Given the plane 2x-3y+7z=4,
a) Find the equation for the line perpendicular to the plane that intersects the plane at the point P=(6,5,1).
b) How far is it from the point (4,-2,3) to the plane?
a)(x-6)/2=(y-5)/(-3)=(z-1)/7
b)The normal equation of the plane
(2/sqrt(62))x-(3/sqrt(62))y+(7/sqrt(62))z-
-4/sqrt(62)=0 62=2^2+(-3)^2+7^2
The distance=(2*4-3*(-2)+7*3-4)/sqrt(62)=
sqrt(31/2)
why do you have = signs in part a?
To find the equation for the line perpendicular to the plane and intersects the plane at a specific point P, we need to find the normal vector of the plane.
a) Finding the normal vector of the plane:
The coefficients of x, y, and z in the equation of the plane, 2x - 3y + 7z = 4, give us the components of the normal vector. So the normal vector to the plane is (2, -3, 7).
b) Calculating the distance from a point to the plane:
To find the distance between a point and the plane, we can use the formula:
distance = |Ax + By + Cz + D| / sqrt(A^2 + B^2 + C^2)
The equation of the plane, 2x - 3y + 7z = 4, matches the form Ax + By + Cz + D = 0, where A = 2, B = -3, C = 7, and D = -4. The coordinates of the point are (4, -2, 3).
Using the formula, we can calculate the distance:
distance = |(2 * 4) + (-3 * -2) + (7 * 3) + (-4)| / sqrt((2^2) + (-3^2) + (7^2))
Simplifying the expression:
distance = |8 + 6 + 21 - 4| / sqrt(4 + 9 + 49)
distance = |31| / sqrt(62)
distance = 31 / sqrt(62)
So the distance from the point (4, -2, 3) to the plane 2x - 3y + 7z = 4 is 31 / sqrt(62) units.