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A position vector defines a point in space relative to an origin. For a given position vector function \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} - \mathbf{k} \), you find a specific point by substituting a value into the function. Here, with \( t = \pi/4 \), you can compute the position by calculating \( \cos(\pi/4) = \sin(\pi/4) = \frac{\sqrt{2}}{2} \). Thus, the position vector becomes \( \mathbf{r}(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \mathbf{i} + \frac{\sqrt{2}}{2} \mathbf{j} - \mathbf{k} \). This tells you exactly where your point is in three-dimensional space, involving both trigonometric identities and direct substitution.

The tangent vector \( \mathbf{T} \) gives the direction in which a curve is heading at any point and is methodically found first by differentiating the position vector \( \mathbf{r}(t) \) with respect to \( t \). For \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} - \mathbf{k} \), its derivative is \( \mathbf{r}'(t) = -\sin t \mathbf{i} + \cos t \mathbf{j} \). Evaluating at \( t = \pi/4 \), we have \( \mathbf{r}'(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2} \mathbf{i} + \frac{\sqrt{2}}{2} \mathbf{j} \). The tangent vector is then normalized by dividing this vector by its own magnitude. Normalizing helps ensure \( \mathbf{T} \) is a unit vector. Normalized, the tangent vector retains its direction but has a magnitude of one.

The normal vector \( \mathbf{N} \), often called the principal normal vector, indicates the direction in which the curve is bending at a given point. It is found by differentiating the tangent vector \( \mathbf{T}(t) \) and normalizing it. For our tangent vector from before, \( \mathbf{T}(t) = -\sin t \mathbf{i} + \cos t \mathbf{j} \), differentiating gives \( \mathbf{T}'(t) = -\cos t \mathbf{i} - \sin t \mathbf{j} \). Evaluate it at \( t = \pi/4 \) resulting in \( \mathbf{T}'(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2} \mathbf{i} - \frac{\sqrt{2}}{2} \mathbf{j} \). After normalization, \( \mathbf{N} = -\frac{\sqrt{2}}{2} \mathbf{i} - \frac{\sqrt{2}}{2} \mathbf{j} \), showing us the unit direction in which the curve turns.

The binormal vector \( \mathbf{B} \) is perpendicular to both the tangent and normal vectors, forming a complete orthogonal basis known as the TNB (Tangent, Normal, Binormal) frame for a curve in space. To find it, you take the cross product of the tangent vector \( \mathbf{T} \) and normal vector \( \mathbf{N} \). When \( \mathbf{T} = -\frac{\sqrt{2}}{2} \mathbf{i} + \frac{\sqrt{2}}{2} \mathbf{j} \) and \( \mathbf{N} = -\frac{\sqrt{2}}{2} \mathbf{i} - \frac{\sqrt{2}}{2} \mathbf{j} \), the cross product results in \( \mathbf{T} \times \mathbf{N} = \mathbf{k} \). This vector, \( \mathbf{B} = \mathbf{k} \), helps in forming the osculating plane's equation with its orthogonality property regarding the preceding vectors.

The osculating plane closely approximates the surface of a curved path at any point. This plane is spanned by the tangent and normal vectors with the binormal vector \( \mathbf{B} \) acting as the normal vector to the plane. For the given vector \( \mathbf{r}(t) \) at \( t = \pi/4 \), the plane is defined with equation \( x\mathbf{i} + y\mathbf{j} + (z + 1)\mathbf{k} = 0 \), simplifying to \( z = -1 \). Essentially, this plane represents how the curve "sits" in three-dimensional space, showing its most significant bending direction. It acts like an immediate best-fit linear approximation to the curve.