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Find the vector and cartesian equations of the plane which passes through the point `(5, 2, 4)`and perpendicular to the line with direction ratios `(2, 3, 1)`. |
Answer» Let `A(5,2,4)` is the point passing through given plane. In vector form, `vecA = 5hati+2hatj+4hatk` Let `vecr` is another vector on that plane, then, line joining `vecr` and `veca` will be, `vecr - vecA` We are given another line with direction cosines (2,3,1) that is perpendicular to the first line. So, dot product of these two lines will be `0`. `:. (vecr - (5hati+2hatj+4hatk)).(2hati+3hatj+hatk) = 0` `=>vecr.(2hati+3hatj+hatk) - 10-6-4 = 0` `=>vecr.(2hati+3hatj+hatk) = 20` We can take, `vecr = ahati+bhatj+chatk ` Then, our equation becomes, `2a+3b+c = 20`, which is required vector equation. Cartesian equation will be, `2x+3y+z = 20` |
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