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MATHEMATICS · Single correct · CBSE · Mathematics · Three-Dimensional Geometry · Three-Dimensional Geometry
Problem The coordinates of a point on the line
x − 1 2 = y + 1 − 3 = z \frac{x-1}{2} = \frac{y+1}{-3} = z 2 x − 1 = − 3 y + 1 = z 4 14 4\sqrt{14} 4 14 ( 1 , − 1 , 0 ) (1, -1, 0) ( 1 , − 1 , 0 ) ( 9 , − 13 , 4 ) (9, -13, 4) ( 9 , − 13 , 4 ) ( 8 14 + 1 , − 12 14 − 1 , 4 14 ) (8\sqrt{14} + 1, -12\sqrt{14} - 1, 4\sqrt{14}) ( 8 14 + 1 , − 12 14 − 1 , 4 14 ) ( − 7 , 11 , − 4 ) (-7, 11, -4) ( − 7 , 11 , − 4 ) ( − 8 14 + 1 , 12 14 − 1 , − 4 14 ) (-8\sqrt{14} + 1, 12\sqrt{14} - 1, -4\sqrt{14}) ( − 8 14 + 1 , 12 14 − 1 , − 4 14 ) Browse the problem library