Consider the following planes: 5x−4y+z=1 4x+y−5z=5 a) Find parametric equations for the line of intersection of the planes. (Use the parameter t.) b) Find the angle between the planes. (Round your answer to one decimal place.)

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Consider the following planes:   5x−4y+z=1  4x+y−5z=5  a) Find parametric equations for the line of intersection of the planes. (Use the parameter t.)  b) Find the angle between the planes. (Round your answer to one decimal place.)

Sharolyn Larson · Accepted Answer

Step 1 Consider the two given planes. 5x−4y+z=1 4x+y−5z=5 a) The objective of this part is to find the parametric equations for the line of intersection of the planes. To find the point of intersection, set z= 0 in the two given equations and solve for x and y. Step 2 1) 5x−4y=1 2) 4x+y=5 Multiply equation (1) by 1 and add with equation (2) multiplied by 4. x−4y=116x+4y=2021x=21x=1 Substitute x=1 in equation (2) to find y. 4x+y=5 4(1)+y=5 y=5−4 y=1 So, one point of intersection is ⟨1, 1, 0⟩ a1x+b1y+cz−d1=0 and a2x+b2y+cz=d2=0 Step 3 Find the cross product of the normal to the planes and it is a vector which is parallel to the line of intersection. ⟨a, b, x⟩×⟨d, e, f⟩=⟨bf−ce, cd−af, ae−bd⟩ The given planes can be written as ⟨5, −4, 1⟩ and ⟨4, 1, −5⟩ So, the cross product is calculated as, ⟨5, −4, 1⟩×⟨4, 1, −5⟩=⟨−4(−5)−1(1), 1(4)−5(−5), 5(1)−(−4(4))⟩ =⟨19, 29, 21⟩ Use the point of intersection (x0, y0, z0) and the vector obtained ⟨a, b, c⟩ to find the parametric equations. x=x0+at y=y0+bt z=z0+ct So, the parametric equations are obtained as: x=1+19t y=1+29t z=21t Step 4 b) The objective is to find the angle between the two given planes. Let θ be the angle between the planes. The formula for angle between planes a1x+b1y+c1z−d1=0 and a2x+b2y+c2z−d2=0 is defined as

Nick Camelot · Accepted Answer

a) To find the parametric equations for the line of intersection of the planes, we can set up a system of equations using the given planes:5x−4y+z=14x+y−5z=5To eliminate one variable, we can multiply the first equation by 4 and the second equation by 5, and then add them together:20x−16y+4z=420x+5y−25z=25Subtracting the first equation from the second equation, we obtain:21y−29z=21Now, we can express the variables in terms of a parameter t. Let&#039;s let y = t. Then, z can be expressed as:z=21t−2129Substituting this value of z back into the first equation, we can solve for x:x=4t+529Therefore, the parametric equations for the line of intersection of the planes are:x=4t+529,y=t,z=21t−2129b) To find the angle between the planes, we can use the formula:θ=arccos(𝐧1·𝐧2‖𝐧1‖·‖𝐧2‖)where 𝐧1 and 𝐧2 are the normal vectors to the planes. The normal vector for a plane of the form Ax+By+Cz=D is given by 𝐧=⟨A,B,C⟩.For the first plane, 5x−4y+z=1, the normal vector is 𝐧1=⟨5,−4,1⟩.For the second plane, 4x+y−5z=5, the normal vector is 𝐧2=⟨4,1,−5⟩.Calculating the dot product and the magnitudes, we have:𝐧1·𝐧2=(5)(4)+(−4)(1)+(1)(−5)=9‖𝐧1‖=52+(−4)2+12=42‖𝐧2‖=42+12+(−5)2=42Plugging these values into the formula, we get:θ=arccos(942·42)=arccos(942)Therefore, the angle between the planes is approximately arccos(942), rounded to one decimal place.

Mr Solver · Accepted Answer

Answer:(a)21t+1919(b)67.4∘Explanation:a) To find the parametric equations for the line of intersection of the planes, we can set up a system of equations using the given plane equations:5x−4y+z=1(1)4x+y−5z=5(2)To eliminate one variable, we can solve this system of equations using the method of substitution. First, let&#039;s solve equation (2) for y:y=5z−4x−5(3)Substituting equation (3) into equation (1), we have:5x−4(5z−4x−5)+z=15x−20z+16x+20+z=121x−19z=−19(4)Now, we can express x and z in terms of a parameter t:Let x=tSubstituting x=t into equation (4), we have:21t−19z=−19z=21t+1919(5)Substituting z=21t+1919 into equation (3), we have:y=5(21t+1919)−4t−5y=105t+9519−4t−5y=105t−7619(6)Therefore, the parametric equations for the line of intersection of the planes are:x=ty=105t−7619z=21t+1919b) To find the angle between the planes, we can use the dot product of their normal vectors. The normal vectors of the planes can be obtained from the coefficients of x, y, and z in the plane equations. The direction ratios of the normal vector for the first plane are (5,−4,1), and for the second plane are (4,1,−5).Let 𝐧1=⟨5,−4,1⟩ be the normal vector for the first plane, and 𝐧2=⟨4,1,−5⟩ be the normal vector for the second plane.The angle θ between the planes is given by:cos(θ)=𝐧1·𝐧2‖𝐧1‖‖𝐧2‖Calculating the dot product and the magnitudes:𝐧1·𝐧2=(5)(4)+(−4)(1)+(1)(−5)=20−4−5=11‖𝐧1‖=52+(−4)2+12=42‖𝐧2‖=42+12+(−5)2=42Substituting these values into the formula for cos(θ):cos(θ)=114242=1142Therefore, the angle between the planes is:θ=arccos(1142)≈67.4∘

Eliza Beth13 · Accepted Answer

a) The first plane equation is:5x−4y+z=1The second plane equation is:4x+y−5z=5To find the line of intersection, we can choose two variables as parameters and express the remaining variable in terms of these parameters.Let&#039;s choose y and z as parameters. We&#039;ll express x in terms of these parameters.Starting with the first plane equation:5x−4y+z=1Solving for x:x=15+45y−15zNow, substituting this expression for x into the second plane equation:4(15+45y−15z)+y−5z=5Simplifying the equation:45+165y−45z+y−5z=5Combining like terms:215y−295z=215Dividing through by 15:21y−29z=21Now, we have a system of two equations:x=15+45y−15z21y−29z=21These equations represent the parametric equations for the line of intersection of the planes.

Consider the following planes: 5x - 4y + z =

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