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. $$ x^{2}+y^{2}+z^{2}=6, \quad x-y-z=0, \quad(2,1,1) $$

Short answer

The symmetric equations for the tangent line at point (2,1,1) are \(\frac{x-2}{-4} = \frac{y-1}{-6} = \frac{z-1}{2}\). The cosine of the angle between the gradient vectors at that point is 0, which means the surfaces are orthogonal at the point (2,1,1).

Step-by-step solution

Calculating normals to the surfaces

The given surfaces are in the form of \(x^{2}+y^{2}+z^{2}=6\) and \( x-y-z=0\). So the gradients of these surfaces are \(\nabla f1 = 2xi + 2yj + 2zk\), \(\nabla f2 = i - j - k\). At the point (2,1,1), the normals are \(\nabla f1 = 4i + 2j + 2k\) and \(\nabla f2 = i - j - k\).

Calculating direction vector for the tangent line

The gradient vectors are normal to the surfaces. Therefore, by taking cross product of the normals we will get the direction vector of the tangent line: \(D = \nabla f1 \times \nabla f2 = [ -4i - 6j + 2k]\).

Writing symmetric equations for the tangent line

After obtaining the direction vector for the tangent line, we can write the symmetric equation of the tangent line at the point (2,1,1): \(\frac{x-2}{-4} = \frac{y-1}{-6} = \frac{z-1}{2}\).

Calculating the angle between the gradient vectors

Using the dot product of the normal vectors we obtained in Step 1, we calculate the angle between the gradient vectors: dot product = \(\nabla f1 \cdot \nabla f2 = (4)(1) + (2)(-1) + (2)(-1) = 0\). Since the dot product is zero, the angle between the gradient vectors is 90 degrees. The cosine of 90 degrees is 0.

Concluding about orthogonality between the surfaces

The cosine value of the angle between the gradient vectors at the point (2,1,1) is 0. This means the vectors, hence the surfaces, are orthogonal at this point.

Key concepts

Symmetric Equations

Symmetric equations play a crucial role in the world of three-dimensional geometry, providing a convenient way to express lines in space. Imagine you're holding a string that stretches infinitely in both directions. The symmetric equations would be the numerical recipe that precisely traces the path this string takes through the air. They are formed by equating the ratios of the differences in the x, y, and z coordinates to a common directional vector. This vector is usually crafted from a point you know the line passes through and its direction or slope.

For instance, if a line crosses through the point (a, b, c) and proceeds in the direction given by the vector (p, q, r), the symmetric equations would be: \[ \frac{x-a}{p} = \frac{y-b}{q} = \frac{z-c}{r} \.\] In our exercise, the tangent line's symmetric equations are meticulously deduced using the known point of intersection (2,1,1) and the direction vector derived from the cross product of gradient vectors of the intersecting surfaces.

Gradient Vectors

Gradient vectors are like the compasses for hills and valleys on the surface of an equation. They show the direction of the steepest ascent if you were to walk on that surface, and their length indicates how steep that ascent would be. In calculus, the gradient vector of a function is constructed by taking partial derivatives with respect to each variable.

Imagine you are mapping a hilly terrain. The gradient vector at a point on the hill would point in the direction where the slope is steepest. In our context, the gradients of the surfaces give us the normals to these surfaces. Specifically, at (2,1,1), the gradients of our functions reveal the perpendicular directions to the tangent planes of the surfaces at that point.

Cross Product

The cross product is a mathematical operation that takes two vectors in three-dimensional space and produces a third vector perpendicular to both. Think of two intersecting roads: the cross product gives you a new direction that shoots straight up into the sky, forming a right angle with the ground.

In our exercise, the cross product is the secret ingredient to find the direction of the tangent line. By crossing the gradient vectors of the two surfaces—equivalent to two roads intersecting at a point—we unearth a new vector that is perpendicular to both and perfect to describe the direction of our elusive tangent line.

Dot Product

The dot product, also known as the scalar product, is a way to multiply two vectors resulting in a single number. It reveals a lot about the relationship between two vectors. For example, it can tell you whether two vectors are pointing in the same direction, are perpendicular, or are obtuse to each other.

Used skillfully in our exercise, the dot product of the gradient vectors is cleverly applied to compute the angle between the surfaces at the given point. It is this value that tells us if the surfaces are casually crossing paths at a graceful angle or if they behave more like the corners of a box, meeting conclusively at ninety degrees—in short, whether they are orthogonal or not.

Orthogonality

Orthogonality is a fancy term for perpendicularity in higher dimensions. When two lines, planes, or gradients are orthogonal, they meet at a perfect right angle—like the letter 'L' or the edges of a neatly wrapped gift box. In three dimensions, orthogonality is a mutual perpendicular stance held by two entities.

In the case of our problem, orthogonality is the grand finale—the final piece of the puzzle. After demonstrating that the dot product of the gradient vectors is zero, which signals an angle of ninety degrees between them, we confirm with triumph that the surfaces are indeed standing proudly, perpendicular to each other at the point of intersection.