NTA Abhyas JEE Main2020MathematicsThree Dimensional GeometryPractice

The image of the line x - 2 2 = y - 3 - 3 = z - 4 1 in the plane x + y + z = 6 can be

Options

Ar → = λ 2 i ^ - 3 j ^ + k ^
Br → = 2 i ^ + j ^ + k ^ + α 2 i ^ - 3 j ^ + k ^
Cr → = j ^ + 2 k ^ + β 2 i ^ - 3 j ^ + k ^
Dr → = i ^ + 2 j ^ + u 2 i ^ - 3 j ^ + k ^

Correct answer

C. r → = j ^ + 2 k ^ + β 2 i ^ - 3 j ^ + k ^

Step-by-step solution

Note that the line is parallel to the plane, hence the image of the line will be a parallel line. Now, let the image of 2,3 , 4 in x + y + z = 6 is x 1 , y 1 , z 1 Then, x 1 - 2 1 = y 1 - 3 1 = z 1 - 4 1 = - 2 2 + 3 + 4 - 6 1 + 1 + 1 i.e. x 1 , y 1 , z 1 ≡ 0,1 , 2 So the required equation is r → = j ^ + 2 k ^ + β 2 i ^ - 3 j ^ + k ^

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