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Vector 
vec(u) has an initial point 
(5,1) and a terminal point 
(-3,4).
Find the magnitude of 
vec(u).
Enter an exact answer as an expression with a square root symbol or enter an approximate answer as a decimal rounded to the nearest hundredth.

|| vec(u)||=◻

Vector u \vec{u} has an initial point (5,1) (5,1) and a terminal point (3,4) (-3,4) .\newlineFind the magnitude of u \vec{u} .\newlineEnter an exact answer as an expression with a square root symbol or enter an approximate answer as a decimal rounded to the nearest hundredth.\newlineu= \|\vec{u}\|=\square

Full solution

Q. Vector u \vec{u} has an initial point (5,1) (5,1) and a terminal point (3,4) (-3,4) .\newlineFind the magnitude of u \vec{u} .\newlineEnter an exact answer as an expression with a square root symbol or enter an approximate answer as a decimal rounded to the nearest hundredth.\newlineu= \|\vec{u}\|=\square
  1. Use Distance Formula: To find the magnitude of the vector u\vec{u}, we need to use the distance formula, which is derived from the Pythagorean theorem. The distance formula for a vector with initial point (x1,y1)(x_1, y_1) and terminal point (x2,y2)(x_2, y_2) is:\newlineu=(x2x1)2+(y2y1)2||\vec{u}|| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
  2. Substitute Given Points: Substitute the given points into the distance formula:\newlineInitial point (x1,y1)=(5,1)(x_1, y_1) = (5, 1)\newlineTerminal point (x2,y2)=(3,4)(x_2, y_2) = (-3, 4)\newlineu=(35)2+(41)2||\vec{u}|| = \sqrt{(-3 - 5)^2 + (4 - 1)^2}
  3. Calculate Differences: Calculate the differences:\newline(35)=8(-3 - 5) = -8\newline(41)=3(4 - 1) = 3\newlineNow, square these differences:\newline(8)2=64(-8)^2 = 64\newline(3)2=9(3)^2 = 9
  4. Add Squares: Add the squares of the differences: 64+9=7364 + 9 = 73
  5. Find Magnitude: Take the square root of the sum to find the magnitude:\newlineu=73||\vec{u}|| = \sqrt{73}

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