🧠:<think>Okay, so I have this problem here about linear equations in different coordinate systems. Let me try to understand it step by step. First, in the Cartesian coordinate system xOy, the equation Ax + By = 0 represents a line that passes through the origin. I remember that in two dimensions, a linear equation like this always represents a straight line, and if there's no constant term (like the 0 here), it must pass through the origin because when x and y are both zero, the equation holds true.Now, the question is asking about the analogous situation in three-dimensional space, specifically in the Oxyz coordinate system. The equation given is Ax + By + Cz = 0, where A, B, and C are not all zero. I need to figure out what this equation represents.Let me think about what a linear equation in three variables represents. In two dimensions, it's a line, but in three dimensions, it's something else. I recall that in three dimensions, a linear equation like Ax + By + Cz + D = 0 represents a plane. But in this case, D is zero because the equation is Ax + By + Cz = 0. So, does that mean it's a plane that passes through the origin?Let me verify that. If I plug in x = 0, y = 0, z = 0 into the equation, I get 0 + 0 + 0 = 0, which is true. So yes, the origin (0,0,0) lies on this plane. That makes sense because without a constant term, the plane must pass through the origin.To make sure I'm not missing anything, let me think about the general form of a plane in three dimensions. It's usually written as Ax + By + Cz + D = 0, where A, B, C are coefficients that determine the orientation of the plane, and D is the constant term that determines its distance from the origin. If D is zero, the plane passes through the origin. So, in this case, since D is zero, it's a plane passing through the origin.I can also think about specific examples to test this. For instance, if A = 1, B = 0, C = 0, then the equation becomes x = 0, which is the y-z plane, passing through the origin. Similarly, if A = 0, B = 1, C = 0, the equation becomes y = 0, which is the x-z plane, also passing through the origin. The same logic applies if C is non-zero; for example, z = 0 is the x-y plane, passing through the origin.What if all three coefficients are non-zero? Let's say A = 1, B = 1, C = 1. The equation becomes x + y + z = 0. This is a plane that passes through the origin because (0,0,0) satisfies the equation. It's tilted relative to the axes, but it still goes through the origin.I think I've got it. The equation Ax + By + Cz = 0 in three-dimensional space represents a plane that passes through the origin. It's similar to how Ax + By = 0 represents a line through the origin in two dimensions. The key difference is the dimensionality: in two dimensions, it's a line; in three dimensions, it's a plane.Just to recap, in two dimensions, a linear equation without a constant term is a line through the origin. Extending that to three dimensions, a linear equation without a constant term is a plane through the origin. The coefficients A, B, and C determine the orientation of the plane, but since there's no constant term, it must pass through the origin.So, putting it all together, the equation Ax + By + Cz = 0 in three-dimensional space represents a plane that passes through the origin.</think>

🧠:<think>Alright, so I have this trigonometric expression to evaluate: tan 70° multiplied by cos 10°, and then all of that multiplied by (√3 tan 20° - 1). Hmm, that looks a bit complicated, but maybe I can break it down step by step.First, let me write it out again to make sure I have it correctly: tan 70° * cos 10° * (√3 tan 20° - 1). Okay, so it's a product of three terms: tan 70°, cos 10°, and a binomial involving tan 20°. I wonder if there are any trigonometric identities that can help simplify this expression.I remember that tan θ can be expressed as sin θ / cos θ, so maybe rewriting tan 70° and tan 20° in terms of sine and cosine might help. Let me try that.So, tan 70° is sin 70° / cos 70°, and tan 20° is sin 20° / cos 20°. Substituting these into the expression, I get:(sin 70° / cos 70°) * cos 10° * (√3 * (sin 20° / cos 20°) - 1).Hmm, that looks a bit better. Maybe I can combine the terms. Let me distribute the cos 10° across the binomial:(sin 70° / cos 70°) * [√3 * (sin 20° / cos 20°) * cos 10° - cos 10°].Wait, that might not be the most straightforward approach. Maybe I should look for angle addition or subtraction formulas or perhaps some product-to-sum identities.Another thought: 70°, 20°, and 10° are all related angles. Maybe there's a way to express some of these angles in terms of others. For example, 70° is 60° + 10°, and 20° is 30° - 10°. I wonder if that helps.Let me try expressing tan 70° as tan(60° + 10°). There's a formula for tan(A + B): [tan A + tan B] / [1 - tan A tan B]. So, tan(60° + 10°) would be [tan 60° + tan 10°] / [1 - tan 60° tan 10°]. Since tan 60° is √3, that becomes [√3 + tan 10°] / [1 - √3 tan 10°].Hmm, interesting. So, tan 70° = [√3 + tan 10°] / [1 - √3 tan 10°]. Maybe I can substitute that back into the original expression.So, replacing tan 70°, the expression becomes:([√3 + tan 10°] / [1 - √3 tan 10°]) * cos 10° * (√3 tan 20° - 1).This still looks complicated, but perhaps there's a way to simplify the denominator. Let me see if I can manipulate the denominator 1 - √3 tan 10°.Wait, tan 10° is sin 10° / cos 10°, so 1 - √3 tan 10° = 1 - √3 (sin 10° / cos 10°) = [cos 10° - √3 sin 10°] / cos 10°. So, the denominator becomes [cos 10° - √3 sin 10°] / cos 10°, which means the entire fraction [√3 + tan 10°] / [1 - √3 tan 10°] becomes [√3 + tan 10°] * [cos 10° / (cos 10° - √3 sin 10°)].So, putting it all together, the expression is:[√3 + tan 10°] * [cos 10° / (cos 10° - √3 sin 10°)] * cos 10° * (√3 tan 20° - 1).Hmm, this is getting a bit messy, but maybe I can simplify further. Let's see if I can combine the cos 10° terms. There are two cos 10° in the numerator now, so that would be cos² 10°.So, the expression becomes:[√3 + tan 10°] * [cos² 10° / (cos 10° - √3 sin 10°)] * (√3 tan 20° - 1).I wonder if the denominator cos 10° - √3 sin 10° can be expressed as a single trigonometric function. There's an identity that a cos θ + b sin θ can be written as R cos(θ - φ), where R = √(a² + b²) and tan φ = b/a.In this case, a = 1 and b = -√3, so R = √(1 + 3) = 2, and tan φ = -√3 / 1 = -√3. So, φ is -60°, since tan(-60°) = -√3.Therefore, cos 10° - √3 sin 10° = 2 cos(10° + 60°) = 2 cos 70°. Ah, that's a useful simplification!So, replacing the denominator, the expression becomes:[√3 + tan 10°] * [cos² 10° / (2 cos 70°)] * (√3 tan 20° - 1).Now, let's look at the numerator: [√3 + tan 10°]. Since tan 10° = sin 10° / cos 10°, this becomes √3 + sin 10° / cos 10° = [√3 cos 10° + sin 10°] / cos 10°.Again, this is similar to the earlier expression. Using the same identity, a cos θ + b sin θ = R cos(θ - φ). Here, a = √3 and b = 1, so R = √(3 + 1) = 2, and tan φ = 1 / √3, which means φ = 30°.Therefore, √3 cos 10° + sin 10° = 2 cos(10° - 30°) = 2 cos(-20°) = 2 cos 20°, since cosine is even.So, [√3 + tan 10°] = [2 cos 20°] / cos 10°.Substituting back into the expression, we have:[2 cos 20° / cos 10°] * [cos² 10° / (2 cos 70°)] * (√3 tan 20° - 1).Simplifying the constants and cosines:The 2 in the numerator and denominator cancels out, leaving:[cos 20° / cos 10°] * [cos² 10° / cos 70°] * (√3 tan 20° - 1).Multiplying the cosines:cos 20° * cos² 10° / (cos 10° * cos 70°) = cos 20° * cos 10° / cos 70°.So, the expression now is:[cos 20° * cos 10° / cos 70°] * (√3 tan 20° - 1).Hmm, let's see. Maybe I can express cos 70° in terms of sin 20°, since cos 70° = sin 20°, because cos(90° - θ) = sin θ.So, cos 70° = sin 20°, which means 1 / cos 70° = 1 / sin 20°.Therefore, the expression becomes:cos 20° * cos 10° / sin 20° * (√3 tan 20° - 1).Now, cos 20° / sin 20° is cot 20°, so:cot 20° * cos 10° * (√3 tan 20° - 1).Hmm, cot 20° is 1 / tan 20°, so let's write that:(1 / tan 20°) * cos 10° * (√3 tan 20° - 1).Multiplying through:cos 10° * (√3 - tan 20° / tan 20°).Wait, that doesn't seem right. Let me check that step again.Actually, distributing (1 / tan 20°) over the binomial:(1 / tan 20°) * √3 tan 20° - (1 / tan 20°) * 1.Simplifying each term:The first term: (1 / tan 20°) * √3 tan 20° = √3.The second term: -(1 / tan 20°) * 1 = -cot 20°.So, the expression simplifies to:cos 10° * (√3 - cot 20°).Hmm, okay. Now, let's see if we can combine √3 and cot 20°. Maybe express cot 20° in terms of sine and cosine:cot 20° = cos 20° / sin 20°.So, the expression becomes:cos 10° * (√3 - cos 20° / sin 20°).To combine the terms inside the parentheses, let's find a common denominator:√3 sin 20° / sin 20° - cos 20° / sin 20° = [√3 sin 20° - cos 20°] / sin 20°.So, the expression is now:cos 10° * [√3 sin 20° - cos 20°] / sin 20°.Hmm, the numerator √3 sin 20° - cos 20° looks familiar. Maybe it can be expressed as a single sine function using the identity a sin θ + b cos θ = R sin(θ + φ), where R = √(a² + b²) and tan φ = b/a.In this case, a = √3 and b = -1, so R = √(3 + 1) = 2, and tan φ = -1 / √3, which means φ = -30°, since tan(-30°) = -1/√3.Therefore, √3 sin 20° - cos 20° = 2 sin(20° - 30°) = 2 sin(-10°) = -2 sin 10°, because sine is an odd function.So, substituting back, the expression becomes:cos 10° * (-2 sin 10°) / sin 20°.Simplifying:-2 cos 10° sin 10° / sin 20°.I recall that sin 2θ = 2 sin θ cos θ, so 2 sin 10° cos 10° = sin 20°. Therefore, cos 10° sin 10° = sin 20° / 2.Substituting that in:-2 * (sin 20° / 2) / sin 20° = -2 * (sin 20° / 2) / sin 20°.Simplifying further:The sin 20° terms cancel out, and the 2 and 2 cancel out, leaving just -1.So, after all that simplification, the value of the original expression is -1.Final Answerboxed{-1}</think>

🧠:<think>Okay, so I have this problem where I'm given a complex number equation: z times (2 minus 3i) equals 6 plus 4i. I need to find the modulus of z. Hmm, modulus of a complex number is like its distance from the origin in the complex plane, right? So, if z is a complex number, say a plus bi, then its modulus is the square root of (a squared plus b squared). Alright, so first, I need to figure out what z is. The equation is z times (2 - 3i) equals 6 + 4i. To solve for z, I should divide both sides by (2 - 3i). That makes sense because if z times something equals something else, dividing both sides by that something should give me z. But wait, dividing by a complex number isn't as straightforward as dividing by a real number. I remember that to simplify expressions like this, I need to multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of (2 - 3i) is (2 + 3i). So, I'll do that to eliminate the imaginary part in the denominator.Let me write this out step by step. So, z equals (6 + 4i) divided by (2 - 3i). To simplify, I'll multiply both the top and bottom by (2 + 3i):z = [(6 + 4i)(2 + 3i)] / [(2 - 3i)(2 + 3i)]Now, I need to multiply out the numerators and the denominators. Let's start with the numerator: (6 + 4i)(2 + 3i). Using the distributive property (FOIL), I get:6*2 + 6*3i + 4i*2 + 4i*3iCalculating each term:6*2 = 126*3i = 18i4i*2 = 8i4i*3i = 12i²Wait, i² is equal to -1, right? So, 12i² is 12*(-1) = -12.So, combining all the terms in the numerator:12 + 18i + 8i - 12Combine like terms:12 - 12 = 018i + 8i = 26iSo, the numerator simplifies to 26i.Now, let's work on the denominator: (2 - 3i)(2 + 3i). Again, using FOIL:2*2 + 2*3i - 3i*2 - 3i*3iCalculating each term:2*2 = 42*3i = 6i-3i*2 = -6i-3i*3i = -9i²Again, i² is -1, so -9i² = -9*(-1) = 9.Now, combining the denominator terms:4 + 6i - 6i + 9The 6i and -6i cancel each other out, leaving:4 + 9 = 13So, the denominator simplifies to 13.Putting it all together, z equals the numerator over the denominator:z = 26i / 13Simplify that:26 divided by 13 is 2, so z = 2i.Wait, so z is 2i? That means it's a purely imaginary number with no real part. So, in the form a + bi, a is 0 and b is 2.Now, to find the modulus of z, which is |z|. As I remembered earlier, the modulus is the square root of (a squared plus b squared). So, plugging in the values:|z| = sqrt(0² + 2²) = sqrt(0 + 4) = sqrt(4) = 2.So, the modulus of z is 2.But let me double-check my steps to make sure I didn't make a mistake. Starting from z = (6 + 4i)/(2 - 3i). I multiplied numerator and denominator by (2 + 3i):Numerator: (6 + 4i)(2 + 3i) = 12 + 18i + 8i + 12i² = 12 + 26i -12 = 26iDenominator: (2 - 3i)(2 + 3i) = 4 + 6i -6i -9i² = 4 + 0 + 9 = 13So, z = 26i / 13 = 2i. That seems correct.Then, modulus is sqrt(0 + 4) = 2. Yep, that checks out.I think I did it right. So, the modulus of z is 2.</think>

🧠:<think>First, I need to find the intercepts of the line ( px + qy = 24 ) with the coordinate axes. For the ( x )-intercept, I set ( y = 0 ):[px = 24 implies x = frac{24}{p}.]For the ( y )-intercept, I set ( x = 0 ):[qy = 24 implies y = frac{24}{q}.]Next, I calculate the area of the triangle formed by these intercepts and the origin. The area ( A ) is given by:[A = frac{1}{2} times frac{24}{p} times frac{24}{q} = frac{576}{pq}.]It is given that the area is 48, so I set up the equation:[frac{576}{pq} = 48.]Solving for ( pq ):[576 = 48pq implies pq = frac{576}{48} = 12.]Therefore, the value of ( pq ) is 12.</think>