Find the magnitude and direction of the vector <3,9>. Round angles to the nearest degree and other values to the nearest tenth. Answers:
9.5; 80°
9.5; 72°
9; 72°
9; 80°
magnitude = sqrt (3^3 + 9^2) = sqrt (90) = 3 sqrt (10) = 9.486832981
tan theta = 9/3 = 3
theta = 71.56505118
To find the magnitude of the vector <3,9>, we can use the Pythagorean theorem. The magnitude is given by the formula:
Magnitude = √(x^2 + y^2)
In this case, x = 3 and y = 9, so we substitute these values into the formula:
Magnitude = √(3^2 + 9^2)
Magnitude = √(9 + 81)
Magnitude = √90
Magnitude ≈ 9.5 (rounded to the nearest tenth)
To find the direction of the vector, we can use trigonometry. The direction is given by the formula:
Direction = arctan(y/x)
In this case, x = 3 and y = 9, so we substitute these values into the formula:
Direction = arctan(9/3)
Direction = arctan(3)
Direction ≈ 71.6° (rounded to the nearest degree)
Therefore, the magnitude of the vector <3,9> is approximately 9.5 and the direction is approximately 72°. Hence, the correct answer is:
9.5; 72°
To find the magnitude and direction of a vector, you can use some basic trigonometry. The magnitude of a vector is the length of the vector, and the direction is the angle that the vector makes with a reference axis.
For the given vector <3,9>, the magnitude can be found using the Pythagorean theorem. The magnitude (length) of a vector is given by the square root of the sum of the squares of its components. So, for this vector:
Magnitude = sqrt(3^2 + 9^2) = sqrt(9 + 81) = sqrt(90) ≈ 9.5.
The direction of the vector can be found by taking the inverse tangent of the ratio of its components. In this case, the direction is with respect to the positive x-axis in a counterclockwise direction. So, we have:
Direction = atan(9/3) ≈ atan(3) ≈ 72°.
Therefore, the correct answer is:
Magnitude = 9.5
Direction = 72°