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  • Find the distance of the point (– 1, – 5, – 10) from the point of intersection of the line r vector = 2 i caret - j caret + 2 k caret + lambda 3 i caret + 4 j caret + 2 k caret and the plane

18  Find the distance of the point (– 1, – 5, – 10) from the point of intersection of the line \overrightarrow{r}=2\widehat{i}-\widehat{j}+2\widehat{k}+ \lambda\left ( 3\widehat{i}+4\widehat{j}+2\widehat{k} \right ) and the plane   \overrightarrow{r}.\left ( \widehat{i}-\widehat{j}+\widehat{k} \right )=5.

Answers (1)

best_answer

Given, 

Equation of a line :

\overrightarrow{r}=2\widehat{i}-\widehat{j}+2\widehat{k}+ \lambda\left ( 3\widehat{i}+4\widehat{j}+2\widehat{k} \right )

Equation of the plane

\overrightarrow{r}.\left ( \widehat{i}-\widehat{j}+\widehat{k} \right )=5

Let's first find out the point of intersection of line and plane.

putting the value of \vec r into the equation of a plane from the equation from line

\left (2\widehat{i}-\widehat{j}+2\widehat{k}+ \lambda\left ( 3\widehat{i}+4\widehat{j}+2\widehat{k} \right ) \right )\cdot (\hat i-\hat j+\hat k)=5

(2+3\lambda)-(4\lambda -1)+(2+2\lambda)=5

\lambda+5=5

\lambda=0

Now, from the equation, any  point p  in line is 

P=(2+3\lambda,4\lambda-1,2+2\lambda)

So the point of intersection is 

P=(2+3*0,4*0-1,2+2*0)=(2,-1,2)

SO, Now,

The distance between the points (-1,-5,-10) and (2,-1,2) is 

d=\sqrt{(2-(-1))^2+(-1-(-5))^2+(2-(-10))^2}=\sqrt{9+16+144}

d=\sqrt{169}=13

Hence the required distance is 13.

Posted by

Pankaj Sanodiya

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