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Find the vector equation of the line passing through the point (1,2,3) and parallel to the planes $\overrightarrow{r}.(\hat{i}-\hat{j}+2\hat{k})=5$ and $\overrightarrow{r}.(3\hat{i}+\hat{j}+\hat{k})=6.$

(Note: This question has been split into 2 questions) This question appeared in 65-1,65-2 and 65-3 versions of the paper in 2013.
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The required line planes through the point having its position vector $\overrightarrow{a}=\overrightarrow{i}+ 2\overrightarrow{j}+ 3\overrightarrow{k}$
The parallel planes are
$\overrightarrow{r}.(\hat {i}-\hat {j}+2\hat {k})=5$ and
$\overrightarrow{r}.(3\hat {i}+\hat {j}+\hat {k})=6$
Hence the required line is parallel to the vector is $\overrightarrow{b}=\overrightarrow{n_1} \times \overrightarrow{n_2}$
$\Rightarrow \overrightarrow{n_1} \times \overrightarrow{n_2}=\begin {vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ 1 & -1 & 2 \\ 3 & 1 & 1 \end{vmatrix}=\overrightarrow{i}(-1-2)-\overrightarrow{j}(1-6)+\overrightarrow{k}(1+3)$
$\overrightarrow{b}=-3\overrightarrow{i}+5 \overrightarrow{j}+4 \overrightarrow{k}$
Hence the equation of the required line is $\overrightarrow{r}=\overrightarrow{a} + \lambda \overrightarrow{b}$
$\overrightarrow{r}=(\overrightarrow{i}+2\overrightarrow{j}+3\overrightarrow{k})+\lambda (-3\overrightarrow{i}+5\overrightarrow{j}+4\overrightarrow{k})$
answered Mar 21, 2013 by meena.p
edited Mar 26, 2013 by balaji.thirumalai

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