Statement 1 - A point on the straight line 2 x + 3 y − 4 z = 5 , 3 x − 2 y + 4 z = 7 can be determined by taking x = k and then solving the two equations for y and z , where k is any real number. Stat

Question

Statement 1 - A point on the straight line $2 x + 3 y - 4 z = 5, 3 x - 2 y + 4 z = 7$ can be determined by taking $x = k$ and then solving the two equations for $y$ and $z$ , where $k$ is any real number.

Statement 2 - If $c^{'} \neq = k c$ , then the straight line $a x + b y + cz + d = 0, ka x + kb y +$ $c^{'} z + d^{'} = 0$ does not intersect the plane $z = α$ , where $α$ is any real number.

TestHub · Accepted Answer

Statement 1: with x = k , 3 y − 4 z = 5 − 2 k and − 2 y + 4 z = 7 − 3 k . Adding: y = 12 − 5 k and z = 4 31 − 13 k ​ , so every real k gives a point. True. Statement 2 : the direction of the line is ( a , b , c ) × ( ka , kb , c ′ ) = ( b ( c ′ − k c ) , a ( k c − c ′ ) , 0 ) , so the line is parallel to every plane z = α . But it lies in the plane z = z 0 ​ for its own constant z 0 ​ , so for α = z 0 ​ it meets the plane in all its points. So the claim that the line does not intersect z = α for any real α is false, and Statement 2 is false.

Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

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