Find the point at which the line intersects the given plane.
$ x = t - 1 $ , $ y = 1 + 2t $ , $ z = 3 - t $ ; $ 3x - y + 2z = 5 $
Hello. So the question is taken from victor's in geometry of the space and we have to find the point at which the line intersect doesn't even play. So the equation of lying and very genesis X. Is equal to T -1. Why is it square to one Plus 2 T. And that is equal to three minus T. And that line passes through the plane three X -Y Plus two. That is equal to five. Okay so we need to find that point. Let me even in order to find that intersection point let us substitute the value of X. Y. That in the equation we get three into three minus one -1. Plus duty Plus two into 3 -T. Is equal to five. We get three T -3 -1- Duty Plus 6- Duty is equal to five. And further solving it will get T. Three t minus two T. S. T t minus two. Tes minus of the and minus three minus minus minus four minus four plus sixties To which is able to fight. So from charity is a 4-1. Let me substitute this value. And the equation so we get the value of x rays. They did X. Is equal to b minus from So that will be minus three minus one. So that would be minus four. Why is equal to one plus duty? So these minus three minus three into two plus one. So that will be minus fight. Okay so one minus six so that will be minus five and that is equal to three minus street. So that will be three minus of minus three. So that will be six. So that given point at which the line intersect with the plane minus four minus five So plus six. Okay, Which is the required points? So hope this clears without any.