1.

The line \(\frac{x-1}2\) = \(\frac{y-2}4\) = \(\frac{z-3}{-3}\) meets the plane 2x + 3y – z = 14 in the pointA. (2, 5, 7) B. (3, 5, 7) C. (5, 7, 3) D. (6, 5, 3)

Answer»

Given : line \(\frac{x-1}2\) = \(\frac{y-2}4\) = \(\frac{z-3}{-3}\) meets plane 2x + 3y – z = 14

 To find: Point of intersection of line and plane. 

Explanation: 

Let the equation of the line be

\(\frac{x-1}2\) = \(\frac{y-2}4\) = \(\frac{z-3}{-3}\) = 2

Therefore, any point on the line is (2λ + 1, 4λ +2, -3λ +3) 

Since this point also lies on the plane, 

2(2λ + 1) + 3(4λ +2) – (-3λ +3) =14 

4λ + 2 + 12λ + 6 + 3λ – 3 = 14 

19λ + 5 = 14

λ = \(\frac{19}{19}\) = 1

Therefore the required point is (3, 5, 7).



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