💡Alright, so I have this problem: Calculate arccos(cos(11)), and everything is in radians. Hmm, okay, let me think about how to approach this. I remember that arccos and cos are inverse functions, but they don't always cancel each other out directly because of their domains and ranges. First, let me recall what the cosine function does. Cosine is a periodic function with a period of 2π, which means cos(x) = cos(x + 2πk) for any integer k. So, cos(11) is the same as cos(11 - 2πk) for some integer k. But arccos has a specific range, right? It only gives outputs between 0 and π radians. So, arccos(cos(x)) will give me an angle in [0, π] that has the same cosine value as x.Okay, so I need to find an angle θ in [0, π] such that cos(θ) = cos(11). Since cosine is periodic, I can subtract multiples of 2π from 11 until I get an angle within [0, π]. Let me try that.First, let's see how many times 2π fits into 11. 2π is approximately 6.28319. So, 11 divided by 6.28319 is roughly 1.75. That means I can subtract 2π once from 11 and still be positive, but subtracting it twice would make it negative. Let me calculate 11 - 2π:11 - 2π ≈ 11 - 6.28319 ≈ 4.71681Hmm, 4.71681 is still greater than π (which is approximately 3.14159). So, that's still outside the range [0, π]. I need to subtract another 2π to get it within that range. Let me try that:11 - 4π ≈ 11 - 12.56637 ≈ -1.56637Wait, that's negative. Arccos only takes values between 0 and π, so a negative angle isn't directly helpful. But I remember that cosine is also an even function, meaning cos(-x) = cos(x). So, cos(-1.56637) = cos(1.56637). Therefore, arccos(cos(-1.56637)) = arccos(cos(1.56637)).But 1.56637 is less than π, so that should be within the principal value range of arccos. So, does that mean arccos(cos(11)) = 1.56637? Let me check.Alternatively, another way to think about it is that arccos(cos(x)) gives the angle in [0, π] that has the same cosine as x. So, if x is greater than π, we subtract multiples of 2π until we get into [0, π], but if subtracting 2π makes it negative, we can use the property that cos(x) = cos(-x) to reflect it back into the positive range.So, starting with x = 11 radians. Since 11 is much larger than 2π, let's find how many full rotations of 2π fit into 11. As I calculated earlier, 11 / (2π) ≈ 1.75, so there's one full rotation (2π) and a remainder. Let's compute 11 - 2π:11 - 2π ≈ 11 - 6.28319 ≈ 4.71681Now, 4.71681 is still greater than π, so we need to subtract another 2π:4.71681 - 2π ≈ 4.71681 - 6.28319 ≈ -1.56638Again, negative. But since cosine is even, cos(-1.56638) = cos(1.56638). So, arccos(cos(-1.56638)) = arccos(cos(1.56638)) = 1.56638, since 1.56638 is within [0, π].Therefore, arccos(cos(11)) = 1.56638 radians. Let me verify this with another approach to make sure I didn't make a mistake.Another way is to recognize that arccos(cos(x)) = |x - 2πk| where k is chosen such that the result is in [0, π]. So, starting with x = 11, we need to find k such that 11 - 2πk is in [0, π].Let's solve for k:0 ≤ 11 - 2πk ≤ πSubtract 11:-11 ≤ -2πk ≤ π - 11Divide by -2π (remembering to reverse the inequalities):11/(2π) ≥ k ≥ (11 - π)/(2π)Calculating the numerical values:11/(2π) ≈ 11 / 6.28319 ≈ 1.75(11 - π)/(2π) ≈ (11 - 3.14159)/6.28319 ≈ (7.85841)/6.28319 ≈ 1.25So, k must be between approximately 1.25 and 1.75. Since k must be an integer, the only possible value is k = 2.Wait, hold on, k = 2 would give:11 - 2π*2 = 11 - 4π ≈ 11 - 12.56637 ≈ -1.56637Which is negative, but as we saw earlier, we can take the absolute value or use the even property of cosine to get 1.56637.So, arccos(cos(11)) = |11 - 4π| ≈ 1.56637 radians.Let me check if this makes sense. If I compute cos(11) and then take arccos of that value, I should get 1.56637.Calculating cos(11):cos(11) ≈ cos(11 radians) ≈ -0.0044257Now, arccos(-0.0044257) should give me an angle in [0, π] where cosine is -0.0044257. Since cosine is negative, the angle must be in the second quadrant, between π/2 and π.Calculating arccos(-0.0044257):Using a calculator, arccos(-0.0044257) ≈ 1.5708 radians, which is approximately π/2. Wait, that doesn't match my earlier result of 1.56637.Hmm, there's a discrepancy here. Let me double-check my calculations.First, cos(11):11 radians is approximately 11 * (180/π) ≈ 630 degrees. Subtracting 360 degrees gives 270 degrees, which is 3π/2. So, 11 radians is equivalent to 3π/2 + (11 - 3π/2). Let's compute 11 - 3π/2:3π/2 ≈ 4.7123911 - 4.71239 ≈ 6.28761Wait, that's still more than 2π, which is 6.28319. So, 6.28761 - 2π ≈ 6.28761 - 6.28319 ≈ 0.00442 radians.So, 11 radians is equivalent to 0.00442 radians in the first rotation. But since cosine has a period of 2π, cos(11) = cos(0.00442). Cos(0.00442) is approximately 0.99999, but since 11 radians is in the fourth quadrant (as 3π/2 < 11 < 2π), cosine should be positive. Wait, but earlier I thought cos(11) was negative. Let me clarify.Wait, 11 radians is more than 3π/2 (which is 4.71239) but less than 2π (which is 6.28319). So, 11 radians is in the fourth quadrant where cosine is positive. Therefore, cos(11) should be positive, not negative. But when I calculated cos(11) earlier, I got approximately -0.0044257, which is negative. That doesn't make sense.Wait, maybe I made a mistake in calculating cos(11). Let me check again.Using a calculator, cos(11) ≈ cos(11 radians) ≈ 0.0044257, which is positive, not negative. So, my earlier calculation was incorrect. I must have confused sine and cosine. So, cos(11) ≈ 0.0044257.Therefore, arccos(cos(11)) = arccos(0.0044257). Since 0.0044257 is positive and very close to 0, arccos(0.0044257) should be very close to π/2, but slightly less. Let me compute it:arccos(0.0044257) ≈ 1.56637 radians.Yes, that matches my earlier result. So, arccos(cos(11)) ≈ 1.56637 radians.But wait, earlier I thought that 11 radians is equivalent to 0.00442 radians in the first rotation, but that's not correct. Let me clarify:11 radians is more than 2π (which is approximately 6.28319). So, 11 - 2π ≈ 4.71681 radians. That's still more than π, so subtract another 2π:4.71681 - 2π ≈ 4.71681 - 6.28319 ≈ -1.56638 radians.But since cosine is even, cos(-1.56638) = cos(1.56638). So, arccos(cos(11)) = arccos(cos(-1.56638)) = arccos(cos(1.56638)) = 1.56638 radians.Therefore, the correct answer is approximately 1.56638 radians.Wait, but earlier I thought that 11 radians is equivalent to 0.00442 radians, but that's not correct because 11 - 2π*1 ≈ 4.71681, which is still greater than π. So, subtracting another 2π gives a negative angle, but using the even property of cosine, we reflect it back to the positive side.So, the correct approach is:1. Start with x = 11 radians.2. Subtract 2π until the result is within [0, 2π]. - 11 - 2π ≈ 4.71681 (still greater than π) - 4.71681 - 2π ≈ -1.56638 (negative)3. Since we have a negative angle, use the even property: cos(-1.56638) = cos(1.56638).4. Now, 1.56638 is within [0, π], so arccos(cos(11)) = 1.56638 radians.Therefore, the final answer is approximately 1.56638 radians.But let me express this exactly in terms of π. Since 11 - 4π ≈ -1.56638, and taking the absolute value gives 1.56638, which is 4π - 11. Wait, no:Wait, 11 - 4π ≈ -1.56638, so |11 - 4π| = 4π - 11 ≈ 1.56638.Therefore, arccos(cos(11)) = 4π - 11.So, the exact value is 4π - 11 radians.Let me verify this:4π - 11 ≈ 12.56637 - 11 ≈ 1.56637 radians, which matches our earlier approximation.Therefore, the exact value is 4π - 11 radians.So, to summarize:arccos(cos(11)) = 4π - 11 radians.This is because 11 radians is more than 2π, so we subtract 4π to bring it within the range [0, π], and since it becomes negative, we take the absolute value using the even property of cosine.I think that makes sense now. So, the final answer is 4π - 11 radians.

💡To find the constant term in the expansion of ((x^{2}+ frac{2}{x^{3}})^{5}), we can use the binomial theorem. The binomial theorem states that:[(a + b)^n = sum_{r=0}^{n} C_{n}^{r} cdot a^{n-r} cdot b^{r}]In this case, (a = x^{2}), (b = frac{2}{x^{3}}), and (n = 5). The general term (T_{r+1}) in the expansion is given by:[T_{r+1} = C_{5}^{r} cdot (x^{2})^{5 - r} cdot left(frac{2}{x^{3}}right)^{r}]Simplifying this, we get:[T_{r+1} = C_{5}^{r} cdot 2^{r} cdot x^{10 - 5r}]To find the constant term, we set the exponent of (x) to zero:[10 - 5r = 0]Solving for (r):[5r = 10 r = 2]Substituting (r = 2) back into the general term:[T_{3} = C_{5}^{2} cdot 2^{2} cdot x^{0} = 10 cdot 4 = 40]Therefore, the constant term in the expansion is 40.

💡Alright, let's tackle this problem step by step. So, we have a function ( h(x) ) defined for all real numbers except zero, and it's an even function. That means ( h(-x) = h(x) ) for all ( x neq 0 ). When ( x > 0 ), the function is given in two parts:- For ( 0 < x leq 4 ), ( h(x) = -frac{x^2}{4} )- For ( x > 4 ), ( h(x) = 4 - 2x )We need to find the range of values for ( t ) such that ( h(t) > h(2) ).First, let's understand what ( h(2) ) is. Since 2 is greater than 0 and less than or equal to 4, we'll use the first part of the piecewise function:( h(2) = -frac{2^2}{4} = -frac{4}{4} = -1 )So, we need to find all ( t ) such that ( h(t) > -1 ).Since ( h(x) ) is an even function, ( h(t) = h(|t|) ). This means we can focus on finding the values of ( |t| ) where ( h(|t|) > -1 ), and then translate that back to the values of ( t ).Let's analyze the function ( h(x) ) for ( x > 0 ):1. For ( 0 < x leq 4 ), ( h(x) = -frac{x^2}{4} )2. For ( x > 4 ), ( h(x) = 4 - 2x )We need to find all ( x > 0 ) such that ( h(x) > -1 ).Let's consider the first interval ( 0 < x leq 4 ):( -frac{x^2}{4} > -1 )Multiply both sides by -4 (remember to reverse the inequality when multiplying by a negative number):( x^2 < 4 )Take square roots:( |x| < 2 )But since ( x > 0 ), this simplifies to:( 0 < x < 2 )Now, let's consider the second interval ( x > 4 ):( 4 - 2x > -1 )Subtract 4 from both sides:( -2x > -5 )Divide both sides by -2 (again, reverse the inequality):( x < frac{5}{2} )But wait, in this interval, ( x > 4 ), and ( frac{5}{2} = 2.5 ), which is less than 4. So, there's no overlap here. Therefore, there are no solutions in this interval.So, combining the results, the values of ( x > 0 ) that satisfy ( h(x) > -1 ) are ( 0 < x < 2 ).Since ( h(x) ) is even, the corresponding values of ( t ) will be symmetric around the y-axis. That is, ( t ) can be in ( (-2, 0) ) or ( (0, 2) ).But we have to remember that ( t neq 0 ) because the function is not defined at ( x = 0 ).Therefore, the range of values for ( t ) is ( (-2, 0) cup (0, 2) ).

💡Okay, so I've got this chemistry problem here, and I need to figure out how many carbon atoms are in a compound. Let me read it again to make sure I understand what's being asked."A compound has 2 H, a certain number of C, and 3 O atoms. The molecular weight of the compound is 62. How many C atoms are in the compound?"Alright, so the compound has hydrogen, carbon, and oxygen atoms. Specifically, there are 2 hydrogen atoms, an unknown number of carbon atoms, and 3 oxygen atoms. The molecular weight, which is the total weight of all the atoms in the molecule, is 62 g/mol. I need to find out how many carbon atoms are in this compound.First, I remember that the molecular weight of a compound is the sum of the atomic weights of all the atoms in it. So, if I can figure out the total weight contributed by hydrogen and oxygen, I can subtract that from the total molecular weight to find out how much weight is left for carbon. Then, I can determine how many carbon atoms that corresponds to.Let me recall the atomic weights of these elements. Hydrogen has an atomic weight of approximately 1 g/mol, carbon is about 12 g/mol, and oxygen is around 16 g/mol. These are standard values, so I can use them for this calculation.Okay, so the compound has 2 hydrogen atoms. Since each hydrogen atom weighs about 1 g/mol, the total weight contributed by hydrogen is 2 * 1 = 2 g/mol.Next, there are 3 oxygen atoms. Each oxygen atom weighs about 16 g/mol, so the total weight from oxygen is 3 * 16 = 48 g/mol.Now, if I add the weights from hydrogen and oxygen together, that gives me the total weight of those two elements in the compound. So, 2 g/mol (from hydrogen) + 48 g/mol (from oxygen) = 50 g/mol.The molecular weight of the entire compound is given as 62 g/mol. If I subtract the combined weight of hydrogen and oxygen from the total molecular weight, I'll get the weight contributed by carbon. So, 62 g/mol (total) - 50 g/mol (H and O) = 12 g/mol.Now, since each carbon atom weighs about 12 g/mol, the number of carbon atoms in the compound is the total weight from carbon divided by the atomic weight of carbon. So, 12 g/mol / 12 g/mol per carbon atom = 1 carbon atom.Wait, that seems straightforward, but let me double-check to make sure I didn't make any mistakes. I'll go through the steps again.1. Calculate the weight from hydrogen: 2 H * 1 g/mol = 2 g/mol.2. Calculate the weight from oxygen: 3 O * 16 g/mol = 48 g/mol.3. Add those together: 2 + 48 = 50 g/mol.4. Subtract from the total molecular weight: 62 - 50 = 12 g/mol.5. Divide by the atomic weight of carbon: 12 / 12 = 1.Everything checks out. So, there is 1 carbon atom in the compound.I wonder if there's another way to approach this problem, maybe by setting up an equation. Let me try that.Let the number of carbon atoms be 'x'. Then, the molecular weight can be expressed as:(2 * 1) + (x * 12) + (3 * 16) = 62Simplifying that:2 + 12x + 48 = 62Combine like terms:50 + 12x = 62Subtract 50 from both sides:12x = 12Divide both sides by 12:x = 1Yep, that also gives me x = 1. So, there's definitely 1 carbon atom in the compound.I think I've got it right. It's always good to verify using different methods to ensure accuracy, especially when dealing with chemistry problems where small mistakes can lead to big errors in understanding.Another thing I might consider is checking if the molecular formula makes sense. If there's 1 carbon atom, 2 hydrogen atoms, and 3 oxygen atoms, the formula would be CH2O3. Wait, that doesn't seem like a common compound I'm familiar with. Let me think about common compounds with similar formulas.For example, carbon dioxide is CO2, which has 1 C, 2 O. Water is H2O, which has 2 H, 1 O. But CH2O3 doesn't ring a bell. Maybe it's a less common compound or perhaps a hypothetical one for the sake of the problem.Alternatively, could there be a mistake in the number of oxygen atoms? If there were 2 oxygen atoms instead of 3, the calculation would be different. Let me see:If O = 2, then:2 H * 1 = 22 O * 16 = 32Total H and O = 2 + 32 = 34Then, molecular weight = 62, so carbon weight = 62 - 34 = 28Number of C atoms = 28 / 12 ≈ 2.33But that's not a whole number, which doesn't make sense because you can't have a fraction of an atom in a molecule. So, that suggests that the original problem likely has 3 oxygen atoms, leading to 1 carbon atom.Alternatively, if there were 4 oxygen atoms:4 O * 16 = 64But then, H and O would already sum to 2 + 64 = 66, which is more than the molecular weight of 62, which isn't possible.So, 3 oxygen atoms seem correct, leading to 1 carbon atom.I think I'm confident in my answer now. There's 1 carbon atom in the compound.

💡Okay, so I have this quadratic equation: ( a x^2 + b x + frac{1}{2} = 0 ). It says that one of the roots is ( -1 ). Hmm, so if ( x = -1 ) is a root, that means when I plug ( x = -1 ) into the equation, it should equal zero. Let me write that down:( a(-1)^2 + b(-1) + frac{1}{2} = 0 )Simplifying that, ( (-1)^2 ) is 1, so:( a(1) + b(-1) + frac{1}{2} = 0 )Which simplifies to:( a - b + frac{1}{2} = 0 )So, if I solve for ( b ), I get:( b = a + frac{1}{2} )Alright, so that's a relationship between ( a ) and ( b ). Now, the problem also mentions the vertex of the quadratic function ( y = a x^2 + b x + frac{1}{2} ) is in the first quadrant. I remember that the vertex of a parabola given by ( y = ax^2 + bx + c ) is at the point ( (-frac{b}{2a}, f(-frac{b}{2a})) ). Since the vertex is in the first quadrant, both the x-coordinate and y-coordinate must be positive.First, let's find the x-coordinate of the vertex:( x = -frac{b}{2a} )We already know ( b = a + frac{1}{2} ), so substitute that in:( x = -frac{a + frac{1}{2}}{2a} )For this x-coordinate to be positive, the numerator and denominator must have the same sign. Let's analyze:The denominator is ( 2a ). The numerator is ( -(a + frac{1}{2}) ). So:( -frac{a + frac{1}{2}}{2a} > 0 )This inequality will hold if both numerator and denominator are positive or both are negative.Case 1: Both numerator and denominator positive.Numerator: ( -(a + frac{1}{2}) > 0 ) implies ( a + frac{1}{2} < 0 ) which implies ( a < -frac{1}{2} )Denominator: ( 2a > 0 ) implies ( a > 0 )But ( a < -frac{1}{2} ) and ( a > 0 ) can't both be true. So this case is impossible.Case 2: Both numerator and denominator negative.Numerator: ( -(a + frac{1}{2}) < 0 ) implies ( a + frac{1}{2} > 0 ) which implies ( a > -frac{1}{2} )Denominator: ( 2a < 0 ) implies ( a < 0 )So combining these, ( -frac{1}{2} < a < 0 )Alright, so ( a ) is between ( -frac{1}{2} ) and 0. Now, let's check the y-coordinate of the vertex to ensure it's positive.The y-coordinate is found by plugging the x-coordinate back into the equation:( y = a left( -frac{b}{2a} right)^2 + b left( -frac{b}{2a} right) + frac{1}{2} )Simplify this step by step.First, compute ( left( -frac{b}{2a} right)^2 ):( left( -frac{b}{2a} right)^2 = frac{b^2}{4a^2} )So,( y = a cdot frac{b^2}{4a^2} + b cdot left( -frac{b}{2a} right) + frac{1}{2} )Simplify each term:First term: ( a cdot frac{b^2}{4a^2} = frac{b^2}{4a} )Second term: ( b cdot left( -frac{b}{2a} right) = -frac{b^2}{2a} )Third term: ( frac{1}{2} )So, putting it all together:( y = frac{b^2}{4a} - frac{b^2}{2a} + frac{1}{2} )Combine the first two terms:( frac{b^2}{4a} - frac{2b^2}{4a} = -frac{b^2}{4a} )So,( y = -frac{b^2}{4a} + frac{1}{2} )We need this y-coordinate to be positive:( -frac{b^2}{4a} + frac{1}{2} > 0 )Let's substitute ( b = a + frac{1}{2} ) into this inequality:( -frac{(a + frac{1}{2})^2}{4a} + frac{1}{2} > 0 )First, expand ( (a + frac{1}{2})^2 ):( (a + frac{1}{2})^2 = a^2 + a + frac{1}{4} )So, substitute back:( -frac{a^2 + a + frac{1}{4}}{4a} + frac{1}{2} > 0 )Let's split the fraction:( -frac{a^2}{4a} - frac{a}{4a} - frac{frac{1}{4}}{4a} + frac{1}{2} > 0 )Simplify each term:( -frac{a}{4} - frac{1}{4} - frac{1}{16a} + frac{1}{2} > 0 )Combine like terms:( -frac{a}{4} - frac{1}{4} + frac{1}{2} - frac{1}{16a} > 0 )Simplify constants:( -frac{1}{4} + frac{1}{2} = frac{1}{4} )So,( -frac{a}{4} + frac{1}{4} - frac{1}{16a} > 0 )Let me write this as:( frac{1}{4} - frac{a}{4} - frac{1}{16a} > 0 )Multiply both sides by 16a to eliminate denominators. But wait, I have to be careful because ( a ) is negative here (since ( -frac{1}{2} < a < 0 )), so multiplying by ( a ) will reverse the inequality sign.But let me see:Wait, actually, let's factor out ( frac{1}{4} ):( frac{1}{4}(1 - a) - frac{1}{16a} > 0 )Alternatively, maybe it's better to find a common denominator.Let me express all terms with denominator 16a:( frac{4a}{16a} - frac{4a^2}{16a} - frac{1}{16a} > 0 )Wait, that might complicate things. Alternatively, let's multiply both sides by 16a, keeping in mind that ( a < 0 ), so inequality flips:Multiplying:( 16a cdot left( frac{1}{4} - frac{a}{4} - frac{1}{16a} right) < 0 )Simplify term by term:( 16a cdot frac{1}{4} = 4a )( 16a cdot (-frac{a}{4}) = -4a^2 )( 16a cdot (-frac{1}{16a}) = -1 )So, putting it together:( 4a - 4a^2 - 1 < 0 )Simplify:( -4a^2 + 4a - 1 < 0 )Multiply both sides by -1 (which flips the inequality again):( 4a^2 - 4a + 1 > 0 )Now, let's factor this quadratic:( 4a^2 - 4a + 1 = (2a - 1)^2 )So, ( (2a - 1)^2 > 0 )Since a square is always non-negative, and it's greater than zero unless the expression inside is zero. So, ( 2a - 1 neq 0 implies a neq frac{1}{2} ). But in our case, ( a ) is between ( -frac{1}{2} ) and 0, so ( a neq frac{1}{2} ) is automatically satisfied. Therefore, the inequality ( (2a - 1)^2 > 0 ) is always true for our ( a ).So, the y-coordinate condition doesn't give us any additional constraints beyond ( -frac{1}{2} < a < 0 ).Alright, so now we have ( -frac{1}{2} < a < 0 ). We need to find the range of ( t = 2a + b ).But we already have ( b = a + frac{1}{2} ). So, substitute that into ( t ):( t = 2a + (a + frac{1}{2}) = 3a + frac{1}{2} )So, ( t = 3a + frac{1}{2} ). Now, since ( a ) is between ( -frac{1}{2} ) and 0, let's find the corresponding range for ( t ).When ( a = -frac{1}{2} ):( t = 3(-frac{1}{2}) + frac{1}{2} = -frac{3}{2} + frac{1}{2} = -1 )When ( a = 0 ):( t = 3(0) + frac{1}{2} = frac{1}{2} )But since ( a ) is strictly greater than ( -frac{1}{2} ) and strictly less than 0, ( t ) will be strictly greater than ( -1 ) and strictly less than ( frac{1}{2} ).So, the range of ( t ) is ( -1 < t < frac{1}{2} ).Looking back at the options:A: ( frac{1}{4} < t < frac{1}{2} )B: ( -1 < t leq frac{1}{4} )C: ( -frac{1}{2} leq t < frac{1}{2} )D: ( -1 < t < frac{1}{2} )So, the correct answer is D.**Final Answer**boxed{D}

💡Okay, so I have this problem where I need to find scalars p, q, and r such that the vector [-3, 6, 2] can be expressed as a linear combination of three given vectors a, b, and c. The vectors a, b, and c are mutually orthogonal, which I remember means that each pair of vectors is perpendicular to each other. That should help simplify things because I can use the properties of orthogonal vectors to find p, q, and r without having to solve a system of equations.First, let me write down the given vectors:a = [1, 1, 1]b = [2, -3, 1]c = [4, 1, -5]And the vector we need to express is:v = [-3, 6, 2]We need to find scalars p, q, r such that:v = p*a + q*b + r*cSince a, b, and c are mutually orthogonal, their dot products with each other are zero. That is:a · b = 0a · c = 0b · c = 0This is super helpful because it means when I take the dot product of both sides of the equation with any one of these vectors, the cross terms will vanish, leaving me with a simple equation to solve for each scalar.Let me recall the formula for the dot product. For two vectors u = [u1, u2, u3] and v = [v1, v2, v3], the dot product is u1*v1 + u2*v2 + u3*v3.So, let's start by taking the dot product of both sides with vector a.First, compute the left-hand side:v · a = [-3, 6, 2] · [1, 1, 1] = (-3)(1) + (6)(1) + (2)(1) = -3 + 6 + 2 = 5Now, compute the right-hand side:(p*a + q*b + r*c) · a = p*(a · a) + q*(b · a) + r*(c · a)But since a, b, and c are mutually orthogonal, b · a = 0 and c · a = 0. So this simplifies to:= p*(a · a) + 0 + 0 = p*(a · a)Compute a · a:a · a = [1, 1, 1] · [1, 1, 1] = 1*1 + 1*1 + 1*1 = 1 + 1 + 1 = 3So, we have:5 = p*3Solving for p:p = 5/3Okay, that was straightforward. Now, let's find q by taking the dot product of both sides with vector b.Compute the left-hand side:v · b = [-3, 6, 2] · [2, -3, 1] = (-3)(2) + (6)(-3) + (2)(1) = -6 - 18 + 2 = -22Wait, let me double-check that:(-3)*2 = -66*(-3) = -182*1 = 2Adding them up: -6 -18 + 2 = (-6 -18) + 2 = (-24) + 2 = -22Hmm, that seems correct.Now, compute the right-hand side:(p*a + q*b + r*c) · b = p*(a · b) + q*(b · b) + r*(c · b)Again, since a, b, c are mutually orthogonal, a · b = 0 and c · b = 0. So this simplifies to:= 0 + q*(b · b) + 0 = q*(b · b)Compute b · b:b · b = [2, -3, 1] · [2, -3, 1] = 2*2 + (-3)*(-3) + 1*1 = 4 + 9 + 1 = 14So, we have:-22 = q*14Solving for q:q = -22/14Simplify the fraction by dividing numerator and denominator by 2:q = -11/7Wait, that seems a bit messy. Let me check my calculations again.Wait, when I computed v · b, I got -22. Let me verify that:v = [-3, 6, 2]b = [2, -3, 1]Dot product:(-3)*2 = -66*(-3) = -182*1 = 2Adding up: -6 -18 + 2 = (-24) + 2 = -22Yes, that's correct.And b · b is 14, so q = -22/14 = -11/7Hmm, okay, that's correct. Maybe it's just a negative fraction.Now, moving on to find r by taking the dot product with vector c.Compute the left-hand side:v · c = [-3, 6, 2] · [4, 1, -5] = (-3)*4 + 6*1 + 2*(-5) = -12 + 6 -10 = (-12 -10) + 6 = (-22) + 6 = -16Wait, let me check:(-3)*4 = -126*1 = 62*(-5) = -10Adding them: -12 + 6 -10 = (-12 -10) +6 = (-22) +6 = -16Yes, that's correct.Now, compute the right-hand side:(p*a + q*b + r*c) · c = p*(a · c) + q*(b · c) + r*(c · c)Again, since a, b, c are mutually orthogonal, a · c = 0 and b · c = 0. So this simplifies to:= 0 + 0 + r*(c · c) = r*(c · c)Compute c · c:c · c = [4, 1, -5] · [4, 1, -5] = 4*4 + 1*1 + (-5)*(-5) = 16 + 1 + 25 = 42So, we have:-16 = r*42Solving for r:r = -16/42Simplify the fraction by dividing numerator and denominator by 2:r = -8/21Okay, so now I have p = 5/3, q = -11/7, and r = -8/21.Wait a second, let me just verify if these values actually satisfy the original equation.Let me compute p*a + q*b + r*c and see if it equals [-3, 6, 2].Compute p*a:p = 5/3, so 5/3 * [1, 1, 1] = [5/3, 5/3, 5/3]Compute q*b:q = -11/7, so -11/7 * [2, -3, 1] = [-22/7, 33/7, -11/7]Compute r*c:r = -8/21, so -8/21 * [4, 1, -5] = [-32/21, -8/21, 40/21]Now, add them all together:First component:5/3 + (-22/7) + (-32/21)Let me convert all to 21 denominators:5/3 = 35/21-22/7 = -66/21-32/21 = -32/21Adding up: 35 -66 -32 = (35 -66) = -31; -31 -32 = -63So, -63/21 = -3Second component:5/3 + 33/7 + (-8/21)Convert to 21 denominators:5/3 = 35/2133/7 = 99/21-8/21 = -8/21Adding up: 35 + 99 -8 = (35 +99)=134; 134 -8=126126/21=6Third component:5/3 + (-11/7) + 40/21Convert to 21 denominators:5/3=35/21-11/7=-33/2140/21=40/21Adding up:35 -33 +40= (35-33)=2; 2+40=4242/21=2So, adding all components together, we get [-3,6,2], which matches the vector v. So, the values p=5/3, q=-11/7, r=-8/21 are correct.Wait, but in my initial calculation for q, I got q=-22/14 which simplifies to -11/7, which is correct. And for r, I had -16/42 which simplifies to -8/21, which is also correct.So, all the steps seem correct, and the verification shows that the solution is accurate.I think that's all. So, the scalars are p=5/3, q=-11/7, and r=-8/21.