[\mathbf{r}.(4\mathbf{i}-12\mathbf{j}-3\mathbf{k})-32=0\].
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