Finding the Magnitude of the Sum of Two Perpendicular Vectors
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Exercise 5.1
$\vec{a}$ directed due north, $|\vec{a}| = 7$ units.
$\vec{b}$ directed due west, $|\vec{b}| = 24$ units.
Find $|\vec{a} + \vec{b}| = ?$
Given: $\vec{AB} = \vec{a}, \vec{BC} = \vec{b}, \vec{AC} = \vec{a} + \vec{b}$.
$|\vec{a}| = 7, |\vec{b}| = 24$
$|\vec{a} + \vec{b}| = |\vec{AC}| = ?$
$C$ is right angle triangle Pythagoras theorem:
$|AB|^2 + |BC|^2 = |AC|^2$
$24^2 + 7^2 = |AC|^2$
$576 + 49 = |AC|^2$
$625 = |AC|^2$
$|AC| = \sqrt{625} = 25$
$|\vec{a} + \vec{b}| = 25$ units.
This question includes visual content: A right-angled triangle labeled with vertices A, B, and C. The base AB is labeled with vector $\vec{a}$, and the vertical leg BC is labeled with vector $\vec{b}$. The hypotenuse AC is labeled with the vector sum $\vec{a} + \vec{b}$.
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Problem Analysis
We are given two vectors, vector a directed due north with a magnitude of 24 units, and vector b directed due west with a magnitude of 7 units.
Since north and west are perpendicular to each other, we can represent their sum as the hypotenuse of a right-angled triangle.
To find the magnitude of the resultant vector, we use the Pythagorean theorem, which states that the square of the hypotenuse equals the sum of the squares of the other two sides.
Now, let's plug in the given values of 24 and 7 into our equation.
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