COMEDK · Maths · 34. Three Dimensional Geometry
If the line \(\dfrac{1-x}{-3}=y=\dfrac{z+2}{2}\) is perpendicular to the line \(\dfrac{3 x-1}{2 b}=3-y=\dfrac{z-1}{a}\), then find the value of \(3 a+3 b\)
- A \(\dfrac{3}{2}\)
- B 4
- C \(\dfrac{1}{2}\)
- D 3
Answer & Solution
Correct Answer
(A) \(\dfrac{3}{2}\)
Step-by-step Solution
Detailed explanation
The first line is given by \(\dfrac{x-1}{3} = \dfrac{y}{1} = \dfrac{z+2}{2}\). The direction vector of this line is \(\vec{v_1} = 3\hat{i} + 1\hat{j} + 2\hat{k}\). The second line is given by \(\dfrac{3x-1}{2b} = \dfrac{y-3}{-1} = \dfrac{z-1}{a}\). Rewriting this in standard…
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