Question

Find the equation of the plane through the point (2,4,3) that is perpendicular to the line r = (2i + 2j + 2k) t(i + 5j − 2k). Write the equation in the form indicated.

a) 7x + y + z = 27
b) 7x - y + z = 27
c) 7x + y + z = 0
d) None of the above

Answer 1

The equation of the plane through the point (2,4,3) and perpendicular to the given line is x + 5y - 2z = 14, which does not match any of the provided options. The correct answer is (d) None of the above.

Step-by-step explanation:

To find the equation of the plane through the point (2,4,3) that is perpendicular to the given line, we must first identify the direction vector of the line as the normal vector of the plane. The line is given by r = (2i + 2j + 2k) + t(i + 5j − 2k). The direction vector of the line can be found as the coefficients of t, which is (1, 5, -2). This is also the normal vector of the plane.

The general equation of a plane in point-normal form is (x − x0)nx + (y − y0)ny + (z − z0)nz = 0, where (x0, y0, z0) is a point on the plane and (nx, ny, nz) is the normal vector. Substituting the point (2,4,3) and the normal vector (1, 5, -2), we get the equation 1(x - 2) + 5(y - 4) - 2(z - 3) = 0. Simplifying, this becomes 1x + 5y - 2z = 1(2) + 5(4) - 2(3), which further simplifies to 1x + 5y - 2z = 14 or x + 5y - 2z = 14.

None of the provided options match this equation, so the correct answer is (d) None of the above.