Question
Mathematics Question on Parallel Lines
The equation of a line passing through the origin and parallel to the line r=3i^+4j^−5k^+t(2i^−j^+7k^), where t is a parameter, is:
(A) 2x=−1y=7z (B) r=m(12i^−6j^+42k^); where m is the parameter (C) r=(12i^−6j^+42k^)+s(0i^−0j^+0k^); where s is the parameter (D) 3x−3=−4y−4=0z+5 (E) 3x=4y=5z
Choose the correct answer from the options given below:
(A) and (B) only
(A), (B) and (C) only
(C), (D) and (E) only
(A) only
(A) and (B) only
Solution
To find the equation of a line passing through the origin and parallel to the given line, we use the direction vector of the line, which is 2i^−j^+7k^.
(A) The symmetric form 2x=−1y=7z represents a line parallel to the given direction vector and passing through the origin.
(B) The vector equation r=m(12i^−6j^+42k^) is obtained by multiplying the direction vector by a scalar and also passes through the origin.
(C) The form r=(12i^−6j^+42k^)+s(0i^−0j^+0k^) is incorrect as it introduces an additional term that is not needed for parallelism through the origin.
(D) The given equation does not pass through the origin due to the constants.
(E) This form does not represent a line parallel to the given vector.