Final answer:
The magnitude of the vector 7(cos 135°i + sin135°j) is 7, and the direction angle is 135°.
Step-by-step explanation:
The question asks us to find the magnitude and direction angle of a vector given in component form as 7(cos 135°i + sin135°j). To find the magnitude, we use the Pythagorean theorem, and to find the direction angle, we use the arctangent function.
First, we calculate the components of the vector:
- x-component: 7×cos(135°) = 7×(-√2/2)
- y-component: 7×sin(135°) = 7×(√2/2)
Then, we find the magnitude of the vector which is the square root of the sum of the squares of its components.
Magnitude = √[(7×(-√2/2))^2 + (7×(√2/2))^2] = √(49/2 + 49/2) = √49 = 7
The direction angle is given by the arctangent of the ratio of the y-component to the x-component. Since both components are equal in magnitude but opposite in sign, the direction angle is 135°, which is the angle in the standard position (measured counterclockwise from the positive x-axis) that the vector makes.
Therefore, the magnitude of the vector is 7, and the direction angle is 135°.