Question

# Find the angle between the given vectors . Round to the nearest tenth of a degree. 1) u= -3i +6j ,v=5i + 2j 2) u= i -j , v=2i +3j Use the dot product

Vectors
Find the angle between the given vectors . Round to the nearest tenth of a degree.
1) u= -3i +6j ,v=5i + 2j
2) u= i -j , v=2i +3j
Use the dot product to determiine wheter the vectors are parallel, orthogonal , or neither .
1) v = 2i +j, w = i-2j
2) v = 4i-j , w=8i-2j
3) v= 3i +3j , w=3i -2j
Find proj w v
1) v = 2i +3j . w = 8i- 6j
2) v= 2i -3j . w =-3i +j

2021-06-03
2021-09-08

1) $$4=-3i+6j;\ v=5i+2j$$

$$\cos\theta=\frac{u\cdot v}{|u||v|}=\frac{(-3i+6j)(5i+2j)}{|(-3i+6j)|\cdot|(5i+2j)|}$$

$$\to\cos\theta=\frac{-15+12}{\sqrt{9+36}\cdot\sqrt{25+4}}=\frac{-3}{3\sqrt{5}\cdot\sqrt{29}}=\frac{-1}{\sqrt{145}}$$

$$\theta=\cos^{-1}(\frac{-1}{\sqrt{145}})=\pi-\cos^{-1}(\frac{1}{\sqrt{145}})=180-88-63^\circ$$

$$\theta=91.372$$

2) $$v=2i+j,\ w=i-ej$$

$$\to v\cdot w=(2i+j)(i-2j)=2-2=0$$

$$v\cdot w=0\Rightarrow$$ V and w are orthogonal

3) $$\text{proj}_wv=\frac{w\cdot v}{v\cdot v}\cdot v$$

$$\text{proj}_wv=\frac{(2i+3j)(8i-6j)}{(2i+3j)(2i+3j)}(2i+3j)$$

$$=\frac{16-18}{4+9}(2i+3j)=-\frac{2}{13}(2i+13j)$$