A) \[\widehat{r}.(-5\hat{i}-3\hat{j}+11\hat{k})+\frac{135}{2}=0\]
B) \[\widehat{r}.\left( \frac{3}{2}\hat{i}+\frac{7}{2}\hat{j}-\frac{9}{2}\hat{k} \right)+\frac{35}{2}=0\]
C) \[\widehat{r}.(4\hat{i}+5\hat{j}-\hat{k})+4=0\]
D) \[\widehat{r}.(-\hat{i}+2\hat{j}+\hat{k})+4=0\]
Correct Answer: A
Solution :
Mid-point M of AB is \[\left( \frac{1}{2}(4\hat{i}+5\hat{j}-10\hat{k}-\hat{i}+2\hat{j}+\hat{k} \right)=\left( \frac{3}{2}\hat{i}+\frac{7}{2}\hat{j}-\frac{9}{2}\hat{k} \right)\] Also, \[\overrightarrow{AB}=-\hat{i}+2\hat{j}+\hat{k}-(4\hat{i}+5\hat{j}-10\hat{k})\] \[=-5\hat{i}-3\hat{j}+11\hat{k}\] So, the plane passing through M and perpendicular to the \[\overrightarrow{AB}\] is \[\left[ \overrightarrow{r}-\left( \frac{3}{2}\hat{i}+\frac{7}{2}\hat{j}-\frac{9}{2}\hat{k} \right) \right].(-5\hat{i}-3\hat{j}+11\hat{k})=0\] \[\Rightarrow \,\,\overrightarrow{r}.(-5\hat{i}-3\hat{j}+11\hat{k})+\frac{135}{2}=0.\]You need to login to perform this action.
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