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'Find the dot product and the angle \ \\theta \ between two vectors \\( \\vec{a}=(\\sqrt{3}, 1), \\vec{b}=(-1,-\\sqrt{3}) \\).'

'To show the condition for points \ \\mathrm{O}, \\mathrm{A}, \\mathrm{B} \ to be collinear, demonstrate the following property:'

'Let $\\mathrm{ABCD}$ be the base of a quadrangular pyramid $\\mathrm{OABCD}$ such that $\\overrightarrow{\\mathrm{OA}}+\\overrightarrow{\\mathrm{OC}}=\\overrightarrow{\\mathrm{OB}}+\\overrightarrow{\\mathrm{OD}}$. For four nonzero real numbers $p, q, r, s$, let points $\\mathrm{P}, \\mathrm{Q}, \\mathrm{R}, \\mathrm{S}$ be defined by $\\overrightarrow{\\mathrm{OP}}=p \\overrightarrow{\\mathrm{OA}}$, $\\overrightarrow{\\mathrm{OQ}}=q \\overrightarrow{\\mathrm{OB}}$, $\\overrightarrow{\\mathrm{OR}}=r \\overrightarrow{\\mathrm{OC}}$, $\\overrightarrow{\\mathrm{OS}}=s \\overrightarrow{\\mathrm{OD}}$. Show that if the four points $\\mathrm{P}, \\mathrm{Q}, \\mathrm{R}, \\mathrm{S}$ lie on the same plane, then $\\frac{1}{p}+\\frac{1}{r}=\\frac{1}{q}+\\frac{1}{s}$.'

'In tetrahedron OABC, let L be the point dividing side AB in the ratio 1:3, M be the point dividing side OC in the ratio 3:1, N be the point dividing segment CL in the ratio 3:2, and P be the intersection of segments LM and ON. If OA=a, OB=b, OC=c, express ON and OP in terms of a, b, and c.'

'Vector Operations (2) \\(2 \\vec{a} + 3 \\vec{b} / / (\\vec{a} - 4 \\vec{b}) , \\vec{a} \\neq \\overrightarrow{0}, \\vec{b} \\neq \\overrightarrow{0} \\) implies that \\\vec{a} / / \\vec{b}\.'

'Vectors a and b on the coordinate plane are not parallel. Let a and b be position vectors corresponding to points A and B, respectively. Also, for positive real numbers x and y, let x a and y b be position vectors corresponding to points P and Q. When line segment PQ divides line segment AB into the ratio 2:1, find the minimum value of xy. All position vectors are considered with respect to the origin O.'

'Vector equation of a line perpendicular to vector n (which is not equal to zero) and passing through point A(vector a) is n·(p-a)=0'

'Definition of Cross Product'

'Definition of dot product, Dot product and components \ \\vec{a} \\neq \\overrightarrow{0}, \\vec{b} \\neq \\overrightarrow{0} \.\nDefinition of dot product\nLet the angle between \ \\vec{a} \ and \ \\vec{b} \ be \\( \\theta\\left(0^{\\circ} \\leqq \\theta \\leqq 180^{\\circ}\\right) \\) , then\n\\\vec{a} \\cdot \\vec{b}=|\\vec{a}||\\vec{b}| \\cos \\theta\\nDot product and components\nIf \\( \\vec{a}=\\left(a_{1}, a_{2}\\right), \\vec{b}=\\left(b_{1}, b_{2}\\right) \\), then\n\\\vec{a} \\cdot \\vec{b}=a_{1} b_{1}+a_{2} b_{2}\\nAlso, let the angle between \ \\vec{a} \ and \ \\vec{b} \ be \ \\theta \, then\n\\\cos \\theta=\\frac{\\vec{a} \\cdot \\vec{b}}{|\\vec{a}||\\vec{b}|}=\\frac{a_{1} b_{1}+a_{2} b_{2}}{\\sqrt{a_{1}{ }^{2}+a_{2}{ }^{2}} \\sqrt{b_{1}{ }^{2}+b_{2}{ }^{2}}}\'

"Proof problem about uniform circular motion\nPoint P moves on a circular path with radius r centered at the origin O, starting from fixed point P₀, such that OP rotates at a rate of ω radians per second.\n(1) Find the magnitude v of P's velocity.\n(2) Show that P's velocity vector and acceleration vector are perpendicular."

'Given vectors a and b satisfying |a|=5,|b|=3,|a-2 b|=7. If the angle between a-2 b and 2 a+b is θ, find the value of cos θ.'

'Find the dot product and the angle $\\theta$ between the two vectors $\\vec{a}=(-2,1,2)$ and $\\vec{b}=(-1,1,0)$.'

'For two non-zero vectors a and b, if there exists only one real number t such that a + tb and a + 3tb are perpendicular, find the angle θ between a and b.'

'For non-zero vectors a and b, such that a+2b and a-2b are orthogonal, and |a+2b|=2|b|.'

'Describe the conditions for vectors a and b to be perpendicular.'

'There is a point \\mathrm{P} inside \\triangle \\mathrm{ABC} such that 2 \\overrightarrow{PA} + 3 \\overrightarrow{PB} + 5 \\overrightarrow{PC} = \\overrightarrow{0}. (1) Where is the point \\mathrm{P} located? (2) Find the ratio of the areas of \\triangle \\mathrm{PBC} : \\triangle \\mathrm{PCA} : \\triangle \\mathrm{PAB}.'

'In a square ABCD with side length 2, find the following dot products.'

'Prove using vectors that the equation 2(AB^2+BC^2)=AC^2+BD^2 holds in the parallelogram ABCD.'

'Given 4 points A(2,1,2), B(-2,2,1), C(-3,-4,2), D(a, b, 5).'

'Basic Example 22 Standard Example 33'

'In tetrahedron OABC, let the midpoint of side OA be P, the midpoint of side BC be Q, the point dividing segment PQ in the ratio 1:2 be R, and the intersection point of line OR and plane ABC be S. If OA=vector a, OB=vector b, OC=vector c, then express OS in terms of vector a, vector b, and vector c.'

'Chapter 2 Vectors in Space - 39\n(1) Let P(x, y, z) and \\overrightarrow{AP}=(x-\\frac{1}{2}, y+\\frac{3}{2}, z-1)\nSince point P lies on line AB, \\overrightarrow{AP}=t\\overrightarrow{AB} for some real number t\n\\overrightarrow{AB}=(\\frac{3}{2}, \\frac{5}{2}, -4), hence\n\\[\\left(x-\\frac{1}{2}, y+\\frac{3}{2}, z-1\\right)=t\\left(\\frac{3}{2}, \\frac{5}{2}, -4\\right)\\]\nThis gives \\left(x-\\frac{1}{2}, y+\\frac{3}{2}, z-1\\right)=\\left(\\frac{3}{2} t, \\frac{5}{2} t, -4 t\\right)\nTherefore, x=\\frac{3}{2} t+\\frac{1}{2}, y=\\frac{5}{2} t-\\frac{3}{2}, z=-4 t+1\nSince point P lies in the yz plane, the x component of \\overrightarrow{OP} is 0\nThus, \\frac{3}{2} t+\\frac{1}{2}=0 gives t=-\\frac{1}{3}\nTherefore, the coordinates of point P are \\left(0, -\\frac{7}{3}, \\frac{7}{3}\\right)\n(2) From (1), \\overrightarrow{OH}=(\\frac{3}{2} t+\\frac{1}{2}, \\frac{5}{2} t-\\frac{3}{2}, -4 t+1). Since AB ⊥ OH, we have \\overrightarrow{AB} \\cdot \\overrightarrow{OH}=0\nTherefore, \\frac{3}{2}\\left(\\frac{3}{2} t+\\frac{1}{2}\\right)+\\frac{5}{2}\\left(\\frac{5}{2} t-\\frac{3}{2}\\right)-4(-4 t+1)=0\nSolving this gives t=\\frac{2}{7}\nThus, the coordinates of point H are \\left(\\frac{13}{14}, -\\frac{11}{14}, -\\frac{1}{7}\\right)'

'In \ \\triangle \\mathrm{OAB} \, given that \ \\mathrm{OA}=2, \\mathrm{OB}=3, \\mathrm{AB}=\\sqrt{7} \ and let the orthocenter be denoted as H. Define \ \\overrightarrow{\\mathrm{OA}}=\\vec{a} \ and \ \\overrightarrow{\\mathrm{OB}}=\\vec{b} \, then answer the following questions:'

'Find the dot product and the angle $\\theta$ between the vectors $\\vec{a}$ and $\\vec{b}$.(1) $\\vec{a}=(3,4), \\vec{b}=(7,1)$(2) $\\vec{a}=(1, \\sqrt{3}), \\vec{b}=(-\\sqrt{3},-1)$(3) $\\vec{a}=(\\sqrt{2},-2), \\vec{b}=(-1, \\sqrt{2})$(4) $\\vec{a}=(-1,2), \\vec{b}=(6,3)$'

'In right triangle ABC, with vectors AB = a, AC = b, and BC = c, find the dot products a⋅b, b⋅c, and c⋅a.'

'Find the value of $x$ that makes $\\vec{a}=(x+2,1)$ and $\\vec{b}=(1,-6)$ perpendicular.'

'For △OAB, let →OP=s→OA+t→OB. Find the range of point P as real numbers s, t satisfying the following conditions: (1) s+t=3 (2) 2s+3t=1, s≥0, t≥0 (3) 2s+3t≤6, s≥0, t≥0'

'\\( 4 \\overrightarrow{\\mathrm{AP}} = \\frac{1}{p^{2} - p + 1}\\{(1 - p) \\overrightarrow{\\mathrm{AB}} + p \\overrightarrow{\\mathrm{AC}}\\} \\)'

'(1) \ \\overrightarrow{\\mathrm{AC}}=\\overrightarrow{\\mathrm{AF}}+\\overrightarrow{\\mathrm{FC}}=\\vec{b}+2 \\vec{a} \\n\ \\vec{b}=\\overrightarrow{\\mathrm{AF}} \ therefore \\( \\vec{a}=\\frac{1}{2}(\\overrightarrow{\\mathrm{AC}}-\\overrightarrow{\\mathrm{AF}}) \\)\nThus \\( \\overrightarrow{\\mathrm{AP}}=2 s \\vec{a}+(3-3 s) \\vec{b} \\)\n\\[\n\egin{array}{l}\n=2 s \\cdot \\frac{1}{2}(\\overrightarrow{\\mathrm{AC}}-\\overrightarrow{\\mathrm{AF}})+(3-3 s) \\overrightarrow{\\mathrm{AF}} \\\\\n=s \\overrightarrow{\\mathrm{AC}}+(3-4 s) \\overrightarrow{\\mathrm{AF}}\n\\end{array}\n\\]\n\nThe conditions for point P to be in the interior of \ \\triangle \\mathrm{ACF} \ are\n\\[\ns>0,3-4 s>0, s+(3-4 s)<1\n\\]\n\nTherefore, \ s>0, s<\\frac{3}{4}, s>\\frac{2}{3} \\nTherefore, the range of the real number \ s \ that we seek is \ \\frac{2}{3}<s<\\frac{3}{4} \\nDividing both sides by \ k \,\n\ \\overrightarrow{\\mathrm{AP}}=s^{\\prime} \\overrightarrow{\\mathrm{AB}}+t^{\\prime} \\overrightarrow{\\mathrm{AC}} \\n\ s^{\\prime}+t^{\\prime}=1, \\quad s^{\\prime} \\geqq 0, t^{\\prime} \\geqq 0 \\nFormulated as \ s^{\\prime}+t^{\\prime}=1, \\quad s^{\\prime} \\geqq 0, t^{\\prime} \\geqq 0 \\n\ \\overrightarrow{\\mathrm{B}^{\\prime} \\mathrm{C}^{\\prime}}=\\overrightarrow{\\mathrm{AC}^{\\prime}}-\\overrightarrow{\\mathrm{AB}^{\\prime}} \\nRefer to dot product and triangle area formula, see Example 5.\n\ \\triangle \\triangle \\mathrm{ADG} \\triangle \\triangle \\mathrm{AEF} \,\n\ \\mathrm{AD}: \\mathrm{AE}=1: 2 \ so\n\ S_{1}: S_{2}=1^{2}: 2^{2} \\n\\( \\varangle \\vec{a}=\\frac{1}{2}(\\overrightarrow{\\mathrm{AC}}-\\vec{b}) \\)\nWhile considering \ \\triangle \\mathrm{ACF} \, transform \ \\overrightarrow{\\mathrm{AP}}=\\overrightarrow{\\mathrm{AC}}+\\square \\overrightarrow{\\mathrm{AF}} \ into this form.\nThe conditions for point P to be in the interior of \ \\triangle \\mathrm{ABC} \ are\n\\n\egin{\overlineray}{l}\n\\overrightarrow{\\mathrm{AP}}=s \\overrightarrow{\\mathrm{AB}}+t \\overrightarrow{\\mathrm{AC}} \\\\\ns>0, \\quad t>0, s+t<1\n\\end{\overlineray}\n\\nConsider taking the intersection.'

'Exercise 5 III → Book p.78\n(\egin{array}{l}4 \\overrightarrow{\\mathrm{AP}}+3 \\overrightarrow{\\mathrm{BP}}+2 \\overrightarrow{\\mathrm{CP}}+\\overrightarrow{\\mathrm{DP}}=\\overrightarrow{0} \\text { from } \\\\4 \\overrightarrow{\\mathrm{AP}}+3(\\overrightarrow{\\mathrm{AP}}-\\overrightarrow{\\mathrm{AB}})+2(\\overrightarrow{\\mathrm{AP}}-\\overrightarrow{\\mathrm{AC}})+(\\overrightarrow{\\mathrm{AP}}-\\overrightarrow{\\mathrm{AD}})=\\overrightarrow{0} \\\\ \\text { Therefore } 10 \\overrightarrow{\\mathrm{AP}}=3 \\overrightarrow{\\mathrm{AB}}+2 \\overrightarrow{\\mathrm{AC}}+\\overrightarrow{\\mathrm{AD}} \\\\ =3 \\overrightarrow{\\mathrm{AB}}+2(\\overrightarrow{\\mathrm{AB}}+\\overrightarrow{\\mathrm{AD}})+\\overrightarrow{\\mathrm{AD}} \\\\ =5 \\overrightarrow{\\mathrm{AB}}+3 \\overrightarrow{\\mathrm{AD}} \\end{array})'

'Independence and dependence one time'

'Point P moves along the side OA, so it can be represented as OP = sOA (0 ≤ s ≤ 1). Also, point Q moves along the side BC, so it can be represented as OQ = (1-t)OB + tOC (0 ≤ t ≤ 1). Find the minimum value of the square of PQ at this time.'

'In the coordinate space with the origin as the center, let A(5,4,-2). What kind of figure does the set of points P(x, y, z) satisfying $|\\overrightarrow{OP}|^{2}-2\\overrightarrow{OA} \\cdot \\overrightarrow{OP}+36=0$ represent? Also, express the equation in terms of x, y, z.'

'Solve the following vector problem. \ a \\overrightarrow{\\mathrm{PA}}+b \\overrightarrow{\\mathrm{PB}}+c \\overrightarrow{\\mathrm{PC}}=\\overrightarrow{0} \ leads to \\(-a \\overrightarrow{\\mathrm{AP}}+b(\\overrightarrow{\\mathrm{AB}}-\\overrightarrow{\\mathrm{AP}})+c(\\overrightarrow{\\mathrm{AC}}-\\overrightarrow{\\mathrm{AP}})=\\overrightarrow{0}\\)'

'Since \ \\overrightarrow{\\mathrm{AB}} \\perp \\overrightarrow{\\mathrm{PH}} \, we have \ \\overrightarrow{\\mathrm{AB}} \\cdot \\overrightarrow{\\mathrm{PH}} = 0 \, which implies \\( 2(2k-9) + 1 \\times (k-6) - 1 \\times (-k) = 0 \\). Therefore, \ k = 4 \'

'Find the angle θ between vectors a and b such that a-(2/5)b is perpendicular to a+b, and a is perpendicular to a-b.'

'(1) Prove:'

'Example 10 Inner product calculation (definition)'

'In tetrahedron ABCD, let M be the midpoint of edge AB and N be the midpoint of edge CD.\n(1) Does there exist a point P that satisfies the equation PA + PB = PC + PD? Provide a proof and answer.'

'Inner product equations in triangle shape problems'

'Explain the method of calculating the dot product of vectors and perform the calculation using a specific example.'

'Example 18 Find the position vector of the orthocenter of a triangle\nIn triangle OAB, with OA=5, OB=6, AB=7, and orthocenter H. Let OA vector be a and OB vector be b, answer the following questions:\n1. Find the dot product a·b.\n2. Express OH vector in terms of a and b.'

'In tetrahedron OABC, let ⃗a=⇀OA, ⃗b=⇀OB, ⃗c=⇀OC. Let the midpoints of segments OA, OB, OC, BC, CA, AB be denoted as L, M, N, P, Q, R respectively, and let ⃗p=⇀LP, ⃗q=⇀MQ, ⃗r=⇀NR.'

'The vector equation of a straight line passing through point A (a vector) and perpendicular to n (not equal to the zero vector) is: n · (p - a) = 0.'

'Calculate the dot product of vectors and explain its geometric meaning.'

'Prove the 10th vector inequality'

'(1) Since \ \\mathrm{AB} \\parallel \\mathrm{DE} \, therefore \ \\overrightarrow{\\mathrm{DE}}=k \\overrightarrow{\\mathrm{AB}} \. Find the real number \ k \, and determine the values of \ a \ and \ b \ when \\( \\overrightarrow{\\mathrm{AB}}=(-3,0,4) \\) and \\( \\overrightarrow{\\mathrm{DE}}=(6, a+1, b+3) \\).'

'Define dot product and components where \ \\vec{a} \\neq \\overrightarrow{0}, \\quad \\vec{b} \\neq \\overrightarrow{0} \.\nThe angle between \ \\vec{a} \ and \ \\vec{b} \ is denoted by \\( \\theta (0^{\\circ} \\leqq \\theta \\leqq 180^{\\circ}) \\).\nThen, the dot product of \ \\vec{a} \ and \ \\vec{b} \ is given by \\\vec{a} \\cdot \\vec{b}=|\\vec{a}||\\vec{b}| \\cos \\theta\\nFor \\( \\vec{a} = (a_1, a_2), \\vec{b} = (b_1, b_2) \\), the dot product of the vectors is \\\vec{a} \\cdot \\vec{b}=a_1 b_1 + a_2 b_2\\nAlso, the cosine of the angle \ \\theta \ is given by \\( \\cos \\theta = \\frac{\\vec{a} \\cdot \\vec{b}}{|\\vec{a}||\\vec{b}|} = \\frac{a_1 b_1 + a_2 b_2}{\\sqrt{a_1^2 + a_2^2} \\sqrt{b_1^2 + b_2^2}}\\]'

'Let P(0, s, 0) and Q(t+1, t+3, -t). Calculate PQ^2 = (t+1)^2 + (t+3-s)^2 + (-t)^2 = s^2 - 2st + 3t^2 - 6s + 8t + 10 = s^2 - 2(t+3)s + 3t^2 + 8t + 10 = {s-(t+3)}^2 - (t+3)^2 + 3t^2 + 8t + 10 = (s-t-3)^2 + 2t^2 + 2t + 1 = (s-t-3)^2 + 2(t+1/2)^2 + 1/2. When s-t-3=0 and t+1/2=0, i.e. s=5/2, t=-1/2, the minimum value is 1/2. Therefore, PQ achieves a minimum value of 1/sqrt(2) when s=5/2, t=-1/2. In other words, when P(0,5/2,0), Q(1/2,5/2,1/2), the minimum value is 1/sqrt(2).'

'Prove that for four points O, A, B, C in space that are not on the same plane, if vector OA=a, vector OB=b, and vector OC=c, then any vector p can be uniquely expressed in the form p=s*a+t*b+u*c (where s, t, u are real numbers).'

'|𝛼 + t𝛽| is greater than or equal to 0, therefore, when |𝛼 + t𝛽|^2 is minimized, |𝛼 + t𝛽| is also minimized. Hence, |𝛼 + t𝛽| takes the minimum value of √26 at t=-1. Another solution is to take point O as the origin, 𝛼 = OA, and 𝛽 = OB. The point C determined by 𝛼 + t𝛽 = OC passes through point A and lies on a line parallel to OB. Therefore, for |𝛼 + t𝛽| to be minimized, (𝛼 + t𝛽) must be perpendicular to 𝛽. In this case, we have (𝛼 + t𝛽)·𝛽 = 0, which leads to solving (2 + t) * 1 + (-4 - t) * (-1) + (-3 + t) * 1 = 0, resulting in 3t + 3 = 0, hence t = -1. At this point, |𝛼 + t𝛽| = |𝛼 - 𝛽| = √(1^2 + (-3)^2 + (-4)^2) = √26. Therefore, |𝛼 + t𝛽| achieves the minimum value of √26 at t=-1.'

'Additional Reference\nReference: Find the cross product \\vec{u} of \\overrightarrow{\\mathrm{OA}} and \\overrightarrow{\\mathrm{OB}}\n\n\\vec{u} = (1 \\cdot 0-(-2)\\cdot 4, (-2)\\cdot 3-2 \\cdot 0, 2 \\cdot 4-1\\cdot 3) = (8, -6, 5)'

'1. Maximum and minimum values of dot product of vectors\n2. Vectors with trajectory, region\n3. Maximum volume of tetrahedron\n4. Treatment of vector equations\n5. Geometric figures in space (spherical surface)\n6. Limit of a point moving on the complex plane\n7. Range of existence of points on the complex plane\n8. Fusion problems of properties of complex numbers and integers\n9. Parametric representation and trajectory\n10. Fusion problems of complex plane, equations, and curves'

'Example 11 | Calculation of Dot Product (Components)'

'(2) continuation'

'In a regular tetrahedron ABCD with edge length 2, find the dot product of vector AB and vector AC.'

'(2) $\\ vec{a} \\ cdot \\ vec{b} = 2 \\ times (-2+ \\ sqrt{3})+(-1) \\ times(1+2 \\ sqrt{3})=-5$ \\ n \\ also $| \\ vec{a}| = \\ sqrt{2^{2}+(-1)^{2}}= \\ sqrt{5}$, \\ n \\ [ | \\ vec{b} | = \\ sqrt{(-2+ \\ sqrt{3})^{2}+(1+2 \\ sqrt{3})^{2}}= \\ sqrt{20}=2 \\ sqrt{5} $\\ n \\ therefore$ \\ cos \\ theta= \\ frac{\\ vec{a} \\ cdot \\ vec{b}}{| \\ vec{a}| | | \\ vec{b}|}= \\ frac{-5}{ \\ sqrt{5} \\ times 2 \\ sqrt{5}}=- \\ frac{1}{2} $\\ n$ 0 ^ { \\ circ} \\ leqq \\ theta \\ leqq 180 ^ { \\ circ} $so$ \\ theta=120 ^ { \\ circ} $'

'Let A(r1,θ1) and B(r2,θ2) [r1 > 0, r2 > 0]. Using the law of cosines, find the distance AB between point A and point B.'

'How do you express that vectors a and b are equal when they have the same magnitude and direction?'

'In general, the vectors in space \ \\overrightarrow{u_{1}}, \\overrightarrow{u_{2}}, \\overrightarrow{u_{3}} \ satisfy the following conditions: \\( \\overrightarrow{u_{i}} \\cdot \\overrightarrow{u_{j}}=\\left\\{\egin{array}{ll}1 & (i=j) \\\\ 0 & (i \\neq j) \\end{array}\\right. \\)'

'Line segment AB and point P. When the following equation holds, what is the position of point P.'

'In the coordinate space with point O as the origin, what kind of figure does the set of points P(x, y, z) satisfying the following conditions represent? Also, express the equations in x, y, z:\n(1) When A(3,-6,2), point P satisfies |→OP|^{2}+2→OP⋅→OA+45=0.\n(2) When A(1,0,0), B(0,2,0), C(0,0,3), point P satisfies →AP⋅(→BP+2→CP)=0.'

'Question 31 | Vector Equation of a Circle\nFor the triangle OAB on the plane and any point P, the following vector equations represent a circle. What kind of circle is it?\n(1) |3 →PA+2 →PB|=5\n(2) →OP⋅(→OP-→AB)=→OA⋅→OB'

'From (1), \ 2 \\overrightarrow{\\mathrm{AB}}+\\overrightarrow{\\mathrm{AC}}=3 \\overrightarrow{\\mathrm{AD}} \ so \|3 \\overrightarrow{\\mathrm{AD}}-\\overrightarrow{\\mathrm{AP}}|=a\'

'For the points O(0,0,0), A(2,1,-2), B(3,4,0), find a vector perpendicular to both vector OA and vector OB with a magnitude of √5.'

'Show the following.'

'(1) What kind of shape does the vector equation |3→OA+2→OB-5→OP|=5 represent for two distinct points A, B, and any point P on the plane? (2) There are points P and triangle ABC on the plane. Find the set of points P that satisfy the condition 2→PA⋅→PB=3→PA⋅→PC.'

'Prove that on a plane, for four distinct points A, B, C, D, and a point O not on the line AB, where OA=a, OB=b. And if OC=3a-2b, OD=-3a+4b, then AB∥CD.'

'Given quadrilateral ABCD and point O, with OA = a, OB = b, OC = c, OD = d. If a + c = b + d and a · c = b · d, determine the shape of this quadrilateral.'

'Find the maximum value of the dot product $\\overrightarrow{OA} \\cdot \\overrightarrow{OB}$ when point A moves on the ellipse $\\frac{x^{2}}{3}+y^{2}=1$. Here, A(x, y) and B(x, y^{2}-2 y, 2 x+y^{3}), with O being the origin.'

'Prove the equation \ \\left|\\frac{1}{2} \\vec{a}-\\frac{1}{3} \\vec{b}\\right|^{2}+\\left|\\frac{1}{2} \\vec{a}+\\frac{1}{3} \\vec{b}\\right|^{2}=\\frac{1}{2}|\\vec{a}|^{2}+\\frac{2}{9}|\\vec{b}|^{2} \'

'Utilization of orthogonal projection vectors'

'Practice(2) Find the angle \ \\theta \ formed by two non-zero vectors \ \\vec{a} \ and \ \\vec{b} \ when there exists a unique real number \ t \ such that \ \\vec{a}+t \\vec{b} \ and \ \\vec{a}+3 t \\vec{b} \ are perpendicular.'

'Given vectors OA and OB. Find the area of triangle QBC if point Q satisfies the condition 256 vector AQ + 3 vector BQ + 2 vector CQ = vector 0.'

'Find a vector \\\vec{p}\ that is perpendicular to both vectors \\(\\vec{a}=(2,1,-2)\\) and \\(\\vec{b}=(3,4,0)\\) and has a magnitude of \\\sqrt{5}\.'

'Orthogonality and Dot Product of Vectors'

'Prove that when \\( (2 \\vec{a}+3 \\vec{b}) / /(\\vec{a}-4 \\vec{b}) \\), then \ \\vec{a} / / \\vec{b} \.'

'The vector equation of a straight line passing through point A(𝑎) and parallel to 𝑑(≠0) is 𝑝=𝑎+𝑡𝑑. Basic knowledge 1, p.343.'

'In triangle OAB, let vec{a} = \\overrightarrow{OA} and vec{b} = \\overrightarrow{OB}, with |\\vec{a}|=3, |\\vec{b}|=5, and \\cos \\angle AOB = \\frac{3}{5}. Find the position vector starting from O where the angle bisector of \\angle AOB intersects the circle with center at B and radius \\sqrt{10}, using vec{a} and vec{b}.'

'Given line segment AB and point P. When the following equation holds, where is point P located? (2) AP-3BP+4BA=0'

'Prove that when A and B are vectors with the origin as the starting point, the vector equation of the bisector of the angle formed by vectors OA=a and OB=b is given by p = t(a/|a| + b/|b|), where t is a variable.'

'Parallel Vectors and Dot Product'

'Solve Example 20 (2) on page 54 using the given information'

'Find the value of t when the angle between two vectors \\( \\vec{a} = (1, t) \\) and \\( \\vec{b} = \\left(1, \\frac{t}{3}\\right) \\) is \ 30^{\\circ} \. Assume t > 0.'

'(1) The condition for $\\vec{a} \\perp \\vec{b}$ is $\\vec{a} \\cdot \\vec{b} = 0$. Here, $\\vec{a} \\cdot \\vec{b} = 3 \\times x + 2 \\times 6 = 3x + 12$. Thus, $3x + 12 = 0$. Therefore, $x = -4$. (2) The condition for $\\vec{a} \\perp \\vec{b}$ is $\\vec{a} \\cdot \\vec{b} = 0$. Here, $\\vec{a} \\cdot \\vec{b} = 3 \\times(-1) + x \\times \\sqrt{3} = \\sqrt{3}x - 3$. Thus, $\\sqrt{3}x - 3 = 0$. Therefore, $x = \\sqrt{3}$.'

'Practice Given line segment AB and point P. When the following equation holds, where is point P located? (1) 3 vector AP + 4 vector BP = 2 vector AB'

'When two vectors a, b satisfy (1) |a + b| = 4 and (2) |a - b| = 3, find the value of a·b.'

'On the plane, from (1), it is given that angle ACB = angle CAD and angle BFC = angle DFA. This implies the form of vectors BC // AD.'

'Practice showing the following in the case where \ \\vec{a}, \\vec{b} \ are non-zero space vectors, \ s, t \ are non-negative real numbers, and \ \\vec{c}=s \\vec{a}+t \\vec{b} \.'

'Dot product of vectors: \\( \\vec{a} = (a_1, a_2, a_3), \\vec{b} = (b_1, b_2, b_3) \\) is given by'

'The vector equation of the plane alpha passing through point A(vector a) and perpendicular to the non-zero vector n is n·(p-vector a)=0 (as discussed in section 1, page 387).'

'Prove the following inequalities.'

'Given a quadrilateral ABCD and a point O, where OA vector is a, OB vector is b, OC vector is c, and OD vector is d. If a + c = b + d and a · c = b · d, determine the shape of this quadrilateral.'

'Given |a| = 3, |b| = 2, |a-2b| = sqrt{17}, find the value of the real number t for which a+b and a+tb are perpendicular.'

'Translate the given text into multiple languages.'

'Find the polar equation of the straight line passing through point \\( A(a, \\alpha) \\) and perpendicular to OA.'

'Dot product of vectors'

'Given the plane ABC determined by three points A(1,1,0), B(3,4,5), and C(1,3,6) in three-dimensional space, if there exists a point P(4,5,z) on the plane, find the value of z.'

In the right triangle $\mathrm{ABC}$ shown in the figure to the right, let \overrightarrow{\mathrm{AB}}=ec{a}, \overrightarrow{\mathrm{AC}}=ec{b}, \overrightarrow{\mathrm{BC}}=ec{c} . Find the dot products ec{a} \cdot ec{b}, ec{b} \cdot ec{c}, ec{c} \cdot ec{a} respectively. Given that |ec{a}|=|\overrightarrow{\mathrm{AB}}|=2,|ec{b}|=|\overrightarrow{\mathrm{AC}}|=2\sqrt{3},|ec{c}|=|\overrightarrow{\mathrm{BC}}|=4 , and the angle between ec{a} and ec{b} is $90^{\circ}$.

TRAINING 19 (3) Let |ec{a}|=1,|ec{b}|=2 . Answer the following questions. (1) When ec{a} \cdot ec{b}=-1 , find the value of |ec{a}-ec{b}| . (2) When |ec{a}+ec{b}|=1 , find the values of ec{a} \cdot ec{b} and |2 ec{a}-3 ec{b}| .

Find the dot product and the angle $heta$ between the following two vectors ec{a}, ec{b} . \[ ec{a} = (1,0,-1), ec{b} = (-1,2,2) \]

Prove that the following equations hold. (1) \( 3 ec{a} \cdot(3 ec{a}-2 ec{b})=9|ec{a}|^{2}-6 ec{a} \cdot ec{b} \) (2) |4 ec{a}-ec{b}|^{2}=16|ec{a}|^{2}-8 ec{a} \cdot ec{b}+|ec{b}|^{2}

Find the dot products $\overrightarrow{\mathrm{OA}} \cdot \overrightarrow{\mathrm{OD}}, \overrightarrow{\mathrm{OB}} \cdot \overrightarrow{\mathrm{OD}}$.

The dot product of vectors $\vec{a}$ and $\vec{b}$ and the angle $\theta$ between them: \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \] \[ \cos \theta =\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} =\frac{a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}}{\sqrt{a_{1}{ }^{2}+a_{2}{ }^{2}+a_{3}{ }^{2}} \sqrt{b_{1}{ }^{2}+b_{2}{ }^{2}+b_{3}{ }^{2}}

Properties of the Dot Product Calculate the dot product of the following vectors and verify the properties of the dot product. ec{a}=\left(2, 3 ight), ec{b}=\left(4, -1 ight) Dot product is 0 Properties of the Dot Product The following properties 1 to 5 hold for the dot product of vectors. 1 ec{a} \cdot ec{a}=|ec{a}|^{2} 2 ec{a} \cdot ec{b}=ec{b} \cdot ec{a} 3 (ec{a}+ec{b}) \cdot ec{c}=ec{a} \cdot ec{c}+ec{b} \cdot ec{c} 4 ec{a} \cdot(ec{b}+ec{c})=ec{a} \cdot ec{b}+ec{a} \cdot ec{c} 5 (k ec{a}) \cdot ec{b}=ec{a} \cdot(k ec{b})=k(ec{a} \cdot ec{b}) where k is a real number. Proof Let ec{a}=\left(a_{1}, a_{2} ight), ec{b}=\left(b_{1}, b_{2} ight), ec{c}=\left(c_{1}, c_{2} ight)。

(1) From $|2 \vec{a}-3 \vec{b}|=10$ we have $\quad|2 \vec{a}-3 \vec{b}|^{2}=100$ Therefore \( \quad(2 \vec{a}-3 \vec{b}) \cdot(2 \vec{a}-3 \vec{b})=100 \) Hence $\quad 4|\vec{a}|^{2}-12 \vec{a} \cdot \vec{b}+9|\vec{b}|^{2}=100$ Given $|\vec{a}|=1,|\vec{b}|=2 \sqrt{2}$, we have \( \quad 4 \times 1^{2}-12 \vec{a} \cdot \vec{b}+9(2 \sqrt{2})^{2}=100 \) That is $4-12 \vec{a} \cdot \vec{b}+72=100$, hence $\vec{a} \cdot \vec{b}=-2$ ! Therefore $\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{-2}{1 \times 2 \sqrt{2}}=-\frac{1}{\sqrt{2}}$ Since $0^{\circ} \leqq \theta \leqq 180^{\circ}$, $\quad \theta=135^{\circ}$

Let $k$ be a real constant. There is a point $\mathrm{P}$ and a triangle $\mathrm{ABC}$ on a certain plane, and the following equation is satisfied. $3 \overrightarrow{\mathrm{PA}}+4 \overrightarrow{\mathrm{PB}}+5 \overrightarrow{\mathrm{PC}}=k \overrightarrow{\mathrm{BC}}$ (1) When point $\mathrm{P}$ is on the line $\mathrm{AB}$, $k=\square$. (2) When point $\mathrm{P}$ is inside triangle $\mathrm{ABC}$, $<k<\square$ holds. However, assume point $\mathrm{P}$ is not on the perimeter of triangle $\mathrm{ABC}$.

Find the angle $heta$ formed by the dot product of the vectors ec{a} and ec{b} .\[ ec{a} = (1,0,1), ec{b} = (2,2,1) \]

Find the value of $x$ when the angle between the two vectors \( \vec{a}=(1,2,-1), \vec{b}=(-1, x, 0) \) is $45^{\circ}$.

Dot product of vectors Angle formed by the dot product of vectors (space)

Find the dot product $\overrightarrow{\mathrm{OA}} \cdot \overrightarrow{\mathrm{OB}}$ of vectors $\overrightarrow{\mathrm{OA}}$ and $\overrightarrow{\mathrm{OB}}$. Take three points $\mathrm{O}, \mathrm{A}, \mathrm{B}$ and let the angle between $\overrightarrow{\mathrm{OA}}$ and $\overrightarrow{\mathrm{OB}}$ be $heta$.

Find the values of $s$ and $t$ when the two vectors \(ec{a}=(s, 3s-1, s-1)\) and \(ec{b}=(t-1, 4, t-3)\) are parallel.

Given vectors \( \vec{a}=\left(a_{1}, a_{2}, a_{3}\right), \vec{b}=\left(b_{1}, b_{2}, b_{3}\right) \) where $a_{1} \neq 0, b_{1} \neq 0$, prove the following: $\vec{a} / / \vec{b} \Longleftrightarrow a_{1} b_{2}-a_{2} b_{1}=a_{1} b_{3}-a_{3} b_{1}=0$

Find the value of $x$ when the angle between the two vectors \( ec{a}=(2,1,1) \) and \( ec{b}=(x, 1,-2) \) is $60^{\circ}$.

The relationship between dot product and work

Prove that vectors are perpendicular using the dot product.

The condition for 13 points to be on a straight line [Collinearity Condition] [=Example 25]. When points A and B are distinct, point C is on line AB ⇔ there exists a real number k such that $\overrightarrow{\mathrm{AC}} = k \overrightarrow{\mathrm{AB}}$. When point C is on the line AB passing through distinct points A and B, $\overrightarrow{\mathrm{AB}} / / \overrightarrow{\mathrm{AC}}$ or $\overrightarrow{\mathrm{AC}} = \overrightarrow{0}$.

In a cube $\mathrm{ABCD}-\mathrm{EFGH}$ with side length 1, find the following dot products. (1) $\overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{HG}}$ (2) $\overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{AF}}$ (3) $\overrightarrow{\mathrm{AF}} \cdot \overrightarrow{\mathrm{AG}}$

Dot product of vectors Shapes and dot product of vectors (space) (1)

TRAINING Practice 1 (4) Let $k$ be a real constant. There is a point $\mathrm{P}$ and triangle $\mathrm{ABC}$ on a certain plane, and they satisfy the following equation. 3 \overrightarrow{\mathrm{PA}}+4 \overrightarrow{\mathrm{PB}}+5 \overrightarrow{\mathrm{PC}}=k \overrightarrow{\mathrm{BC}} (1) When point $\mathrm{P}$ is on the line $\mathrm{AB}$, $k=\square$. (2) When point $\mathrm{P}$ is inside triangle $\mathrm{ABC}$, $<k<\square$. Assume that point $\mathrm{P}$ is not on the edge of triangle $\mathrm{ABC}$.

Determine the value of $x$ that makes the two vectors $\vec{a}, \vec{b}$ parallel. (1) \( \vec{a}=(x,-2), \vec{b}=(2,1) \) (2) \( \vec{a}=(-9, x), \vec{b}=(x,-1) \)

Find the area S of triangle OAB in the following cases. (1) When |\overrightarrow{\mathrm{OA}}|=\sqrt{2},|\overrightarrow{\mathrm{OB}}|=\sqrt{3}, \overrightarrow{\mathrm{OA}} \cdot \overrightarrow{\mathrm{OB}}=2

Calculate the components of the dot product of vectors (space)

(2) Since \( (\vec{a}-3 \vec{b}) \perp(2 \vec{a}+\vec{b}) \), we have \( \quad(\vec{a}-3 \vec{b}) \cdot(2 \vec{a}+\vec{b})=0 \) Therefore \( \quad \vec{a} \cdot(2 \vec{a}+\vec{b})-3 \vec{b} \cdot(2 \vec{a}+\vec{b})=0 \) Thus, $\quad 2|\vec{a}|^{2}-5 \vec{a} \cdot \vec{b}-3|\vec{b}|^{2}=0$ Given $|\vec{a}|=2,|\vec{b}|=1$, therefore $\quad 2 \cdot 2^{2}-5 \vec{a} \cdot \vec{b}-3 \cdot 1^{2}=0$ (1) Hence $\vec{a} \cdot \vec{b}=1$, therefore $\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{1}{2 \times 1}=\frac{1}{2}$ $\Leftrightarrow|\vec{p}|$ is treated as $|\vec{p}|^{2}$. Since $0^{\circ} \leqq \theta \leqq 180^{\circ}$, therefore $\quad \theta=60^{\circ}$

Find the following dot products. (1) \overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{ED}}, (2) \overrightarrow{\mathrm{AF}} \cdot \overrightarrow{\mathrm{BG}}, (3) \overrightarrow{\mathrm{BH}} \cdot \overrightarrow{\mathrm{DF}}

Therefore, let $heta$ be the angle between $\overrightarrow{\mathrm{OC}}$ and $\overrightarrow{\mathrm{MN}}$, then \[ \cos \theta=\frac{\overrightarrow{\mathrm{OC}} \cdot \overrightarrow{\mathrm{MN}}}{|\overrightarrow{\mathrm{OC}}||\overrightarrow{\mathrm{MN}}|}=\frac{1}{2} \div\left(1 \times \frac{1}{\sqrt{2}}\right)=\frac{\sqrt{2}}{2}\] Since $0^{\circ} \leqq \theta \leqq 180^{\circ}$, $\quad \theta=45^{\circ}$ 〔 Let $\theta$ be the angle between non-zero vectors $\vec{p}$ and $\vec{q}$, then $\cos \theta=\frac{\vec{p} \cdot \vec{q}}{|\vec{p}||\vec{q}|}$.

Please calculate the dot product of the following vectors $\vec{a}$ and $\vec{b}$:\n\n $\vec{a} = \overrightarrow{OA}, \vec{b} = \overrightarrow{OB}$, with the angle $\theta = 60^{\circ}$ between the vectors, and | $\vec{a}$ | = 5, | $\vec{b}$ | = 3

(1) Find the value of $x$ such that \( \vec{a}=(5,1) \) and \( \vec{b}=(2, x) \) are perpendicular. (2) Find the unit vector $\vec{e}$ that is perpendicular to \( \vec{c}=(\sqrt{3}, 1) \).

Given vectors \( ec{a}=\left(a_{1}, a_{2}, a_{3} ight) \) and \( ec{b}=\left(b_{1}, b_{2}, b_{3} ight) \) where $a_{1} eq 0, b_{1} eq 0$, prove that the following holds: ec{a} / / ec{b} \Longleftrightarrow a_{1} b_{2} - a_{2} b_{1} = a_{1} b_{3} - a_{3} b_{1} = 0

Please calculate the dot product of the following two vectors: Vector \(\vec{a} = (3, 4)\) and Vector \(\vec{b} = (1, 2)\)

For the vectors shown in the figure on the right, list all pairs of vector numbers as follows. (1) Vectors with equal magnitude (2) Vectors with the same direction (3) Equal vectors (4) Opposite vectors

In the triangle $riangle \mathrm{ABC}$ with vertices at points \( \mathrm{A}(4, 3, -3), \mathrm{B}(3, 1, 0), \mathrm{C}(5, -2, 1) \), find the inner product $\overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{BC}}$ and the measure of angle $\mathrm{ABC}$ denoted by $heta$.

Angle between vectors and perpendicularity condition Find the angle between the vectors ec{a}=\left(1, 0 ight), ec{b}=\left(0, 1 ight) and prove that these vectors are perpendicular. Let the angle between two non-zero vectors ec{a}=\left(a_{1}, a_{2} ight), ec{b}=\left(b_{1}, b_{2} ight) be $heta$. Then, \cos heta=rac{ec{a} \cdot ec{b}}{|ec{a}||ec{b}|}=rac{a_{1} b_{1}+a_{2} b_{2}}{\sqrt{a_{1}^{2}+a_{2}^{2}} \sqrt{b_{1}^{2}+b_{2}^{2}}} where $0^{\circ} \leqq heta \leqq 180^{\circ}$