MAT1503 Assignment 5
(COMPLETE ANSWERS) 2024
- DUE 10 September 2024
CONTACT:
,MAT1503 Assignment 5 (COMPLETE ANSWERS) 2024 -
DUE 10 September 2024
Question 1: 12 Marks (1.1) Let U and V be the planes
given by: (2) U : λx + 5y − 2λz − 3 = 0, V : −λx + y + 2z
+ 1 = 0. Determine for which value(s) of λ the planes U
and V are: (a) orthogonal, (2) (b) Parallel. (2) (1.2) Find an
equation for the plane that passes through the origin (0,
0, 0) and is parallel to the (3) plane −x + 3y − 2z = 6.
(1.3) Find the distance between the point (−1,−2, 0) and
the plane 3x − y + 4z = −2. (3)
Let's go through each part of the question step by step.
Question 1.1: Determining the Value(s) of λ for the Planes U and V
Given the planes:
U:λx+5y−2λz−3=0U: \lambda x + 5y - 2\lambda z - 3 = 0U:λx+5y−2λz−3=0
V:−λx+y+2z+1=0V: -\lambda x + y + 2z + 1 = 0V:−λx+y+2z+1=0
(a) Orthogonal Planes:
Two planes are orthogonal if the dot product of their normal vectors is zero.
The normal vector of plane UUU is nU=(λ,5,−2λ)\mathbf{n_U} = (\lambda, 5, -2\
lambda)nU=(λ,5,−2λ).
The normal vector of plane VVV is nV=(−λ,1,2)\mathbf{n_V} = (-\lambda, 1, 2)nV
=(−λ,1,2).
The dot product of these vectors is:
nU⋅nV=λ(−λ)+5(1)+(−2λ)(2)=−λ2+5−4λ\mathbf{n_U} \cdot \mathbf{n_V} = \lambda(-\lambda)
+ 5(1) + (-2\lambda)(2) = -\lambda^2 + 5 - 4\lambdanU⋅nV=λ(−λ)+5(1)+(−2λ)(2)=−λ2+5−4λ
For the planes to be orthogonal:
−λ2−4λ+5=0-\lambda^2 - 4\lambda + 5 = 0−λ2−4λ+5=0
This quadratic equation can be solved using the quadratic formula:
, λ=−(−4)±(−4)2−4(−1)(5)2(−1)\lambda = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(-1)(5)}}{2(-
1)}λ=2(−1)−(−4)±(−4)2−4(−1)(5) λ=4±16+20−2\lambda = \frac{4 \pm \sqrt{16 + 20}}{-
2}λ=−24±16+20 λ=4±36−2=4±6−2\lambda = \frac{4 \pm \sqrt{36}}{-2} = \frac{4 \pm 6}{-
2}λ=−24±36=−24±6 λ=10−2=−5orλ=−2−2=1\lambda = \frac{10}{-2} = -5 \quad \text{or} \quad
\lambda = \frac{-2}{-2} = 1λ=−210=−5orλ=−2−2=1
So, the planes are orthogonal when λ=−5\lambda = -5λ=−5 or λ=1\lambda = 1λ=1.
(b) Parallel Planes:
Two planes are parallel if their normal vectors are scalar multiples of each other.
We check if:
(λ,5,−2λ)=k(−λ,1,2)(\lambda, 5, -2\lambda) = k(-\lambda, 1, 2)(λ,5,−2λ)=k(−λ,1,2)
For some scalar kkk, this gives us three equations:
λ=−kλ,5=k,−2λ=2k\lambda = -k\lambda, \quad 5 = k, \quad -2\lambda = 2kλ=−kλ,5=k,−2λ=2k
From the first equation:
k=−1k = -1k=−1
From the second equation:
k=5k = 5k=5
These two equations give a contradiction, so there are no values of λ\lambdaλ that make the
planes parallel.
Question 1.2: Equation of the Plane Passing Through the Origin and Parallel to
the Given Plane
The equation of a plane parallel to −x+3y−2z=6-x + 3y - 2z = 6−x+3y−2z=6 and passing through
the origin will have the same normal vector, so the equation will be:
−1(x−0)+3(y−0)−2(z−0)=0-1(x - 0) + 3(y - 0) - 2(z - 0) = 0−1(x−0)+3(y−0)−2(z−0)=0
Simplifying, we get:
−x+3y−2z=0-x + 3y - 2z = 0−x+3y−2z=0
Question 1.3: Distance Between the Point and the Plane
(COMPLETE ANSWERS) 2024
- DUE 10 September 2024
CONTACT:
,MAT1503 Assignment 5 (COMPLETE ANSWERS) 2024 -
DUE 10 September 2024
Question 1: 12 Marks (1.1) Let U and V be the planes
given by: (2) U : λx + 5y − 2λz − 3 = 0, V : −λx + y + 2z
+ 1 = 0. Determine for which value(s) of λ the planes U
and V are: (a) orthogonal, (2) (b) Parallel. (2) (1.2) Find an
equation for the plane that passes through the origin (0,
0, 0) and is parallel to the (3) plane −x + 3y − 2z = 6.
(1.3) Find the distance between the point (−1,−2, 0) and
the plane 3x − y + 4z = −2. (3)
Let's go through each part of the question step by step.
Question 1.1: Determining the Value(s) of λ for the Planes U and V
Given the planes:
U:λx+5y−2λz−3=0U: \lambda x + 5y - 2\lambda z - 3 = 0U:λx+5y−2λz−3=0
V:−λx+y+2z+1=0V: -\lambda x + y + 2z + 1 = 0V:−λx+y+2z+1=0
(a) Orthogonal Planes:
Two planes are orthogonal if the dot product of their normal vectors is zero.
The normal vector of plane UUU is nU=(λ,5,−2λ)\mathbf{n_U} = (\lambda, 5, -2\
lambda)nU=(λ,5,−2λ).
The normal vector of plane VVV is nV=(−λ,1,2)\mathbf{n_V} = (-\lambda, 1, 2)nV
=(−λ,1,2).
The dot product of these vectors is:
nU⋅nV=λ(−λ)+5(1)+(−2λ)(2)=−λ2+5−4λ\mathbf{n_U} \cdot \mathbf{n_V} = \lambda(-\lambda)
+ 5(1) + (-2\lambda)(2) = -\lambda^2 + 5 - 4\lambdanU⋅nV=λ(−λ)+5(1)+(−2λ)(2)=−λ2+5−4λ
For the planes to be orthogonal:
−λ2−4λ+5=0-\lambda^2 - 4\lambda + 5 = 0−λ2−4λ+5=0
This quadratic equation can be solved using the quadratic formula:
, λ=−(−4)±(−4)2−4(−1)(5)2(−1)\lambda = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(-1)(5)}}{2(-
1)}λ=2(−1)−(−4)±(−4)2−4(−1)(5) λ=4±16+20−2\lambda = \frac{4 \pm \sqrt{16 + 20}}{-
2}λ=−24±16+20 λ=4±36−2=4±6−2\lambda = \frac{4 \pm \sqrt{36}}{-2} = \frac{4 \pm 6}{-
2}λ=−24±36=−24±6 λ=10−2=−5orλ=−2−2=1\lambda = \frac{10}{-2} = -5 \quad \text{or} \quad
\lambda = \frac{-2}{-2} = 1λ=−210=−5orλ=−2−2=1
So, the planes are orthogonal when λ=−5\lambda = -5λ=−5 or λ=1\lambda = 1λ=1.
(b) Parallel Planes:
Two planes are parallel if their normal vectors are scalar multiples of each other.
We check if:
(λ,5,−2λ)=k(−λ,1,2)(\lambda, 5, -2\lambda) = k(-\lambda, 1, 2)(λ,5,−2λ)=k(−λ,1,2)
For some scalar kkk, this gives us three equations:
λ=−kλ,5=k,−2λ=2k\lambda = -k\lambda, \quad 5 = k, \quad -2\lambda = 2kλ=−kλ,5=k,−2λ=2k
From the first equation:
k=−1k = -1k=−1
From the second equation:
k=5k = 5k=5
These two equations give a contradiction, so there are no values of λ\lambdaλ that make the
planes parallel.
Question 1.2: Equation of the Plane Passing Through the Origin and Parallel to
the Given Plane
The equation of a plane parallel to −x+3y−2z=6-x + 3y - 2z = 6−x+3y−2z=6 and passing through
the origin will have the same normal vector, so the equation will be:
−1(x−0)+3(y−0)−2(z−0)=0-1(x - 0) + 3(y - 0) - 2(z - 0) = 0−1(x−0)+3(y−0)−2(z−0)=0
Simplifying, we get:
−x+3y−2z=0-x + 3y - 2z = 0−x+3y−2z=0
Question 1.3: Distance Between the Point and the Plane
