Find point of intersection of plane 3x-2y+7z-31=0 with line that passes through origin and is perpendicular to

Question

the plane

Answer 1

The normal vector to the plane is also the directional vector to the line perpendicular to it. The plane

3x - 2y + 7z - 31 = 0

has normal vector n = <3, -2, 7>

This vector, along with a point on the line, namely (0, 0, 0) defines the line r.

r(t) = <0, 0, 0> + tn = <0, 0, 0> + t<3, -2, 7> = t<3, -2, 7>
where t is a scalar that ranges over the real numbers

The line and plane will be equal at the point of intersection, so set the equations equal and solve for t.

3x - 2y + 7z - 31 = 0

3(3t) - 2(-2t) + 7(7t) - 31 = 0
9t + 4t + 49t = 31
62t = 31
t = 1/2

r(1/2) = (1/2)<3, -2, 7> = <3/2, -1, 7/2>

The point of intersection of the line r(t) and the plane is
(3/2, -1, 7/2).

Answer 2

If the line that passes through the origin and is perpendicular to the plane is vector a, then:

the point of intersection = a[a • (0, 0, 31/7)] / |a|²

Answer 3

as a results of fact the mandatory line is perpendicular to 3x - 2y= -4, the fabricated from slopes of the two strains might desire to be -a million. as a result, slope of the mandatory line would be = -2/3 Now, the usually occurring equation of a line is, : y = mx+c replace the fee of m as a result, y = (-2/3)x + c all of us understand that the factor (5,5) is in this line. Substituting the values of x and y, we are in a position to discover the fee of c. 5= (-10/3) + c as a result, c = -25/3 as a result, the basically perfect equation of the line would be : y = (-2/3)x - (25/3) that's, 3y = -2x -25 or 2x + 3y + 25 = 0 :)

Answer 4

this is very chim.
sori man hard luck