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Math  /  Geometry

QuestionVector v\mathbf{v} has initial point P(11,12)P(11,12) and terminal point Q(19,−1)Q(19,-1). Vector w\mathbf{w} has initial point R(7,12)R(7,12) and terminal point S(−1,−1)S(-1,-1).
Part: 0/30 / 3 □\square
Part 1 of 3 (a) Find the magnitude of v\mathbf{v}. Give the exact answer in simplest form. ∥v∥=\|\mathbf{v}\|= □\square □\sqrt{\square}

Studdy Solution

STEP 1

1. Vector v\mathbf{v} is defined by its initial point P(11,12)P(11,12) and terminal point Q(19,−1)Q(19,-1).
2. The magnitude of a vector v=⟨x2−x1,y2−y1⟩\mathbf{v} = \langle x_2 - x_1, y_2 - y_1 \rangle is given by ∥v∥=(x2−x1)2+(y2−y1)2\|\mathbf{v}\| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.

STEP 2

1. Determine the components of vector v\mathbf{v}.
2. Apply the formula for the magnitude of a vector.
3. Simplify the expression to find the exact magnitude.

STEP 3

Determine the components of vector v\mathbf{v}:
v=⟨x2−x1,y2−y1⟩=⟨19−11,−1−12⟩ \mathbf{v} = \langle x_2 - x_1, y_2 - y_1 \rangle = \langle 19 - 11, -1 - 12 \rangle
v=⟨8,−13⟩ \mathbf{v} = \langle 8, -13 \rangle

STEP 4

Apply the formula for the magnitude of vector v\mathbf{v}:
∥v∥=(x2−x1)2+(y2−y1)2 \|\mathbf{v}\| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
∥v∥=82+(−13)2 \|\mathbf{v}\| = \sqrt{8^2 + (-13)^2}

STEP 5

Simplify the expression to find the exact magnitude:
∥v∥=82+(−13)2 \|\mathbf{v}\| = \sqrt{8^2 + (-13)^2}
∥v∥=64+169 \|\mathbf{v}\| = \sqrt{64 + 169}
∥v∥=233 \|\mathbf{v}\| = \sqrt{233}
The magnitude of vector v\mathbf{v} is:
233 \boxed{\sqrt{233}}

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