Question

bindu (3,-1,1 ) se hokar jane vale samtal ka sdish samikaran gyat karo jo santali r.(2i+3j+k)=5 or r.(i+5j-2k)=1 ki patichedi rekha se hokar jata hai​

Answer 1

Answer:

Step-by-step explanation:

Given, planes are

$\tt{\vec{r}\cdot(2\hat{i}+3\hat{j}+\hat{k})=5}\\\tt{\vec{r}\cdot(\hat{i}+5\hat{j}-2\hat{k})=1}$

Its cartesian forms are

$\tt{P_{1}\colon2x+3y+z-5=0}\\\tt{P_{2}\colon\,x+5y-2z-1=0}$

Any plane passing through the intersection of $\sf{P_{1}}$ and $\sf{P_{2}}$ is

$\tt{P\colon(2x+3y+z-5)+\lambda(x+5y-2z-1)=0}$

$\tt{\implies\,P\colon(2+\lambda)x+(3+5\lambda)y+(1-2\lambda)z-(5+\lambda)=0}$

It is given that (3,-1,1) lies on 'P'

So,

$\tt{\implies\,(2+\lambda)(3)+(3+5\lambda)(-1)+(1-2\lambda)(1)-(5+\lambda)=0}$

$\tt{\implies\,6+3\lambda-3-5\lambda+1-2\lambda-5-\lambda=0}$

$\tt{\implies\,3\lambda-5\lambda-2\lambda-\lambda-3+1-5+6=0}$

$\tt{\implies\,-5\lambda-1=0}$

$\tt{\implies\,\lambda=-\dfrac{1}{5}}$

Put this value in P, in order to obtain the required equation of the plane

So,

$\tt{\implies\,\bigg(2-\dfrac{1}{5}\bigg)x+\bigg(3-5\cdot\dfrac{1}{5}\bigg)y+\bigg(1+2\cdot\dfrac{1}{5}\bigg)z-\bigg(5-\dfrac{1}{5}\bigg)=0}$

$\tt{\implies\,\bigg(\dfrac{10-1}{5}\bigg)x+(3-1)y+\bigg(\dfrac{5+2}{5}\bigg)z-\bigg(\dfrac{25-1}{5}\bigg)=0}$

$\tt{\implies\,\bigg(\dfrac{9}{5}\bigg)x+2y+\bigg(\dfrac{7}{5}\bigg)z-\bigg(\dfrac{24}{5}\bigg)=0}$

$\tt{\implies\,9x+10y+7z-24=0}$