Chapter 7: Problem 42
Find a unit vector in the direction of the given vector. $$\mathbf{v}=\langle 40,-9\rangle$$
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Chapter 7: Problem 42
Find a unit vector in the direction of the given vector. $$\mathbf{v}=\langle 40,-9\rangle$$
These are the key concepts you need to understand to accurately answer the question.
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Find the dot product \langle 11,12\rangle\(\cdot\langle-2,3\rangle\) Solution: Multiply the outer and inner components. $$ \langle 11,12\rangle \cdot\langle-2,3\rangle=(11)(3)+(12)(-2) $$ Simplify. \(\langle 11,12\rangle \cdot\langle-2,3\rangle=9\) This is incorrect. What mistake was made?
The projection of v onto \(\mathbf{u}\) is defined by proju \(\mathbf{v}=\left(\frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}|^{2}}\right) \mathbf{u}\) This vector is depicted below. Heuristically, this is the "shadow" of v on u. (GRAPH CAN'T COPY) a. Compute proju \(2 \mathbf{u}\) b. What is proju \(c \mathbf{u}\) for any \(c>0 ?\)
A car that weighs 2500 pounds is parked on a hill in San Francisco with a slant of \(40^{\circ}\) from the horizontal. How much force will keep it from rolling down the hill?
Find the Cartesian equation for \(r=\frac{a \sin (2 \theta)}{\cos ^{3} \theta-\sin ^{3} \theta}\).
Convert (-2,-2) to polar coordinates. Solution: Label \(x\) and \(y.\) \(x=-2, y=-2\) Find \(r . \quad r=\sqrt{x^{2}+y^{2}}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}\) Find \(\theta . \quad \tan \theta=\frac{-2}{-2}=1\) \(\theta=\tan ^{-1}(1)=\frac{\pi}{4}\) Write the point in polar coordinates. \(\quad\left(2 \sqrt{2}, \frac{\pi}{4}\right)\) This is incorrect. What mistake was made?
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