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b) Express the Cartesian coordinates (−12,−5)(-12,-5) in polar coordinates $(r, \thet...
Jul 1, 2024
Solution by Steps
step 1
Calculate the radius rr using the formula r=x2+y2r = \sqrt{x^2 + y^2}
step 2
Substitute x=−12x = -12 and y=−5y = -5 into the formula: r=(−12)2+(−5)2r = \sqrt{(-12)^2 + (-5)^2}
step 3
Simplify the expression: r=144+25=169=13r = \sqrt{144 + 25} = \sqrt{169} = 13
step 4
Calculate the angle θ\theta using the formula θ=tan⁡−1(yx)\theta = \tan^{-1}\left(\frac{y}{x}\right)
step 5
Substitute x=−12x = -12 and y=−5y = -5 into the formula: θ=tan⁡−1(−5−12)\theta = \tan^{-1}\left(\frac{-5}{-12}\right)
step 6
Simplify the expression: θ=tan⁡−1(512)\theta = \tan^{-1}\left(\frac{5}{12}\right)
step 7
Since the point (−12,−5)(-12, -5) is in the third quadrant, add 180∘180^\circ to the angle: θ=tan⁡−1(512)+180∘\theta = \tan^{-1}\left(\frac{5}{12}\right) + 180^\circ
step 8
Calculate the final angle: θ≈22.62∘+180∘=202.62∘\theta \approx 22.62^\circ + 180^\circ = 202.62^\circ
Answer
The polar coordinates are (13,202.62∘)(13, 202.62^\circ)
Key Concept
Conversion from Cartesian to Polar Coordinates
Explanation
To convert Cartesian coordinates to polar coordinates, calculate the radius using r=x2+y2r = \sqrt{x^2 + y^2} and the angle using θ=tan⁡−1(yx)\theta = \tan^{-1}\left(\frac{y}{x}\right). Adjust the angle based on the quadrant.
Question c) Find the polar equation of (x2+y2)2=4xy\left(x^{2}+y^{2}\right)^{2}=4 x y.
step 1
Recall the polar coordinate transformations: x=rcos⁡θx = r \cos \theta and y=rsin⁡θy = r \sin \theta
step 2
Substitute xx and yy in the given equation: ((rcos⁡θ)2+(rsin⁡θ)2)2=4(rcos⁡θ)(rsin⁡θ)\left((r \cos \theta)^{2} + (r \sin \theta)^{2}\right)^{2} = 4 (r \cos \theta) (r \sin \theta)
step 3
Simplify the left side: (r2(cos⁡2θ+sin⁡2θ))2\left(r^2 (\cos^2 \theta + \sin^2 \theta)\right)^{2}
step 4
Use the Pythagorean identity cos⁡2θ+sin⁡2θ=1\cos^2 \theta + \sin^2 \theta = 1: (r2⋅1)2=4r2cos⁡θsin⁡θ\left(r^2 \cdot 1\right)^{2} = 4 r^2 \cos \theta \sin \theta
step 5
Simplify further: r4=4r2cos⁡θsin⁡θr^4 = 4 r^2 \cos \theta \sin \theta
step 6
Divide both sides by r2r^2: r2=4cos⁡θsin⁡θr^2 = 4 \cos \theta \sin \theta
step 7
Use the double-angle identity sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2 \sin \theta \cos \theta: r2=2sin⁡2θr^2 = 2 \sin 2\theta
Answer
The polar equation is r2=2sin⁡2θr^2 = 2 \sin 2\theta
Key Concept
Conversion from Cartesian to Polar Equations
Explanation
To convert a Cartesian equation to a polar equation, use the transformations x=rcos⁡θx = r \cos \theta and y=rsin⁡θy = r \sin \theta, and simplify using trigonometric identities.
Solution by Steps
Question 1(a): Find $\mathbf{u} \cdot \mathbf{v}$
step 1
Calculate the dot product of vectors u=⟨4,−1,1⟩\mathbf{u} = \langle 4, -1, 1 \rangle and v=⟨2,3,−1⟩\mathbf{v} = \langle 2, 3, -1 \rangle using the formula u⋅v=u1v1+u2v2+u3v3\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3
step 2
Substitute the values: u⋅v=4⋅2+(−1)⋅3+1⋅(−1)\mathbf{u} \cdot \mathbf{v} = 4 \cdot 2 + (-1) \cdot 3 + 1 \cdot (-1)
step 3
Simplify the expression: u⋅v=8−3−1=4\mathbf{u} \cdot \mathbf{v} = 8 - 3 - 1 = 4
Answer
The dot product u⋅v\mathbf{u} \cdot \mathbf{v} is 4.
Key Concept
Dot Product
Explanation
The dot product of two vectors is calculated by multiplying corresponding components and summing the results.
Question 1(b): Find the angle $\theta$ between $\mathbf{u}$ and $\mathbf{v}$
step 1
Use the formula cos⁡θ=u⋅v∥u∥∥v∥\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|} to find the angle θ\theta
step 2
Calculate the magnitudes ∥u∥\|\mathbf{u}\| and ∥v∥\|\mathbf{v}\| using ∥u∥=u12+u22+u32\|\mathbf{u}\| = \sqrt{u_1^2 + u_2^2 + u_3^2} and ∥v∥=v12+v22+v32\|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2}
step 3
Substitute the values: ∥u∥=42+(−1)2+12=16+1+1=18=32\|\mathbf{u}\| = \sqrt{4^2 + (-1)^2 + 1^2} = \sqrt{16 + 1 + 1} = \sqrt{18} = 3\sqrt{2} and ∥v∥=22+32+(−1)2=4+9+1=14\|\mathbf{v}\| = \sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14}
step 4
Substitute u⋅v\mathbf{u} \cdot \mathbf{v}, ∥u∥\|\mathbf{u}\|, and ∥v∥\|\mathbf{v}\| into the formula: cos⁡θ=432⋅14=4328=43⋅27=237\cos \theta = \frac{4}{3\sqrt{2} \cdot \sqrt{14}} = \frac{4}{3\sqrt{28}} = \frac{4}{3 \cdot 2\sqrt{7}} = \frac{2}{3\sqrt{7}}
step 5
Calculate θ\theta using θ=arccos⁡(237)\theta = \arccos \left( \frac{2}{3\sqrt{7}} \right)
Answer
The angle θ\theta between u\mathbf{u} and v\mathbf{v} is arccos⁡(237)\arccos \left( \frac{2}{3\sqrt{7}} \right).
Key Concept
Angle Between Vectors
Explanation
The angle between two vectors can be found using the dot product and the magnitudes of the vectors.
Solution by Steps
step 1
Given vectors u=⟨3α,1,−4α⟩\mathbf{u} = \langle 3\alpha, 1, -4\alpha \rangle and v=⟨α,4,2⟩\mathbf{v} = \langle \alpha, 4, 2 \rangle, we need to find the values of α\alpha such that u\mathbf{u} and v\mathbf{v} are orthogonal
step 2
Two vectors are orthogonal if their dot product is zero. The dot product of u\mathbf{u} and v\mathbf{v} is given by: u⋅v=(3α)(α)+(1)(4)+(−4α)(2)\mathbf{u} \cdot \mathbf{v} = (3\alpha)(\alpha) + (1)(4) + (-4\alpha)(2)
step 3
Simplify the dot product: 3α2+4−8α=03\alpha^2 + 4 - 8\alpha = 0
step 4
Solve the quadratic equation 3α2−8α+4=03\alpha^2 - 8\alpha + 4 = 0 using the quadratic formula α=−b±b2−4ac2a\alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=3a = 3, b=−8b = -8, and c=4c = 4
step 5
Calculate the discriminant: (−8)2−4(3)(4)=64−48=16(-8)^2 - 4(3)(4) = 64 - 48 = 16
step 6
Find the roots: α=8±166=8±46\alpha = \frac{8 \pm \sqrt{16}}{6} = \frac{8 \pm 4}{6}
step 7
Simplify the roots: α=2\alpha = 2 or α=23\alpha = \frac{2}{3}
Answer
α=2\alpha = 2 or α=23\alpha = \frac{2}{3}
Key Concept
Orthogonal Vectors
Explanation
Two vectors are orthogonal if their dot product is zero. By setting the dot product to zero and solving the resulting quadratic equation, we find the values of α\alpha.
Question 3
step 1
Given a point (2,1,−1)(2,1,-1) and a line x−33=y+15=z2\frac{x-3}{3} = \frac{y+1}{5} = \frac{z}{2}, we need to find the equation of the plane perpendicular to the line
step 2
The direction vector of the line is d=⟨3,5,2⟩\mathbf{d} = \langle 3, 5, 2 \rangle
step 3
The normal vector to the plane is the same as the direction vector of the line, n=⟨3,5,2⟩\mathbf{n} = \langle 3, 5, 2 \rangle
step 4
The equation of the plane is given by n⋅(r−r0)=0\mathbf{n} \cdot (\mathbf{r} - \mathbf{r_0}) = 0, where r0=(2,1,−1)\mathbf{r_0} = (2,1,-1) and r=(x,y,z)\mathbf{r} = (x,y,z)
step 5
Substitute the normal vector and point into the plane equation: 3(x−2)+5(y−1)+2(z+1)=03(x-2) + 5(y-1) + 2(z+1) = 0
step 6
Simplify the equation: 3x−6+5y−5+2z+2=0⇒3x+5y+2z−9=03x - 6 + 5y - 5 + 2z + 2 = 0 \Rightarrow 3x + 5y + 2z - 9 = 0
Answer
3x+5y+2z−9=03x + 5y + 2z - 9 = 0
Key Concept
Equation of a Plane
Explanation
The equation of a plane can be found using a point on the plane and a normal vector. The normal vector is perpendicular to the plane and can be derived from the direction vector of a given line.
Question 4(a)
step 1
Given points A(1,2,3)A(1,2,3), B(1,−1,0)B(1,-1,0), and C(2,4,7)C(2,4,7), we need to find the normal vector to the plane ABCABC
step 2
Find vectors AB\mathbf{AB} and AC\mathbf{AC}: AB=⟨1−1,−1−2,0−3⟩=⟨0,−3,−3⟩\mathbf{AB} = \langle 1-1, -1-2, 0-3 \rangle = \langle 0, -3, -3 \rangle and AC=⟨2−1,4−2,7−3⟩=⟨1,2,4⟩\mathbf{AC} = \langle 2-1, 4-2, 7-3 \rangle = \langle 1, 2, 4 \rangle
step 3
The normal vector n\mathbf{n} is the cross product of AB\mathbf{AB} and AC\mathbf{AC}: n=AB×AC\mathbf{n} = \mathbf{AB} \times \mathbf{AC}
step 4
Calculate the cross product: n=⟨(−3)(4)−(−3)(2),(−3)(1)−(0)(4),(0)(2)−(−3)(1)⟩=⟨−12+6,−3,3⟩=⟨−6,−3,3⟩\mathbf{n} = \langle (-3)(4) - (-3)(2), (-3)(1) - (0)(4), (0)(2) - (-3)(1) \rangle = \langle -12 + 6, -3, 3 \rangle = \langle -6, -3, 3 \rangle
Answer
n=⟨−6,−3,3⟩\mathbf{n} = \langle -6, -3, 3 \rangle
Key Concept
Normal Vector to a Plane
Explanation
The normal vector to a plane can be found by taking the cross product of two vectors lying on the plane.
Question 4(b)
step 1
Given points A(1,2,3)A(1,2,3), B(1,−1,0)B(1,-1,0), and C(2,4,7)C(2,4,7), we need to find the parametric equation of the line parallel to ABCABC and passing through BB
step 2
The direction vector of the line is the same as the normal vector found in part (a), d=⟨−6,−3,3⟩\mathbf{d} = \langle -6, -3, 3 \rangle
step 3
The parametric equation of the line passing through B(1,−1,0)B(1,-1,0) is given by r(t)=r0+td\mathbf{r}(t) = \mathbf{r_0} + t\mathbf{d}, where r0=(1,−1,0)\mathbf{r_0} = (1,-1,0)
step 4
Substitute the point and direction vector: r(t)=⟨1,−1,0⟩+t⟨−6,−3,3⟩\mathbf{r}(t) = \langle 1, -1, 0 \rangle + t \langle -6, -3, 3 \rangle
step 5
Write the parametric equations: x=1−6tx = 1 - 6t, y=−1−3ty = -1 - 3t, z=3tz = 3t
Answer
x=1−6tx = 1 - 6t, y=−1−3ty = -1 - 3t, z=3tz = 3t
Key Concept
Parametric Equation of a Line
Explanation
The parametric equation of a line can be found using a point on the line and a direction vector parallel to the line.
Question 4(c)
step 1
Given points A(1,2,3)A(1,2,3), B(1,−1,0)B(1,-1,0), and D(0,1,1)D(0,1,1), we need to find the equation of the plane ABDABD
step 2
Find vectors AB\mathbf{AB} and AD\mathbf{AD}: AB=⟨1−1,−1−2,0−3⟩=⟨0,−3,−3⟩\mathbf{AB} = \langle 1-1, -1-2, 0-3 \rangle = \langle 0, -3, -3 \rangle and AD=⟨0−1,1−2,1−3⟩=⟨−1,−1,−2⟩\mathbf{AD} = \langle 0-1, 1-2, 1-3 \rangle = \langle -1, -1, -2 \rangle
step 3
The normal vector n\mathbf{n} is the cross product of AB\mathbf{AB} and AD\mathbf{AD}: n=AB×AD\mathbf{n} = \mathbf{AB} \times \mathbf{AD}
step 4
Calculate the cross product: n=⟨(−3)(−2)−(−3)(−1),(0)(−2)−(−3)(−1),(0)(−1)−(−3)(−1)⟩=⟨6−3,0−3,0−3⟩=⟨3,−3,−3⟩\mathbf{n} = \langle (-3)(-2) - (-3)(-1), (0)(-2) - (-3)(-1), (0)(-1) - (-3)(-1) \rangle = \langle 6 - 3, 0 - 3, 0 - 3 \rangle = \langle 3, -3, -3 \rangle
step 5
The equation of the plane is given by n⋅(r−r0)=0\mathbf{n} \cdot (\mathbf{r} - \mathbf{r_0}) = 0, where r0=(1,2,3)\mathbf{r_0} = (1,2,3) and r=(x,y,z)\mathbf{r} = (x,y,z)
step 6
Substitute the normal vector and point into the plane equation: 3(x−1)−3(y−2)−3(z−3)=03(x-1) - 3(y-2) - 3(z-3) = 0
step 7
Simplify the equation: 3x−3−3y+6−3z+9=0⇒3x−3y−3z+12=0⇒x−y−z+4=03x - 3 - 3y + 6 - 3z + 9 = 0 \Rightarrow 3x - 3y - 3z + 12 = 0 \Rightarrow x - y - z + 4 = 0
Answer
x−y−z+4=0x - y - z + 4 = 0
Key Concept
Equation of a Plane
Explanation
The equation of a plane can be found using a point on the plane and a normal vector, which can be derived from the cross product of two vectors lying on the plane.
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