Vector Components

1. Find the components of the vector between two points by subtracting the coordinates of the initial point from the terminal point. 2. For 2D vectors, the component form is $\vec{v} = (x_2 - x_1, y_2 - y_1)$. 3. For 3D vectors, the component form is $\vec{v} = (x_2 - x_1, y_2 - y_1, z_2 - z_1)$. --- **Exercise 1:** (a) Points (1, 5) and (4, 1) $$\vec{v} = (4 - 1, 1 - 5) = (3, -4)$$ (b) Points (0, 0, 4) and (2, 3, 0) $$\vec{v} = (2 - 0, 3 - 0, 0 - 4) = (2, 3, -4)$$ --- **Exercise 2:** (a) Points (-3, 3) and (2, 3) $$\vec{v} = (2 - (-3), 3 - 3) = (5, 0)$$ (b) Points (3, 0, 4) and (0, 4, 4) $$\vec{v} = (0 - 3, 4 - 0, 4 - 4) = (-3, 4, 0)$$ --- **Exercise 8:** Given initial point $P(-1, 3, -5)$ and vector $v = (6, 7, -3)$. (a) Vector $u$ has the same direction as $v$, so $u = kv$ for some $k > 0$. Terminal point $Q = P + u = (-1 + 6k, 3 + 7k, -5 - 3k)$. (b) Vector $u$ is oppositely directed to $v$, so $u = -kv$ for some $k > 0$. Terminal point $Q = P + u = (-1 - 6k, 3 - 7k, -5 + 3k)$. --- **Exercise 9:** Given $u = (4, -1)$, $v = (0, 5)$, $w = (-3, -3)$. (a) $u + w = (4 + (-3), -1 + (-3)) = (1, -4)$ (b) $v - 3u = (0 - 3*4, 5 - 3*(-1)) = (-12, 8)$ (c) $2(u - 5w) = 2((4 - 5*(-3), -1 - 5*(-3))) = 2(4 + 15, -1 + 15) = 2(19, 14) = (38, 28)$ (d) $3v - 2(u + 2w) = 3(0, 5) - 2((4 + 2*(-3), -1 + 2*(-3))) = (0, 15) - 2(4 - 6, -1 - 6) = (0, 15) - 2(-2, -7) = (0, 15) - (-4, -14) = (4, 29)$ --- **Exercise 10:** Given $u = (-3, 1, 2)$, $v = (4, 0, -8)$, $w = (6, -1, -4)$. (a) $v - w = (4 - 6, 0 - (-1), -8 - (-4)) = (-2, 1, -4)$ (b) $6u + 2v = 6(-3, 1, 2) + 2(4, 0, -8) = (-18, 6, 12) + (8, 0, -16) = (-10, 6, -4)$ (c) $-3(v - 8w) = -3((4 - 8*6, 0 - 8*(-1), -8 - 8*(-4))) = -3(4 - 48, 0 + 8, -8 + 32) = -3(-44, 8, 24) = (132, -24, -72)$ (d) $(2u - 7w) - (8v + u) = (2(-3,1,2) - 7(6,-1,-4)) - (8(4,0,-8) + (-3,1,2)) = ((-6,2,4) - (42,-7,-28)) - ((32,0,-64) + (-3,1,2)) = (-48,9,32) - (29,1,-62) = (-77,8,94)$ --- **Exercise 11:** Given $u = (-3, 2, 1, 0)$, $v = (4, 7, -3, 2)$, $w = (5, -2, 8, 1)$. (a) $v - w = (4 - 5, 7 - (-2), -3 - 8, 2 - 1) = (-1, 9, -11, 1)$ (b) $-u + (v - 4w) = (-(-3), -2, -1, 0) + (4 - 4*5, 7 - 4*(-2), -3 - 4*8, 2 - 4*1) = (3, -2, -1, 0) + (-16, 15, -35, -2) = (-13, 13, -36, -2)$ (c) $6(u - 3v) = 6((-3 - 3*4, 2 - 3*7, 1 - 3*(-3), 0 - 3*2)) = 6((-3 - 12, 2 - 21, 1 + 9, 0 - 6)) = 6((-15, -19, 10, -6)) = (-90, -114, 60, -36)$ (d) $(6v - w) - (4u + v) = (6*4 - 5, 6*7 - (-2), 6*(-3) - 8, 6*2 - 1) - (4*(-3) + 4, 4*2 + 7, 4*1 - 3, 4*0 + 2) = (24 - 5, 42 + 2, -18 - 8, 12 - 1) - (-12 + 4, 8 + 7, 4 - 3, 0 + 2) = (19, 44, -26, 11) - (-8, 15, 1, 2) = (27, 29, -27, 9)$ --- **Exercise 12:** Given $u = (1, -2, 3, 5, 0)$, $v = (0, 4, -1, 1, 2)$, $w = (7, 1, -4, -2, 3)$. (a) $v + w = (0 + 7, 4 + 1, -1 - 4, 1 - 2, 2 + 3) = (7, 5, -5, -1, 5)$ (b) $3(2u - v) = 3((2*1 - 0, 2*(-2) - 4, 2*3 - (-1), 2*5 - 1, 2*0 - 2)) = 3((2, -8, 7, 9, -2)) = (6, -24, 21, 27, -6)$ (c) $(3u - v) - (2u + 4w) = ((3*1 - 0, 3*(-2) - 4, 3*3 - (-1), 3*5 - 1, 3*0 - 2)) - ((2*1 + 4*7, 2*(-2) + 4*1, 2*3 + 4*(-4), 2*5 + 4*(-2), 2*0 + 4*3)) = (3, -10, 10, 14, -2) - (30, 0, -10, 2, 12) = (-27, -10, 20, 12, -14)$ (d) $\frac{1}{2}(w - 5v + 2u) + v = \frac{1}{2}((7, 1, -4, -2, 3) - 5(0, 4, -1, 1, 2) + 2(1, -2, 3, 5, 0)) + (0, 4, -1, 1, 2)$ Calculate inside parentheses: $w - 5v + 2u = (7, 1, -4, -2, 3) - (0, 20, -5, 5, 10) + (2, -4, 6, 10, 0) = (7, 1 - 20, -4 + 5, -2 - 5, 3 - 10) + (2, -4, 6, 10, 0) = (7, -19, 1, -7, -7) + (2, -4, 6, 10, 0) = (9, -23, 7, 3, -7)$ Half of this is $(4.5, -11.5, 3.5, 1.5, -3.5)$ plus $v$ gives $(4.5, -7.5, 2.5, 2.5, -1.5)$ --- **Exercise 13:** Given $3u + v - 2w = 3x + 2w$, solve for $x$. Rearranged: $$3x = 3u + v - 2w - 2w = 3u + v - 4w$$ Divide both sides by 3: $$x = u + \frac{v}{3} - \frac{4w}{3}$$ Using vectors from Exercise 11: $$x = (-3, 2, 1, 0) + \frac{1}{3}(4, 7, -3, 2) - \frac{4}{3}(5, -2, 8, 1)$$ Calculate each component: $$x = (-3 + \frac{4}{3} - \frac{20}{3}, 2 + \frac{7}{3} + \frac{8}{3}, 1 - 1 - \frac{32}{3}, 0 + \frac{2}{3} - \frac{4}{3}) = (-3 - \frac{16}{3}, 2 + 5, 1 - 1 - \frac{32}{3}, 0 - \frac{2}{3}) = (-\frac{25}{3}, 7, -\frac{32}{3}, -\frac{2}{3})$$ --- **Exercise 14:** Given $2u = v + x = 7x + w$, solve for $x$. From $v + x = 7x + w$, rearranged: $$v - w = 7x - x = 6x$$ So: $$x = \frac{v - w}{6}$$ Also, $2u = v + x$, so: $$x = 2u - v$$ Equate both expressions for $x$: $$2u - v = \frac{v - w}{6}$$ Multiply both sides by 6: $$12u - 6v = v - w$$ Rearranged: $$12u - 7v + w = 0$$ Using vectors from Exercise 12: $$12(1, -2, 3, 5, 0) - 7(0, 4, -1, 1, 2) + (7, 1, -4, -2, 3) = (12, -24, 36, 60, 0) - (0, 28, -7, 7, 14) + (7, 1, -4, -2, 3) = (19, -51, 25, 51, -11) \neq 0$$ No solution unless vectors satisfy this condition. --- **Final answers:** 1a: $(3, -4)$ 1b: $(2, 3, -4)$ 2a: $(5, 0)$ 2b: $(-3, 4, 0)$ 8a: $Q = (-1 + 6k, 3 + 7k, -5 - 3k)$ 8b: $Q = (-1 - 6k, 3 - 7k, -5 + 3k)$ 9a: $(1, -4)$ 9b: $(-12, 8)$ 9c: $(38, 28)$ 9d: $(4, 29)$ 10a: $(-2, 1, -4)$ 10b: $(-10, 6, -4)$ 10c: $(132, -24, -72)$ 10d: $(-77, 8, 94)$ 11a: $(-1, 9, -11, 1)$ 11b: $(-13, 13, -36, -2)$ 11c: $(-90, -114, 60, -36)$ 11d: $(27, 29, -27, 9)$ 12a: $(7, 5, -5, -1, 5)$ 12b: $(6, -24, 21, 27, -6)$ 12c: $(-27, -10, 20, 12, -14)$ 12d: $(4.5, -7.5, 2.5, 2.5, -1.5)$ 13: $\left(-\frac{25}{3}, 7, -\frac{32}{3}, -\frac{2}{3}\right)$ 14: $x = \frac{v - w}{6}$ and $x = 2u - v$ with condition $12u - 7v + w = 0$ for solution.