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The magnitude of a vector is essentially its "length" or "size". Think of it as the distance from the starting point to the ending point of a vector in a coordinate system. To find this, we use the Pythagorean theorem. For a vector \( \mathbf{a} = -8\mathbf{i} + 15\mathbf{j} \), this means combining the square of each component. In formula terms, the magnitude \( |\mathbf{a}| \) is: \[ |\mathbf{a}| = \sqrt{(-8)^2 + (15)^2} = \sqrt{64 + 225} = \sqrt{289} = 17. \]

• First, square each component: \(-8^2=64\) and \(15^2=225\).
• Add these results together: \(64+225=289\).
• Finally, take the square root of the sum \(\sqrt{289}=17\).

The vector’s magnitude helps in understanding how "long" the vector is, regardless of direction.

The direction of a vector indicates which way it is pointing, relative to an origin point. Once the magnitude is known, finding a "unit vector" in the same direction is possible. A unit vector has a magnitude of 1, meaning it's typically a scaled-down version of the original vector. To determine this, divide each component of vector \( \mathbf{a} \) by its magnitude. For the vector \( \mathbf{a} = -8\mathbf{i} + 15\mathbf{j} \): \[ \mathbf{u} = \frac{\mathbf{a}}{|\mathbf{a}|} = \frac{-8\mathbf{i} + 15\mathbf{j}}{17} = -\frac{8}{17}\mathbf{i} + \frac{15}{17}\mathbf{j}. \]

• Take \(-8\) (\(\mathbf{i}\) direction) and divide it by the magnitude \(17\) to get \(-\frac{8}{17}\).
• Take \(15\) (\(\mathbf{j}\) direction) and divide it by the magnitude \(17\) to get \(\frac{15}{17}\).

The unit vector \(\mathbf{u}\) is in the same direction but has a unifying size of 1, making it useful for modeling direction without considering size.

Finding a vector in the opposite direction involves flipping the direction. To achieve this, multiply each component of the unit vector by \(-1\). This transformation switches all components to point in the reversed direction. The given unit vector \( \mathbf{u} = -\frac{8}{17}\mathbf{i} + \frac{15}{17}\mathbf{j} \) is switched by: \[ \mathbf{v} = -\left(-\frac{8}{17}\mathbf{i} + \frac{15}{17}\mathbf{j}\right) = \frac{8}{17}\mathbf{i} - \frac{15}{17}\mathbf{j}. \]

• Change \(-\frac{8}{17}\) to \(\frac{8}{17}\) for the \(\mathbf{i}\) component.
• Change \(\frac{15}{17}\) to \(-\frac{15}{17}\) for the \(\mathbf{j}\) component.

This process results in vector \(\mathbf{v}\), perfectly mirroring the original vector direction into an opposite course, essential for calculations that require reversing orientations.