A) \[6\mathbf{i}-10\mathbf{j}+3\mathbf{k}\]
B) \[\frac{1}{5}(6\mathbf{i}-10\mathbf{j}+3\mathbf{k})\]
C) \[-6\mathbf{i}+10\mathbf{j}-3\mathbf{k}\]
D) None of these
Correct Answer: B
Solution :
The vector equation of the line joining the points \[\mathbf{i}-2\mathbf{j}+\mathbf{k}\] and \[-2\mathbf{j}+3\mathbf{k}\]is \[\mathbf{r}=(\mathbf{i}-2\mathbf{j}+\mathbf{k})+\lambda (-\mathbf{i}+2\mathbf{k})\] ?..(i) The vector equation of the plane through the origin, \[4\mathbf{j}\]and \[2\mathbf{i}+\mathbf{k}\] is \[\mathbf{r}\,.\,(4\mathbf{i}-8\mathbf{k})=0\] ?..(ii) (Using \[\mathbf{r}.(\mathbf{a}\times \mathbf{b}+\mathbf{b}\times \mathbf{c}+\mathbf{c}\times \mathbf{a})=[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]\]) The position vector of any point on (i) is \[(\mathbf{i}-2\mathbf{j}+\mathbf{k})\]\[+\lambda (-\mathbf{i}+2\mathbf{k})\]. If it lies on (ii), then \[((\mathbf{i}-2\mathbf{j}+\mathbf{k})+\lambda (-\mathbf{i}+2\mathbf{k})).(4\mathbf{i}-8\mathbf{k})=0\] Þ \[-4-20\lambda =0\Rightarrow \lambda =-\frac{1}{5}\] Putting the value of \[\lambda \] in \[(\mathbf{i}-2\mathbf{j}+\mathbf{k})+\lambda (-\mathbf{i}+2\mathbf{k})\], we get the position vector of the required point as \[\frac{1}{5}(6\mathbf{i}-10\mathbf{j}+3\mathbf{k})\].
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