Question

Find the distance between the point (1,3,2) and the plane r = i-j+k + λ(i)+ μ(j-k)

Answer 1

\[r=i-j+k+\lambda~ i+\mu~j-\mu~k\]
\[or~ x~ i+y~j+zk=\left( 1+\lambda \right)i+\left( \mu-1 \right)j+\left( 1-\mu\right)k\]
\[x=1+\lambda~~~...(1) ,y=\mu-1...(2),z=1-\mu ...(3)\]
adding (1),(2) and (3)
\[x+y+z=1+\lambda+\mu-1+1-\mu =1+\lambda~or~x+y+z-1-\lambda=0...(4)\]
perpendicular distance of (1,3,2) from the plane (4)
\[d=\frac{ \left| 1+3+2-1-\lambda \right| }{ \sqrt{1^{2}+1^{2}+1^{2}} }=\frac{ \left| 5-\lambda \right| }{\sqrt{3} }\]
NOTE:-If plane is ax+by+cz+d=0
then perpendicular distance from (x1,y1,z1) is
\[D=\frac{ \left| ax1+by1+cz1+d \right| }{\sqrt{a^2+b^2+c^2}}\]

Answer 2

Then how do I go on with this formula to find the distance?

Answer 3

i have solved this question.