Question

In Exercises 61–64, find the magnitude ||v||, to the nearest hundredth, and the direction angle θ, to the nearest tenth of a degree, for each given vector v. v = (4i - 2j) - (4i - 8j)

Accepted Answer

First, simplify the given vector expression by subtracting the components of the vectors: $$\mathbf{v} = (4\mathbf{i} - 2\mathbf{j}) - (4\mathbf{i} - 8\mathbf{j}). $$This means subtract the $$i$$ components and the $$j$$ components separately. Calculate the $$i$$ component of $$\mathbf{v} by $$subtracting: $$4 - 4 = 0. $$Calculate the $$j$$ component of $$\mathbf{v} by $$subtracting: $$-2 - (-8) = -2 + 8 = 6. So$$, $$\mathbf{v} = 0\mathbf{i} + 6\mathbf{j}. $$Find the magnitude $$||\mathbf{v}||$$ using the formula for the magnitude of a vector: $$||\mathbf{v}|| = \sqrt{(v_x)^2$ + (v_y)^2$}$$, where $$v_x$$ and $$v_y$$ are the components of $$\mathbf{v}. $$Substitute the components into the magnitude formula: $$||\mathbf{v}|| = \sqrt{0^2$ + 6^2$} = \sqrt{36}. $$This will give the length of the vector. To find the direction angle $$	heta$$, use the formula $$	heta = 	an$$^{-1}$$\left(\frac{v_y}{v_x}ight). $$Since $$v_x = 0$$, consider the position of the vector on the coordinate plane to determine $$	heta$$ correctly.

In Exercises 61–64, find the magnitude ||v||, to the nearest hundredth, and the direction angle θ, to the nearest tenth of a degree, for each given vector v. v = (4i - 2j) - (4i - 8j)

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Key Concepts

Vector Subtraction

Magnitude of a Vector

Direction Angle of a Vector