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Find the point where the line `(x-1)/2=(y-2)/-3=(z+3)/4` meets the plane `2x+4y-z=1`.A. (3,-1,1)B. (3,1,1)C. (1,1,3)D. (1,3,1) |
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Answer» Correct Answer - A Let point be (a , b , c) then 2a + 4b - c = 1 `" " …. (i)` and a = 2k + 1 , b = -3k + 2 and c = 4k - 3 (where k is constant ) On substituting these values in Eq. (i) , we get 2(2k + 1) + 4(-3 k + 2)- (4k - 3) = 1 `implies k = 1` Hence , required point is (3 , -1 , 1) . |
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