JEE Mains · Maths · STD 12 - 11. three dimension geometry
The equation of the plane containing the straight line \(\frac{x}{2} = \frac{y}{3} = \frac{z}{4}\) and perpendicular to the plane containing the straight lines \(\frac{x}{3} = \frac{y}{4} = \frac{z}{2}\) and \(\frac{x}{4} = \frac{y}{2} = \frac{z}{3}\) is
- A \(x + 2y - 2z = 0\)
- B \(x - 2y + z = 0\)
- C \(5x + 2y - 4z = 0\)
- D \(3x + 2y - 3z = 0\)
Answer & Solution
Correct Answer
(B) \(x - 2y + z = 0\)
Step-by-step Solution
Detailed explanation
Vector along the normal to the plane containing the lines \(\frac{x}{3}=\frac{y}{4}=\frac{z}{2}\) and \(\frac{x}{4}=\frac{y}{2}=\frac{z}{3}\) \(\left( {8{\rm{ }}\hat i - \hat j - 10{\rm{ }}\hat k} \right).\) Vector perpendicular to the vectors \(2\hat i + 3\hat j + 4\hat k\) and…
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