A) \[\mathbf{r}=(\mathbf{i}+2\mathbf{j}-\mathbf{k})+\lambda (-\mathbf{i}+5\mathbf{j}-3\mathbf{k})\]
B) \[\mathbf{r}=\mathbf{i}+3\mathbf{j}+2\mathbf{k}+\lambda (\mathbf{i}-5\mathbf{j}+3\mathbf{k})\]
C) \[\mathbf{r}=\mathbf{i}+3\mathbf{j}+2\mathbf{k}+\lambda (\mathbf{i}+5\mathbf{j}+3\mathbf{k})\]
D) \[\mathbf{r}=\mathbf{i}+3\mathbf{j}+2\mathbf{k}+\lambda (-\mathbf{i}+5\mathbf{j}-3\mathbf{k})\]
Correct Answer: D
Solution :
The required line passes through the point \[\mathbf{i}+3\mathbf{j}+2\mathbf{k}\] and is perpendicular to the lines \[\mathbf{r}=(\mathbf{i}+2\mathbf{j}-\mathbf{k})+\lambda (2\mathbf{i}+\mathbf{j}+\mathbf{k})\] and \[\mathbf{r}=(2\mathbf{i}+6\mathbf{j}+\mathbf{k})+\mu (\mathbf{i}+2\mathbf{j}+3\mathbf{k})\], therefore it is parallel to the vector \[\mathbf{b}=(2\mathbf{i}+\mathbf{j}+\mathbf{k})\times (\mathbf{i}+2\mathbf{j}+3\mathbf{k})\]= \[(\mathbf{i}-5\mathbf{j}+3\mathbf{k})\] Hence, the equation of the required line is \[\mathbf{r}=(\mathbf{i}+3\mathbf{j}+2\mathbf{k})+\lambda '(\mathbf{i}-5\mathbf{j}+3\mathbf{k})\] Þ \[\mathbf{r}=(\mathbf{i}+3\mathbf{j}+2\mathbf{k})+\lambda (-\mathbf{i}+5\mathbf{j}-3\mathbf{k})\], where \[\lambda =-\lambda '\].
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