Question:
What is the equation of the plane passing through the point (1, 1, 1) and perpendicular to the line whose direction ratio is (3, 2, 1)?
What is the equation of the plane passing through the point (1, 1, 1) and perpendicular to the line whose direction ratio is (3, 2, 1)?
Options:
- a) x + 2y + 3z = 6
- b) 3x + 2y + z = 6
- c) x + y + z = 3
- d) 3x + 2y + z = 0
- a) x + 2y + 3z = 6
- b) 3x + 2y + z = 6
- c) x + y + z = 3
- d) 3x + 2y + z = 0
correct answer : b)
Explanation:
Point (1, 1, 1) and d.rs of normal
are <3, 2, 1>
\ Equation of the plane is
3(x – 1) + 2(y – 1) + 1(z – 1) = 0
Þ 3x + 2y + z = 6
Point (1, 1, 1) and d.rs of normal
are <3, 2, 1>
\ Equation of the plane is
3(x – 1) + 2(y – 1) + 1(z – 1) = 0
Þ 3x + 2y + z = 6