answer：Okay, so I have this problem where I'm given two vectors, **a** and **b**, and I need to find the projection **p** of both vectors onto some vector **v**. The catch is that **p**, **a**, and **b** all have to be collinear. Hmm, collinear means they lie on the same straight line, right? So, **p** must be a point on the line that passes through both **a** and **b**.First, let me write down the vectors:**a** = [2, -2, 3]**b** = [1, 4, 1]I remember that the projection of a vector **a** onto another vector **v** is given by:projᵥ**a** = ( (**a** · **v**) / ||**v**||² ) **v**Similarly, the projection of **b** onto **v** would be:projᵥ**b** = ( (**b** · **v**) / ||**v**||² ) **v**Since both projections result in the same vector **p**, that means:projᵥ**a** = projᵥ**b** = **p**So, setting them equal:( (**a** · **v**) / ||**v**||² ) **v** = ( (**b** · **v**) / ||**v**||² ) **v**Since **v** is not the zero vector, we can divide both sides by **v** (or more accurately, since they are scalar multiples, their scalar coefficients must be equal):(**a** · **v**) / ||**v**||² = (**b** · **v**) / ||**v**||²Which simplifies to:**a** · **v** = **b** · **v**So, (**a** - **b**) · **v** = 0That means the vector (**a** - **b**) is orthogonal to **v**. Let me compute (**a** - **b**):**a** - **b** = [2 - 1, -2 - 4, 3 - 1] = [1, -6, 2]So, [1, -6, 2] · **v** = 0Therefore, **v** must be orthogonal to [1, -6, 2]. Hmm, so **v** lies in the plane perpendicular to [1, -6, 2]. But I'm not sure if that's directly useful yet.Wait, but **p** is the projection of both **a** and **b** onto **v**, so **p** must lie on the line through **a** and **b**. So, **p** is a point on the line connecting **a** and **b**, and it's also the projection of both vectors onto **v**.So, maybe I can parametrize the line through **a** and **b** and find the point **p** on that line such that **p** is the projection of both **a** and **b** onto some vector **v**.Let me parametrize the line. Let's let **p** = **a** + t(**b** - **a**), where t is a scalar parameter.Compute **b** - **a**:**b** - **a** = [1 - 2, 4 - (-2), 1 - 3] = [-1, 6, -2]So, **p** = [2, -2, 3] + t[-1, 6, -2] = [2 - t, -2 + 6t, 3 - 2t]So, **p** is [2 - t, -2 + 6t, 3 - 2t]Now, since **p** is the projection of both **a** and **b** onto **v**, that means **p** is the closest point on the line spanned by **v** to both **a** and **b**. But wait, since **p** is on the line through **a** and **b**, and it's the projection of both **a** and **b** onto **v**, that implies that **v** must be orthogonal to the direction vector of the line through **a** and **b**, which is [-1, 6, -2].Wait, no, actually, the projection of **a** onto **v** is **p**, so **a** - **p** must be orthogonal to **v**. Similarly, **b** - **p** must be orthogonal to **v**. Therefore, both (**a** - **p**) and (**b** - **p**) are orthogonal to **v**.But since (**a** - **p**) and (**b** - **p**) are both orthogonal to **v**, their difference should also be orthogonal to **v**. Let me compute (**a** - **p**) - (**b** - **p**) = **a** - **b**. So, **a** - **b** is orthogonal to **v**, which we already knew.But maybe another approach is to realize that since **p** is the projection of both **a** and **b** onto **v**, the vector **p** must lie on the line through **a** and **b**, and also on the line spanned by **v**. So, the intersection of these two lines is **p**.But I'm not sure if that's the right way to think about it.Alternatively, since **p** is the projection of both **a** and **b** onto **v**, the vector **p** must satisfy:**p** = projᵥ**a** = projᵥ**b**Which implies that **p** is the same for both, so the projection of **a** and **b** onto **v** is the same point **p**. That suggests that **v** is such that both **a** and **b** project to the same point on **v**. So, **v** must be aligned in such a way that both **a** and **b** have the same component along **v**.But since **p** is on the line through **a** and **b**, maybe we can find **p** by finding the point on the line where the projection of **a** and **b** onto the direction of the line is the same.Wait, that might be a bit circular.Alternatively, maybe I can set up the equations for the projection.Let me denote **v** as [v₁, v₂, v₃]. Then, the projection of **a** onto **v** is:(**a** · **v**) / ||**v**||² * **v**Similarly for **b**. Since both projections are equal to **p**, we have:(**a** · **v**) / ||**v**||² * **v** = (**b** · **v**) / ||**v**||² * **v**Which simplifies to (**a** · **v**) = (**b** · **v**), as before.So, (**a** - **b**) · **v** = 0Which gives us the equation:[1, -6, 2] · [v₁, v₂, v₃] = 0So, v₁ - 6v₂ + 2v₃ = 0So, **v** lies in the plane defined by this equation.But we also know that **p** is on the line through **a** and **b**, so **p** = [2 - t, -2 + 6t, 3 - 2t]And since **p** is the projection of **a** onto **v**, we have:**p** = projᵥ**a** = ( (**a** · **v**) / ||**v**||² ) **v**Similarly, **p** = ( (**a** · **v**) / ||**v**||² ) **v**But **p** is also equal to [2 - t, -2 + 6t, 3 - 2t]So, we can write:( (**a** · **v**) / ||**v**||² ) **v** = [2 - t, -2 + 6t, 3 - 2t]Similarly, since **p** is also the projection of **b** onto **v**, we have:( (**b** · **v**) / ||**v**||² ) **v** = [2 - t, -2 + 6t, 3 - 2t]But since (**a** · **v**) = (**b** · **v**) = let's say k, then:k / ||**v**||² * **v** = **p**So, **p** is a scalar multiple of **v**. Therefore, **p** and **v** are colinear, meaning **p** is in the direction of **v**.But **p** is also on the line through **a** and **b**, which has direction vector [-1, 6, -2]. So, **p** is a point where the line through **a** and **b** intersects the line spanned by **v**.Wait, but since **p** is the projection, it's the closest point on **v** to **a** and **b**. So, the vector from **a** to **p** is orthogonal to **v**, and similarly for **b** to **p**.So, (**a** - **p**) · **v** = 0And (**b** - **p**) · **v** = 0But since (**a** - **p**) and (**b** - **p**) are both orthogonal to **v**, their difference, which is (**a** - **b**), is also orthogonal to **v**, which we already established.So, maybe I can use the fact that (**a** - **p**) is orthogonal to **v**, and since **p** is on the line through **a** and **b**, I can express **p** as **a** + t(**b** - **a**), which is what I did earlier.So, **p** = [2 - t, -2 + 6t, 3 - 2t]Now, since (**a** - **p**) is orthogonal to **v**, and (**a** - **p**) = t(**b** - **a**) = t[-1, 6, -2]So, t[-1, 6, -2] · **v** = 0But we already know that [1, -6, 2] · **v** = 0, which is the same as [-1, 6, -2] · **v** = 0, just scaled by -1.So, t[-1, 6, -2] · **v** = 0But since [-1, 6, -2] · **v** = 0, this equation is automatically satisfied for any t. So, that doesn't give us new information.Hmm, maybe I need another approach.Since **p** is the projection of both **a** and **b** onto **v**, and **p** lies on the line through **a** and **b**, perhaps **p** is the point where the line through **a** and **b** intersects the line spanned by **v**.But since **v** is a vector, the line spanned by **v** is just the line through the origin in the direction of **v**. So, **p** is the intersection point of the line through **a** and **b** and the line through the origin in the direction of **v**.But **p** is also the projection of **a** and **b** onto **v**, so it's the closest point on **v**'s line to both **a** and **b**.Wait, but if **p** is the projection of both **a** and **b** onto **v**, then **p** must be the same for both, meaning that **a** and **b** are equidistant from **p** along the direction orthogonal to **v**.But since **p** is on the line through **a** and **b**, maybe **p** is the midpoint or some specific point on that line.Alternatively, perhaps I can set up the equations for the projection.Let me denote **p** = [2 - t, -2 + 6t, 3 - 2t]Since **p** is the projection of **a** onto **v**, we have:**p** = ( (**a** · **v**) / ||**v**||² ) **v**Similarly, **p** = ( (**b** · **v**) / ||**v**||² ) **v**But since (**a** · **v**) = (**b** · **v**) = k, as before, then:**p** = (k / ||**v**||² ) **v**So, **p** is a scalar multiple of **v**. Therefore, **p** and **v** are colinear.But **p** is also on the line through **a** and **b**, which has direction vector [-1, 6, -2]. So, **p** is the intersection of the line through **a** and **b** and the line through the origin in the direction of **v**.But since **p** is the projection, it's the closest point on **v**'s line to **a** and **b**.Wait, maybe I can use the fact that **p** is the projection, so the vector from **a** to **p** is orthogonal to **v**.So, (**a** - **p**) · **v** = 0Similarly, (**b** - **p**) · **v** = 0But since **p** is on the line through **a** and **b**, **p** = **a** + t(**b** - **a**), so:(**a** - **p**) = -t(**b** - **a**) = -t[-1, 6, -2] = [t, -6t, 2t]So, [t, -6t, 2t] · **v** = 0Which is the same as t[1, -6, 2] · **v** = 0But we already know that [1, -6, 2] · **v** = 0, so this equation is satisfied for any t.Hmm, so this doesn't help us find t.Maybe I need to use the fact that **p** is a scalar multiple of **v**. So, **p** = s**v** for some scalar s.But **p** is also equal to [2 - t, -2 + 6t, 3 - 2t]So, [2 - t, -2 + 6t, 3 - 2t] = s[v₁, v₂, v₃]But we also have [1, -6, 2] · [v₁, v₂, v₃] = 0So, v₁ - 6v₂ + 2v₃ = 0So, we have:2 - t = s v₁-2 + 6t = s v₂3 - 2t = s v₃And v₁ - 6v₂ + 2v₃ = 0So, we have four equations:1) 2 - t = s v₁2) -2 + 6t = s v₂3) 3 - 2t = s v₃4) v₁ - 6v₂ + 2v₃ = 0We can try to express v₁, v₂, v₃ in terms of s and t.From equation 1: v₁ = (2 - t)/sFrom equation 2: v₂ = (-2 + 6t)/sFrom equation 3: v₃ = (3 - 2t)/sNow, substitute these into equation 4:(2 - t)/s - 6*(-2 + 6t)/s + 2*(3 - 2t)/s = 0Multiply through by s to eliminate denominators:(2 - t) - 6*(-2 + 6t) + 2*(3 - 2t) = 0Now, expand each term:2 - t + 12 - 36t + 6 - 4t = 0Combine like terms:(2 + 12 + 6) + (-t - 36t - 4t) = 020 - 41t = 0So, 20 = 41tt = 20/41Wait, but in the original problem, the user's solution had t = 16/39. Did I make a mistake?Let me check my calculations.From equation 4:v₁ - 6v₂ + 2v₃ = 0Substituting:(2 - t)/s - 6*(-2 + 6t)/s + 2*(3 - 2t)/s = 0Multiply by s:(2 - t) - 6*(-2 + 6t) + 2*(3 - 2t) = 0Compute each term:2 - t + 12 - 36t + 6 - 4tCombine constants: 2 + 12 + 6 = 20Combine t terms: -t -36t -4t = -41tSo, 20 - 41t = 0 => t = 20/41Hmm, but the user's solution had t = 16/39. Maybe I made a mistake in setting up the equations.Wait, let me go back.I set **p** = s**v**, which is correct because **p** is a scalar multiple of **v**.Then, **p** = [2 - t, -2 + 6t, 3 - 2t] = s[v₁, v₂, v₃]So, v₁ = (2 - t)/s, v₂ = (-2 + 6t)/s, v₃ = (3 - 2t)/sThen, substituting into v₁ - 6v₂ + 2v₃ = 0:(2 - t)/s - 6*(-2 + 6t)/s + 2*(3 - 2t)/s = 0Multiply by s:(2 - t) -6*(-2 + 6t) + 2*(3 - 2t) = 0Compute:2 - t + 12 - 36t + 6 - 4t = 02 + 12 + 6 = 20-t -36t -4t = -41tSo, 20 -41t =0 => t=20/41But the user's solution had t=16/39. So, perhaps I made a mistake in the setup.Wait, maybe I should have used the fact that **p** is the projection, so **p** = projᵥ**a** = ( (**a** · **v**) / ||**v**||² ) **v**But **p** is also equal to [2 - t, -2 + 6t, 3 - 2t]So, let me write:(**a** · **v**) / ||**v**||² * **v** = [2 - t, -2 + 6t, 3 - 2t]Let me denote k = (**a** · **v**) / ||**v**||², so **p** = k**v**So, k**v** = [2 - t, -2 + 6t, 3 - 2t]But **v** is a vector, so we can write:v₁ = (2 - t)/kv₂ = (-2 + 6t)/kv₃ = (3 - 2t)/kNow, we also have the condition that (**a** - **b**) · **v** = 0, which is [1, -6, 2] · **v** = 0So, substituting **v** from above:1*(2 - t)/k + (-6)*(-2 + 6t)/k + 2*(3 - 2t)/k = 0Multiply through by k:(2 - t) + 6*(2 - 6t) + 2*(3 - 2t) = 0Wait, let me compute each term:1*(2 - t) = 2 - t-6*(-2 + 6t) = 12 - 36t2*(3 - 2t) = 6 - 4tSo, adding them up:(2 - t) + (12 - 36t) + (6 - 4t) = 0Combine constants: 2 + 12 + 6 = 20Combine t terms: -t -36t -4t = -41tSo, 20 -41t = 0 => t=20/41Same result as before. So, t=20/41But the user's solution had t=16/39. So, perhaps the user made a mistake?Wait, let me check the user's solution.The user wrote:The line through **a** and **b** can be represented by:**p** = [2, -2, 3] + t([-1, 6, -2]) = [2 - t, -2 + 6t, 3 - 2t]For collinearity, **p** must be orthogonal to the direction vector [-1, 6, -2]:[2 - t, -2 + 6t, 3 - 2t] · [-1, 6, -2] = 0Expanding:(-2 + t) -6*(-2 + 6t) + 2*(3 - 2t) = 0Wait, let me compute that:(2 - t)(-1) + (-2 + 6t)(6) + (3 - 2t)(-2)= (-2 + t) + (-12 + 36t) + (-6 + 4t)= (-2 -12 -6) + (t +36t +4t)= (-20) + (41t) = 0So, 41t -20 =0 => t=20/41Wait, but the user wrote:(-2 + t) -6*(-2 + 6t) + 2*(3 - 2t) = 0Which is:(-2 + t) +12 -36t +6 -4t =0Which is:(-2 +12 +6) + (t -36t -4t) = 16 -39t=0 => t=16/39Wait, that's different. So, the user expanded the dot product incorrectly.Because the dot product is:(2 - t)(-1) + (-2 +6t)(6) + (3 -2t)(-2)= (-2 + t) + (-12 +36t) + (-6 +4t)= (-2 -12 -6) + (t +36t +4t)= (-20) +41t=0 => t=20/41But the user did:(-2 + t) -6*(-2 +6t) +2*(3 -2t)= (-2 +t) +12 -36t +6 -4t= (-2 +12 +6) + (t -36t -4t)=16 -39t=0 => t=16/39So, the user incorrectly expanded the dot product. Instead of multiplying each component correctly, they seem to have distributed the multiplication incorrectly.So, the correct value of t is 20/41, not 16/39.Therefore, the correct **p** is:[2 - t, -2 +6t, 3 -2t] with t=20/41Compute each component:2 -20/41 = (82 -20)/41 =62/41-2 +6*(20/41)= -2 +120/41= (-82 +120)/41=38/413 -2*(20/41)=3 -40/41= (123 -40)/41=83/41So, **p**= [62/41, 38/41,83/41]Wait, but let me double-check the calculations.Compute 2 - t:2 =82/41, so 82/41 -20/41=62/41-2 +6t:-2= -82/41, 6t=120/41, so -82/41 +120/41=38/413 -2t:3=123/41, 2t=40/41, so 123/41 -40/41=83/41Yes, so **p**= [62/41, 38/41,83/41]But the user's answer was [58/39,40/39,55/39], which is different.So, I think the user made a mistake in expanding the dot product, leading to an incorrect value of t.Therefore, the correct **p** is [62/41,38/41,83/41]But let me confirm by another method.Alternatively, since **p** is the projection of both **a** and **b** onto **v**, and **p** lies on the line through **a** and **b**, we can find **p** by finding the point on the line where the vector from **a** to **p** is orthogonal to the direction vector of the line.Wait, no, because **p** is the projection onto **v**, not onto the line through **a** and **b**.Wait, perhaps I can use the formula for the projection of a point onto a line.But in this case, **p** is the projection of **a** and **b** onto **v**, which is a vector, not a line.Wait, but **v** is a vector, so the projection of **a** onto **v** is a point on the line spanned by **v**.So, **p** is the intersection point of the line through **a** and **b** and the line spanned by **v**.But since **p** is the projection, it's the closest point on **v**'s line to **a** and **b**.So, perhaps we can find **p** by finding the point on the line through **a** and **b** that is closest to the origin, since **v**'s line passes through the origin.Wait, no, because **v** is arbitrary, not necessarily the direction of the line through **a** and **b**.Alternatively, perhaps I can set up the equations for the projection.Let me denote **p** = [x, y, z]Since **p** is the projection of **a** onto **v**, we have:**p** = ( (**a** · **v**) / ||**v**||² ) **v**Similarly, **p** = ( (**b** · **v**) / ||**v**||² ) **v**So, (**a** · **v**) = (**b** · **v**) = kThus, **p** = (k / ||**v**||² ) **v**But **p** is also on the line through **a** and **b**, so:**p** = **a** + t(**b** - **a**) = [2 - t, -2 +6t,3 -2t]So, [2 - t, -2 +6t,3 -2t] = (k / ||**v**||² ) **v**But **v** is a vector, so we can write:v₁ = (2 - t) * ||**v**||² /kv₂ = (-2 +6t) * ||**v**||² /kv₃ = (3 -2t) * ||**v**||² /kBut this seems complicated.Alternatively, since **p** is the projection, the vector from **a** to **p** is orthogonal to **v**.So, (**a** - **p**) · **v** =0Similarly, (**b** - **p**) · **v** =0But **p** is on the line through **a** and **b**, so **p** = **a** + t(**b** - **a**) = [2 - t, -2 +6t,3 -2t]So, (**a** - **p**) = t(**b** - **a**) = t[-1,6,-2]So, t[-1,6,-2] · **v** =0But we also have (**a** - **b**) · **v** =0, which is [-1,6,-2] · **v** =0So, t[-1,6,-2] · **v** =0 is automatically satisfied because [-1,6,-2] · **v**=0So, this doesn't give us new information.Alternatively, since **p** is the projection, the vector **p** is the closest point on **v**'s line to **a** and **b**.So, the distance from **a** to **p** is equal to the distance from **b** to **p**.Wait, no, because **p** is the projection, it's the closest point on **v**'s line to **a**, and also the closest to **b**. So, the distances from **a** and **b** to **p** are both minimal, but they don't have to be equal.But since **p** is on the line through **a** and **b**, maybe the distances from **a** and **b** to **p** are related.Alternatively, perhaps I can use the fact that **p** is the projection, so the vector **p** is the same for both **a** and **b**.So, (**a** · **v**) / ||**v**||² = (**b** · **v**) / ||**v**||²Which implies (**a** · **v**) = (**b** · **v**)So, (**a** - **b**) · **v** =0Which is the same as before.So, I think the correct approach is to parametrize **p** as [2 - t, -2 +6t,3 -2t], then set up the equation that (**a** - **p**) is orthogonal to **v**, but since (**a** - **p**) is t[-1,6,-2], and we know that [-1,6,-2] · **v**=0, this doesn't help.Alternatively, since **p** is a scalar multiple of **v**, we can write **p** = s**v**So, [2 - t, -2 +6t,3 -2t] = s[v₁, v₂, v₃]And we have v₁ -6v₂ +2v₃=0So, substituting:v₁ = (2 - t)/sv₂ = (-2 +6t)/sv₃ = (3 -2t)/sThen, (2 - t)/s -6*(-2 +6t)/s +2*(3 -2t)/s=0Multiply by s:(2 - t) -6*(-2 +6t) +2*(3 -2t)=0Compute:2 - t +12 -36t +6 -4t=02+12+6=20-t-36t-4t=-41tSo, 20-41t=0 => t=20/41Therefore, **p**= [2 -20/41, -2 +6*(20/41),3 -2*(20/41)] = [62/41,38/41,83/41]So, the correct **p** is [62/41,38/41,83/41]But the user's solution had t=16/39, leading to **p**= [58/39,40/39,55/39]So, I think the user made a mistake in expanding the dot product, leading to an incorrect t.Therefore, the correct answer should be **p**= [62/41,38/41,83/41]

answer：To determine the full capacity of the fuel tank, let's denote the total capacity as ( x ) gallons.Initially, the tank is ( frac{5}{6} ) full, which means it contains ( frac{5}{6}x ) gallons of fuel.After using 15 gallons, the tank is ( frac{2}{3} ) full, meaning it contains ( frac{2}{3}x ) gallons of fuel.The difference between the initial and final amounts of fuel is the 15 gallons used. Therefore, we can set up the equation:[frac{5}{6}x - frac{2}{3}x = 15]To simplify the equation, find a common denominator for the fractions:[frac{5}{6}x - frac{4}{6}x = 15]Subtracting the fractions gives:[frac{1}{6}x = 15]To solve for ( x ), multiply both sides of the equation by 6:[x = 15 times 6 = 90]Thus, the full capacity of the tank is 90 gallons.

answer：First, I recognize that 9 can be expressed as (3^2). This allows me to rewrite the original equation (9^x = 3^{12}) in terms of the same base.By substituting (9) with (3^2), the equation becomes ((3^2)^x = 3^{12}).Next, I apply the power of a power rule, which states that ((a^m)^n = a^{m cdot n}). Applying this rule to the left side of the equation, I get (3^{2x} = 3^{12}).Since the bases are the same, I can set the exponents equal to each other: (2x = 12).Finally, I solve for (x) by dividing both sides of the equation by 2, resulting in (x = 6).

answer：Okay, so I have this math problem here: I need to calculate the result of ( x^{8} cdot x^{2} ), and then choose the correct answer from the options given, which are A: ( x^{4} ), B: ( x^{6} ), C: ( x^{10} ), and D: ( x^{16} ). Alright, let's break this down step by step.First, I remember that when you multiply two expressions with the same base, like ( x ) in this case, you can add their exponents. So, the general rule is ( x^{a} cdot x^{b} = x^{a+b} ). That seems straightforward. So applying that rule here, ( x^{8} cdot x^{2} ) should be ( x^{8+2} ), which is ( x^{10} ). Hmm, that seems pretty clear. So, according to that, the answer should be C: ( x^{10} ).But wait, maybe I should double-check to make sure I'm not missing anything. Sometimes, in math problems, there can be exceptions or special cases, especially if the base isn't just a simple variable like ( x ). But in this case, the base is the same in both terms, so the rule should apply without any issues. Let me think about it another way to confirm.If I expand ( x^{8} ) and ( x^{2} ), what do I get? Well, ( x^{8} ) is ( x ) multiplied by itself 8 times: ( x cdot x cdot x cdot x cdot x cdot x cdot x cdot x ). Similarly, ( x^{2} ) is ( x cdot x ). So when I multiply these two together, I'm essentially multiplying ( x ) by itself 8 times and then 2 more times, which totals 10 times. So that gives me ( x^{10} ), which again points to option C.Is there any chance that I might have confused the rule? For example, sometimes people confuse adding exponents with multiplying them, especially when dealing with powers raised to powers. But in this case, it's just a straightforward multiplication of two terms with the same base, so adding the exponents is definitely the right approach.Let me also consider if there's any possibility that the problem is trying to trick me in some way. Maybe the exponents are negative or something? But no, both exponents here are positive, so that shouldn't be an issue. Also, the base is the same, so there's no need to worry about different bases or anything like that.Another thing to think about is if I've seen similar problems before and how they were solved. From what I recall, whenever you multiply like bases with exponents, you add the exponents. Dividing them would involve subtracting the exponents, but that's not the case here. So, yeah, adding 8 and 2 to get 10 seems correct.Just to be absolutely thorough, let's plug in a number for ( x ) and see if the result makes sense. Let's say ( x = 2 ). Then, ( x^{8} ) would be ( 2^{8} = 256 ), and ( x^{2} ) would be ( 2^{2} = 4 ). Multiplying these together gives ( 256 cdot 4 = 1024 ). Now, ( x^{10} ) when ( x = 2 ) is ( 2^{10} = 1024 ), which matches the result of the multiplication. So that checks out.Alternatively, if I had chosen a different base, say ( x = 3 ), ( x^{8} = 6561 ) and ( x^{2} = 9 ). Multiplying these gives ( 6561 cdot 9 = 59049 ). And ( x^{10} ) when ( x = 3 ) is ( 3^{10} = 59049 ), which again matches. So this further confirms that the rule works as expected.I guess another way to think about it is in terms of the properties of exponents. The rule ( x^{a} cdot x^{b} = x^{a+b} ) is one of the fundamental exponent rules. It's derived from the fact that when you multiply two groups of the same base, you're essentially increasing the number of times the base is multiplied by itself. So, if you have 8 instances of ( x ) and then 2 more, you end up with 10 instances, hence ( x^{10} ).I also remember that this rule is part of the laws of exponents, which include other rules like ( (x^{a})^{b} = x^{ab} ) and ( x^{a} / x^{b} = x^{a-b} ). It's important to keep these rules straight to avoid mistakes. In this problem, since we're dealing with multiplication, the addition of exponents is the correct approach.One thing that sometimes trips people up is when the exponents are variables themselves, but in this case, the exponents are constants, so it's simpler. If the exponents were variables, say ( x^{m} cdot x^{n} ), the result would still be ( x^{m+n} ), following the same rule.I also wonder if there's a way to visualize this. Maybe using area or volume models? For example, if ( x^{8} ) represents the area of a square with sides ( x^{4} ), and ( x^{2} ) represents another dimension, multiplying them could give a volume or a higher-dimensional space. But honestly, that might be overcomplicating things for this problem. The straightforward exponent rule is sufficient here.Another angle is to think about the problem in terms of logarithms, but that might not be necessary. Logarithms are useful for solving equations where the variable is in the exponent, but here we're just simplifying an expression, so basic exponent rules are enough.I could also consider if there's any connection to polynomial multiplication. When multiplying two polynomials, you distribute the terms, but in this case, since both terms are monomials with the same base, it's a simpler case where you just add the exponents. If the bases were different, you wouldn't be able to combine them like that, but here, since they're the same, it works out.Thinking about this in terms of orders of magnitude might help some people. If ( x ) is a large number, say ( x = 10 ), then ( x^{8} ) is ( 100,000,000 ) and ( x^{2} ) is ( 100 ). Multiplying them gives ( 10,000,000,000 ), which is ( 10^{10} ). Again, that aligns with the rule of adding exponents.I also recall that exponents are shorthand for repeated multiplication, so ( x^{8} ) is shorthand for multiplying ( x ) eight times, and ( x^{2} ) is multiplying ( x ) two times. When you multiply these together, you're effectively multiplying ( x ) a total of ten times, hence ( x^{10} ).It's also worth noting that this rule is consistent across different mathematical contexts, whether you're dealing with algebra, calculus, or other areas. Understanding and applying exponent rules correctly is foundational for more advanced topics.I suppose another way to verify is by using the definition of exponents. The exponent indicates how many times the base is used as a factor in multiplication. So, ( x^{8} ) has eight factors of ( x ), and ( x^{2} ) has two factors of ( x ). When you multiply these together, you're combining all these factors, resulting in ten factors of ( x ), which is ( x^{10} ).I could also think about this in terms of binary operations or computer science, where exponents are used in algorithms and data structures. For instance, in analyzing the time complexity of an algorithm, understanding how operations scale with exponents is crucial. Here, knowing that multiplying ( x^{8} ) by ( x^{2} ) results in ( x^{10} ) helps in understanding how the complexity increases.Additionally, considering the properties of logarithms, if I were to take the logarithm of both sides, I could see that ( log(x^{8} cdot x^{2}) = log(x^{10}) ), which simplifies to ( log(x^{8}) + log(x^{2}) = log(x^{10}) ). Using the logarithm power rule, this becomes ( 8log(x) + 2log(x) = 10log(x) ), which is consistent. So, even through logarithms, the result holds true.I might also think about how this applies in real-world scenarios, like calculating areas or volumes. For example, if you have a square with side length ( x^{4} ), its area is ( x^{8} ). If you then scale each side by ( x^{2} ), the new area would be ( (x^{4} cdot x^{2})^{2} = (x^{6})^{2} = x^{12} ). Wait, that's a different exponent, but it still follows the same rule of adding exponents when multiplying like bases.Wait, actually, in that example, I'm scaling the side length, which affects the area differently. The area scales with the square of the side length, so if the side length becomes ( x^{6} ), the area becomes ( x^{12} ). But that's a different problem because it involves scaling in two dimensions. In our original problem, we're just multiplying two single-dimensional terms, so it's simpler.Another thought: if I were to graph ( y = x^{8} ) and ( y = x^{2} ), their product would be ( y = x^{10} ). The shape of the graph would change accordingly, but the key point is that the exponents add when multiplying the functions. This is consistent with how functions behave under multiplication.I could also use a calculator to verify with actual numbers, but since I already did that with ( x = 2 ) and ( x = 3 ), and both gave consistent results, I feel confident that the rule applies correctly here.In summary, by applying the exponent rule for multiplying like bases, adding the exponents 8 and 2 gives us 10, resulting in ( x^{10} ). This is confirmed by expanding the exponents, plugging in numbers, and considering the properties of exponents in different mathematical contexts. Therefore, the correct answer should be option C: ( x^{10} ).

answer：Okay, so I have this problem where I need to find the distance between two parallel lines in 3D space. The lines are given in parametric form, which I remember is something like a point plus a scalar multiple of a direction vector. Let me write them down again to make sure I have them right.The first line is:[begin{pmatrix} 3 -4 1 end{pmatrix} + t begin{pmatrix} 2 -14 0 end{pmatrix}]And the second line is:[begin{pmatrix} 2 -7 4 end{pmatrix} + s begin{pmatrix} 2 -14 0 end{pmatrix}]Alright, so both lines have the same direction vector, which is (begin{pmatrix} 2 -14 0 end{pmatrix}). That makes sense because they're parallel. If they had different direction vectors, they might not be parallel or could even intersect.Now, I need to find the distance between these two lines. I remember that for parallel lines, the distance is the length of the vector that connects a point on one line to a point on the other line, but this vector has to be perpendicular to the direction vector of the lines. Hmm, okay, so I need to find such a vector and then compute its magnitude.First, let me pick a point on each line. For the first line, when (t = 0), the point is (begin{pmatrix} 3 -4 1 end{pmatrix}). For the second line, when (s = 0), the point is (begin{pmatrix} 2 -7 4 end{pmatrix}). So, the vector connecting these two points is:[begin{pmatrix} 3 -4 1 end{pmatrix} - begin{pmatrix} 2 -7 4 end{pmatrix} = begin{pmatrix} 1 3 -3 end{pmatrix}]Let me call this vector (mathbf{v}). So, (mathbf{v} = begin{pmatrix} 1 3 -3 end{pmatrix}).Now, I need to find the component of this vector that is perpendicular to the direction vector (mathbf{d} = begin{pmatrix} 2 -14 0 end{pmatrix}). I think this involves using the projection formula. The projection of (mathbf{v}) onto (mathbf{d}) is given by:[text{proj}_{mathbf{d}} mathbf{v} = left( frac{mathbf{v} cdot mathbf{d}}{mathbf{d} cdot mathbf{d}} right) mathbf{d}]Let me compute the dot product (mathbf{v} cdot mathbf{d}):[1 times 2 + 3 times (-14) + (-3) times 0 = 2 - 42 + 0 = -40]And the dot product (mathbf{d} cdot mathbf{d}):[2 times 2 + (-14) times (-14) + 0 times 0 = 4 + 196 + 0 = 200]So, the projection is:[text{proj}_{mathbf{d}} mathbf{v} = left( frac{-40}{200} right) begin{pmatrix} 2 -14 0 end{pmatrix} = left( -0.2 right) begin{pmatrix} 2 -14 0 end{pmatrix} = begin{pmatrix} -0.4 2.8 0 end{pmatrix}]Okay, so that's the projection. Now, to find the component of (mathbf{v}) that's perpendicular to (mathbf{d}), I subtract this projection from (mathbf{v}):[mathbf{v}_{perp} = mathbf{v} - text{proj}_{mathbf{d}} mathbf{v} = begin{pmatrix} 1 3 -3 end{pmatrix} - begin{pmatrix} -0.4 2.8 0 end{pmatrix} = begin{pmatrix} 1 + 0.4 3 - 2.8 -3 - 0 end{pmatrix} = begin{pmatrix} 1.4 0.2 -3 end{pmatrix}]Now, the distance between the two lines is the magnitude of this perpendicular vector. Let me compute that:[|mathbf{v}_{perp}| = sqrt{(1.4)^2 + (0.2)^2 + (-3)^2} = sqrt{1.96 + 0.04 + 9} = sqrt{11}]So, the distance is (sqrt{11}). Let me just double-check my calculations to make sure I didn't make any mistakes.First, the vector connecting the two points: 3-2=1, -4-(-7)=3, 1-4=-3. That seems right. Then, the dot product with the direction vector: 1*2 + 3*(-14) + (-3)*0 = 2 -42 + 0 = -40. Correct.The projection scalar: -40/200 = -0.2. Multiply by direction vector: 2*(-0.2) = -0.4, -14*(-0.2)=2.8, 0*(-0.2)=0. So, projection vector is (-0.4, 2.8, 0). Subtracting that from the original vector: 1 - (-0.4)=1.4, 3 - 2.8=0.2, -3 -0=-3. So, the perpendicular vector is (1.4, 0.2, -3). Its magnitude: sqrt(1.96 + 0.04 + 9) = sqrt(11). Yep, that all checks out.I think I did everything correctly. So, the distance between the two parallel lines is (sqrt{11}).**Final Answer**The distance between the two parallel lines is boxed{sqrt{11}}.

answer：Okay, so I have this problem about horses racing around a circular track. There are 10 horses, each named Horse 1 through Horse 10. Each horse has a lap time equal to its name, so Horse 1 takes 1 minute per lap, Horse 2 takes 2 minutes, and so on up to Horse 10, which takes 10 minutes per lap. At time 0, all the horses are together at the starting point. They start running, and we know that the least time S when all 10 horses meet again at the starting point is 2520 minutes. That makes sense because 2520 is the least common multiple (LCM) of the numbers 1 through 10. But now, the question is asking for the least time T where at least 5 of the horses are again at the starting point. So, I need to find the smallest T such that T is a multiple of the lap times of at least 5 horses. First, I should think about how to approach this. Since we're dealing with multiples of lap times, we need to find the LCM of some subset of 5 lap times. The goal is to find the smallest such LCM. I remember that the LCM of a set of numbers is the smallest number that is a multiple of each of them. So, to minimize T, I should look for a subset of 5 lap times whose LCM is as small as possible. Let me list the lap times: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. I think starting with the smaller numbers might give a smaller LCM. Let's try the first five horses: 1, 2, 3, 4, 5. Calculating the LCM of these numbers:- LCM of 1 and 2 is 2.- LCM of 2 and 3 is 6.- LCM of 6 and 4 is 12 (since 4 is 2² and 6 is 2×3, so LCM is 2²×3=12).- LCM of 12 and 5 is 60 (since 5 is prime and doesn't divide 12, so multiply 12×5=60).So, the LCM of 1, 2, 3, 4, 5 is 60. Is there a smaller T? Let's check if another combination of 5 horses could have a smaller LCM. What if I include Horse 6 instead of Horse 5? So, horses 1, 2, 3, 4, 6. Calculating LCM:- LCM of 1, 2, 3, 4 is 12.- LCM of 12 and 6 is still 12 because 6 divides 12.So, the LCM here is 12. Wait, that's smaller than 60. But does that mean T could be 12? Let me check how many horses meet at 12 minutes.At 12 minutes:- Horse 1: 12/1 = 12 laps, so yes.- Horse 2: 12/2 = 6 laps, yes.- Horse 3: 12/3 = 4 laps, yes.- Horse 4: 12/4 = 3 laps, yes.- Horse 6: 12/6 = 2 laps, yes.- Horse 12: Wait, there's no Horse 12. So, only horses 1, 2, 3, 4, 6 meet at 12 minutes. That's 5 horses. So, T could be 12. But wait, earlier I thought T was 60, but now I see that 12 is a smaller T where 5 horses meet. But hold on, is 12 the smallest possible? Let's check smaller multiples. What about 6 minutes? Let's see which horses meet at 6 minutes:- Horse 1: 6/1 = 6 laps, yes.- Horse 2: 6/2 = 3 laps, yes.- Horse 3: 6/3 = 2 laps, yes.- Horse 6: 6/6 = 1 lap, yes.- Horse 4: 6/4 = 1.5 laps, which is not an integer, so no.- Horse 5: 6/5 = 1.2 laps, no.- Horse 7: 6/7 ≈ 0.857 laps, no.- Similarly, others won't meet.So, at 6 minutes, only horses 1, 2, 3, 6 meet. That's 4 horses, which is less than 5. So, 6 minutes isn't enough.What about 8 minutes? Let's check:- Horse 1: 8/1 = 8 laps, yes.- Horse 2: 8/2 = 4 laps, yes.- Horse 4: 8/4 = 2 laps, yes.- Horse 8: 8/8 = 1 lap, yes.- Horse 3: 8/3 ≈ 2.666 laps, no.- Horse 5: 8/5 = 1.6 laps, no.- Others won't meet.So, at 8 minutes, horses 1, 2, 4, 8 meet. That's 4 horses again. Not enough.What about 10 minutes? Let's check:- Horse 1: 10/1 = 10 laps, yes.- Horse 2: 10/2 = 5 laps, yes.- Horse 5: 10/5 = 2 laps, yes.- Horse 10: 10/10 = 1 lap, yes.- Horse 4: 10/4 = 2.5 laps, no.- Horse 3: 10/3 ≈ 3.333 laps, no.- Others won't meet.So, at 10 minutes, horses 1, 2, 5, 10 meet. That's 4 horses. Still not enough.What about 12 minutes? As I calculated earlier, horses 1, 2, 3, 4, 6 meet. That's 5 horses. So, T could be 12.Is there a smaller T than 12 where at least 5 horses meet? Let's check 9 minutes:- Horse 1: 9/1 = 9 laps, yes.- Horse 3: 9/3 = 3 laps, yes.- Horse 9: 9/9 = 1 lap, yes.- Horse 2: 9/2 = 4.5 laps, no.- Horse 4: 9/4 = 2.25 laps, no.- Others won't meet.So, only 3 horses meet at 9 minutes. Not enough.How about 15 minutes? Let's see:- Horse 1: 15/1 = 15 laps, yes.- Horse 3: 15/3 = 5 laps, yes.- Horse 5: 15/5 = 3 laps, yes.- Horse 15: Doesn't exist.- Horse 2: 15/2 = 7.5 laps, no.- Horse 4: 15/4 = 3.75 laps, no.- Others won't meet.So, only 3 horses meet at 15 minutes. Not enough.Wait, maybe I should consider another combination of 5 horses. For example, horses 2, 4, 6, 8, 10. Let's find their LCM.Calculating LCM of 2, 4, 6, 8, 10:- Prime factors: - 2 = 2 - 4 = 2² - 6 = 2×3 - 8 = 2³ - 10 = 2×5- The LCM is the product of the highest powers of all primes present: 2³, 3, 5. So, LCM = 8×3×5 = 120.That's larger than 12, so 12 is still better.Another combination: horses 1, 2, 3, 4, 6 as before, LCM 12.What about horses 1, 2, 3, 4, 5? LCM 60, which is larger than 12.What about horses 1, 2, 3, 4, 7? Let's calculate their LCM.- LCM of 1, 2, 3, 4 is 12.- LCM of 12 and 7 is 84 (since 7 is prime and doesn't divide 12).So, LCM is 84, which is larger than 12.Similarly, including 7, 8, 9, or 10 will likely increase the LCM beyond 12.So, it seems that 12 is the smallest T where at least 5 horses meet. Wait, but earlier I thought T was 60, but then realized 12 is smaller. So, I need to confirm if 12 is indeed the correct answer.Let me list the horses that meet at 12 minutes:- Horse 1: 12/1 = 12 laps.- Horse 2: 12/2 = 6 laps.- Horse 3: 12/3 = 4 laps.- Horse 4: 12/4 = 3 laps.- Horse 6: 12/6 = 2 laps.That's 5 horses. Perfect.Is there any smaller T where at least 5 horses meet? Let's check T=12 is indeed the smallest.We saw that at T=6, only 4 horses meet. At T=8, 4 horses. At T=10, 4 horses. At T=12, 5 horses. So, yes, 12 is the smallest.Therefore, T=12. The sum of the digits of T is 1 + 2 = 3.Wait, but the options given are A)2, B)3, C)4, D)5, E)6. So, 3 is an option, B.But wait, earlier I thought T=60, but then realized 12 is smaller. So, the answer should be 3.But hold on, let me double-check. Is there a T smaller than 12 where at least 5 horses meet?At T=12, 5 horses meet. At T=6, 4 horses. At T=4, let's see:- Horse 1: 4/1=4 laps.- Horse 2: 4/2=2 laps.- Horse 4: 4/4=1 lap.- Others: 4/3≈1.333, 4/5=0.8, etc. So, only 3 horses meet.At T=3:- Horse 1: 3/1=3 laps.- Horse 3: 3/3=1 lap.- Others: 3/2=1.5, 3/4=0.75, etc. So, only 2 horses.At T=2:- Horse 1: 2/1=2 laps.- Horse 2: 2/2=1 lap.- Others: 2/3≈0.666, etc. So, only 2 horses.At T=1:- Only Horse 1.So, indeed, the smallest T where at least 5 horses meet is 12 minutes. The sum of the digits of 12 is 1 + 2 = 3.But wait, the answer choices include 3 as option B. So, I think that's the answer.However, I'm a bit confused because initially, I thought 60 was the answer, but then realized 12 is smaller. So, I need to make sure I didn't miss any smaller T.Wait, another thought: what about T=24? Let's see how many horses meet there.At T=24:- Horse 1: 24 laps.- Horse 2: 12 laps.- Horse 3: 8 laps.- Horse 4: 6 laps.- Horse 6: 4 laps.- Horse 8: 3 laps.- Horse 12: Doesn't exist.- Horse 24: Doesn't exist.So, horses 1, 2, 3, 4, 6, 8 meet. That's 6 horses. But since we're looking for the least T, 12 is still smaller.Another angle: maybe a different combination of 5 horses could have a smaller LCM than 12. For example, horses 2, 4, 6, 8, 10 have LCM 120, which is larger. Horses 1, 2, 3, 4, 6 have LCM 12. Horses 1, 2, 3, 4, 5 have LCM 60. Horses 1, 2, 3, 4, 7 have LCM 84. So, 12 is indeed the smallest.Therefore, T=12, sum of digits is 1+2=3. So, answer is B)3.Wait, but in the initial problem statement, it says "at least 5 of the horses". So, 5 or more. At T=12, exactly 5 horses meet. Is there a T smaller than 12 where more than 5 horses meet? Let's check.At T=12, 5 horses meet. At T=24, 6 horses meet, but 24 is larger than 12. So, 12 is still the smallest T where at least 5 meet.Another check: T=12 is the first time when 5 horses meet. Before that, at T=6, only 4 meet. So, yes, 12 is the answer.Therefore, the sum of the digits of T=12 is 1+2=3.