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The tangent vector is an essential concept in the study of curves in three-dimensional space. It's a vector that touches a curve at a given point and shows the direction in which the curve is heading at that particular point. To compute the tangent vector for a curve represented by a position vector \( \mathbf{r}(t) \), you first find the derivative, \( \mathbf{r}'(t) \). This gives you a vector that points along the curve. For example, in the problem, the position vector is \( \mathbf{r}(t) = e^t \mathbf{i} + e^t \cos t \mathbf{j} + e^t \sin t \mathbf{k} \). The derivative \( \mathbf{r}'(t) \) is calculated as \( e^t \mathbf{i} + (e^t \cos t - e^t \sin t) \mathbf{j} + (e^t \sin t + e^t \cos t) \mathbf{k} \).

• Evaluate at \( t=0 \): \( \mathbf{i} + \mathbf{j} \)
• Normalize to get the unit tangent vector \( \mathbf{T}(t) \)

The unit tangent vector is significant because it removes the effect of the speed along the curve, leaving only the direction.

The osculating plane at a point on a curve in 3D space is the plane that "kisses" the curve at that point and shares the same first and second derivative vectors of the curve at that point. It is the plane that best approximates the curve near that point. The osculating plane can be defined using the tangent vector \( \mathbf{T}(t) \) and the principal normal vector \( \mathbf{N}(t) \). The equation of the osculating plane involves the position vector and the binormal vector \( \mathbf{B}(t) \).

• The normal vector to the osculating plane is the binormal vector \( \mathbf{B}(t) \), found as the cross-product \( \mathbf{T}(t) \times \mathbf{N}(t) \).
• For the given \( \mathbf{r}(t) \), at \( t = 0 \), the equational form is \( \sqrt{6}(y - z) = 0 \).

The equation provides the relationship between the coordinates \( y \) and \( z \), indicating how they lie in the plane.

The normal plane is constructed to be perpendicular to the curve's tangent direction at a given point. This plane captures how the curve bends around the point and is spanned by the principal normal vector \( \mathbf{N}(t) \) and the binormal vector \( \mathbf{B}(t) \).

• To define it, consider the vectors \( \mathbf{N}(t) \) and \( \mathbf{B}(t) \), both perpendicular to the tangent vector \( \mathbf{T}(t) \).
• In this exercise, the equation of the normal plane at \( t = 0 \) is given by \( \sqrt{3}(x - y + z - 1) = 0 \).

This equation shows the spatial relationship of the curve as viewed in the plane, focusing primarily on the curve's bending rather than its orientation.

The rectifying plane is formed by the tangent vector \( \mathbf{T}(t) \) and the binormal vector \( \mathbf{B}(t) \). Unlike the osculating and normal planes, the rectifying plane primarily profiles how the curve twists rather than bends at a point.

• The tangent vector \( \mathbf{T}(t) \) and binormal vector \( \mathbf{B}(t) \) define the plane.
• The equation for the rectifying plane, for this problem at \( t = 0 \), is derived as \( x + y - 2 = 0 \).

This equation highlights the relation between \( x \) and \( y \), indicating their compensation for the curve's twisting in spatial terms.