Question

The angle between the lines \(3x = 2y = -z\) and \(-x = 6y = -4z\) is

• A. \(\pi/3\)
• B. \(\pi/4\)
• C. \(\pi/2\)
• D. \(\pi/6\)

Hint:

If the dot product of direction vectors is zero, then the lines are perpendicular and the angle is \(\pi/2\).

Answer 1

The Correct Option is C

Concept:
The angle between two lines is found using the angle between their direction vectors. If \(\vec{d}_1\) and \(\vec{d}_2\) are direction vectors, then: \[ \cos\theta=\frac{\vec{d}_1\cdot\vec{d}_2}{|\vec{d}_1||\vec{d}_2|} \] ip

Step 1: Find direction ratios of the first line.
From \[ 3x=2y=-z=t \] we get \[ x=\frac{t}{3},\quad y=\frac{t}{2},\quad z=-t \] So direction ratios are proportional to: \[ \left(\frac13,\frac12,-1\right) \] Multiplying by \(6\), a direction vector is: \[ \vec{d}_1=(2,3,-6) \] ip

Step 2: Find direction ratios of the second line.
From \[ -x=6y=-4z=s \] we get \[ x=-s,\quad y=\frac{s}{6},\quad z=-\frac{s}{4} \] So direction ratios are proportional to: \[ \left(-1,\frac16,-\frac14\right) \] Multiplying by \(12\), a direction vector is: \[ \vec{d}_2=(-12,2,-3) \] ip

Step 3: Find the dot product.
\[ \vec{d}_1\cdot\vec{d}_2 = (2)(-12)+(3)(2)+(-6)(-3) \] \[ =-24+6+18=0 \] ip

Step 4: Conclude the angle.
Since the dot product is zero, the lines are perpendicular. Therefore, \[ \theta=\frac{\pi}{2} \] ip Hence, the correct answer is:
\[ \boxed{(C)\ \frac{\pi}{2}} \]