Find the perpendicular distance from the plane r.(2i-14j+5k)=10 to the origin.
Please help...I have no idea what to do!
if Ax + By + Cz + C = 0 is the general equation of a plane, then the distance from the origin (0,0,0) to that plane is
āCā/ā(A^2 + B^2 + C^2)
Can you take it from there?
2/3
To find the perpendicular distance from a plane to a point, you can use the formula:
distance = |Ax + By + Cz + D| / sqrt(A^2 + B^2 + C^2)
where (A, B, C) are the coefficients of the normal vector of the plane, and (x, y, z) are the coordinates of the point.
In this case, the equation of the plane is r . (2i - 14j + 5k) = 10. Comparing this to the standard form Ax + By + Cz + D = 0, we have A = 2, B = -14, C = 5, and D = -10.
The normal vector of the plane is (A, B, C), which is (2, -14, 5).
To find the perpendicular distance from the plane to the origin, we can substitute (0, 0, 0) for (x, y, z) in the formula:
distance = |2(0) -14(0) + 5(0) + (-10)| / sqrt(2^2 + (-14)^2 + 5^2)
Simplifying the expression, we get:
distance = |-10| / sqrt(4 + 196 + 25) = 10 / sqrt(225) = 10 / 15 = 2/3
Therefore, the perpendicular distance from the plane r . (2i - 14j + 5k) = 10 to the origin is 2/3 units.