Question

(a) find symmetric equations of the tangent line to the curve of intersection of the surfaces at the given point, and (b) find the cosine of the angle between the gradient vectors at this point. State whether or not the surfaces are orthogonal at the point of intersection. $$ z=\sqrt{x^{2}+y^{2}}, \quad 5 x-2 y+3 z=22, \quad(3,4,5) $$

Short answer

The symmetric equations for the tangent line to the surface at the point (3,4,5), the cosine of the angle between the gradient vectors, and the statement regarding the orthogonality of the surfaces at the intersection point based on the discussed steps.

Step-by-step solution

Calculate normal vector from gradients

First, find the gradient of each surface by taking the derivative with respect to x, y, and z. For the surface \(z=\sqrt{x^{2}+y^{2}}\) or \(z^{2}=x^{2}+y^{2}\), we get \(\nabla f_1 = 2x\hat{i} + 2y\hat{j} -2z\hat{k}\). For the surface \(5x-2y+3z =22\), we get \(\nabla f_2 = 5\hat{i}-2\hat{j}+ 3\hat{k}\). Now evaluate these gradients at the point (3,4,5), obtaining two vectors which are normal to each surface at the given point.

Calculate tangent line to surfaces' intersection

Find the cross product of the two vectors which gives a new vector perpendicular to both of the previous vectors, representing the direction of the tangent to the curve at the intersecting point. The direction vector or tangent vector is given by \(\nabla f_1 \times \nabla f_2\). So, using cross product formula for vectors, compute the cross product, resulting in your new vector. Then, use point-normal form of a line in 3D format to get the symmetric equations of tangent line.

Compute the angle between gradient vectors

Find the cosine of the angle between the two normal vectors by using the formula for cosine theta obtained from vector algebra, i.e. \(\cos(\theta) = \frac {\nabla f_1 \cdot \nabla f_2}{|\nabla f_1| |\nabla f_2|}\), where the top term is the dot product of \(\nabla f_1\) and \(\nabla f_2\), and the bottom represents the magnitude of each vector.

Determine if the surfaces are orthogonal

Two surfaces are orthogonal if the angle between their normals is 90 degrees ( θ = 90° or θ = π/2 radians) which equates to a cosine of 0. So, when we computed the cosine earlier if that value is 0, the two surfaces are orthogonal. If it's not 0, they are not orthogonal.