Discussion
If the equation of the straight line passing through the point(a,1,3) and parallel to the vector \(\frac{2}{3}\hat{i} + \frac{3}{2}\hat{j} + \hat{k}\) is \(\dfrac{3x + 6}{b} = \dfrac{2y - 2}{3} = \dfrac{z - 3}{1}\), then the value of a + b is equal to
Answer: E
a + b = 0.
Rewrite the line in symmetric form by dividing out the coefficients of x and y:
\[ \frac{x + 2}{b/3} = \frac{y - 1}{3/2} = \frac{z - 3}{1} \]
The line passes through (-2, 1, 3), so a = -2. Its direction \(\left(\frac{b}{3}, \frac{3}{2}, 1\right)\) must match \(\left(\frac{2}{3}, \frac{3}{2}, 1\right)\), and since the last two components already agree, \(\frac{b}{3} = \frac{2}{3}\) and b = 2.
So a + b = -2 + 2 = 0. Reading a as +2 from '3x + 6' (sign slip) gives 4, and reading b as \(\frac{2}{3}\) without the factor 3 gives a non-integer.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development