Test Index

VITEEE 2010 Paper

Question : 114 of 120

Marks: +1, -0

The distance between the line and the plane

Solution:

The given line is

\vec{r}=\hat{2i}-\hat{2j}+\hat{3k}+\lambda(\hat{i}-\hat{j}+\hat{4k})

On comparing it with

\vec{r}=\vec{a}+t\vec{b}

, we get

\vec{a}=\hat{2i}-\hat{2j}+\hat{3k},\vec{b}=\hat{i}-\hat{j}+\hat{4k}

Also, the plane is

\vec{r}\cdot(\hat{i}+\hat{5j}+\hat{k})=5

On comparing it with

\vec{r}\cdot\vec{n}=d

, we get

\vec{n}=\hat{i}+\hat{5j}+\hat{k} \text{ and } d=5

Since,

\vec{b}\cdot\vec{n}=(\hat{i}-\hat{j}+\hat{4k})\cdot(\hat{i}+\hat{5j}+\hat{k})

=1-5+4=0

∴ Given line is parallel to the given plane.

Now, distance between the line and the plane is given by required distance..

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