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junior college 1 | H2 Maths
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Need help for part i! Thanks!
length of this normal vector
= √(2² + 1² + 2²)
= √9
= 3
Change equation to the form of the unit normal vector,
r.(3(⅔ ⅓ ⅔)) = 36
3r.(⅔ ⅓ ⅔) = 36
r.(⅔ ⅓ ⅔) = 36/3 = 12
Perpendicular distance = 12 units
See 1 Answer
The equation of the line has the fixed point B. The direction vector must be perpendicular to the plane so he's just using the plane's normal vector (2 1 2)
The point of intersection of this line and the plane will satisfy the equations of both.
Once this point is found, the vector from B to this point (which lies on the line, so this vector is perpendicular to the plane) can be found by subtracting OB from the position vector of this point. Its magnitude is simply the perpendicular distance from B to the point