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Find the equation of the plane passing through the line of intersection of the planes $\overrightarrow{r}.(\hat{i}+3\hat{j})-6=0$ and $\overrightarrow{r}.(3\hat{i}-\hat{j}-4\hat{k})=0$,whose perpendicular distance from origin is unity.

(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.

1 Answer

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Let the plane$\overrightarrow{r}.\overrightarrow{n_1}=d_1$ be $\overrightarrow{r}.(\overrightarrow{i}+3\overrightarrow{j})=6 \rightarrow d_1=\frac{6}{\sqrt {10}}$
Let the other plane be $\overrightarrow{r}.\overrightarrow{n_2}=d_2$ be $\overrightarrow{r}.(\overrightarrow{3i}-\overrightarrow{j}-4\overrightarrow{k})=0 \rightarrow d_2=0$
$\textbf{Step 2}$:
Therefore $\overrightarrow{r}.(n_1+\lambda \overrightarrow{n_2})=d_1+\lambda d_2$ represents the plane passing through the intersection of the above two planes.
Given $d_1+\lambda d_2=1$ $\Rightarrow \lambda =\frac{\sqrt {10}}{6}$
$\textbf{Step 3}$:
Therefore the required equation is
$\overrightarrow{r}.\bigg[(\overrightarrow{i}+3\overrightarrow{j})+\frac{\sqrt {10}}{6}(3\overrightarrow{i}-\overrightarrow{j}-4\overrightarrow{k})\bigg]=1$
$\overrightarrow{r}.\bigg[6\overrightarrow{i}+18\overrightarrow{j}+3\sqrt {10}\overrightarrow{i}-10\overrightarrow{j}-4 \sqrt {10} \overrightarrow{k}\bigg]=6$
$\Rightarrow \overrightarrow{r}.[\overrightarrow{i}(6+3 \sqrt {10})+\overrightarrow{j}(18-\sqrt {10})-4 \sqrt {10}\overrightarrow{k})=6$
answered Mar 21, 2013 by meena.p
edited Mar 26, 2013 by balaji.thirumalai

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