,MAT1503 Assignment 5 (COMPLETE ANSWERS)
2024 - DUE 10 September 2024 ; 100% TRUSTED
Complete, trusted solutions and explanations.
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 break down each part of the question step-by-step:
1.1 (a) Orthogonal Planes
To determine for which value(s) of λ\lambdaλ the planes UUU
and VVV are orthogonal, we need to check the dot product of
their normal vectors.
The planes are given by: 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
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).
Two planes are orthogonal if their normal vectors are
orthogonal. This means their dot product should be zero:
,nU⋅nV=(λ,5,−2λ)⋅(−λ,1,2)\mathbf{n}_U \cdot \mathbf{n}_V = (\
lambda, 5, -2\lambda) \cdot (-\lambda, 1, 2)nU⋅nV
=(λ,5,−2λ)⋅(−λ,1,2)
Calculate the dot product:
nU⋅nV=λ(−λ)+5⋅1+(−2λ)⋅2\mathbf{n}_U \cdot \mathbf{n}_V = \
lambda(-\lambda) + 5 \cdot 1 + (-2\lambda) \cdot 2nU⋅nV
=λ(−λ)+5⋅1+(−2λ)⋅2 =−λ2+5−4λ= -\lambda^2 + 5 - 4\
lambda=−λ2+5−4λ
Set the dot product to zero:
−λ2+5−4λ=0-\lambda^2 + 5 - 4\lambda = 0−λ2+5−4λ=0
Rearrange into standard quadratic form:
λ2+4λ−5=0\lambda^2 + 4\lambda - 5 = 0λ2+4λ−5=0
Solve this quadratic equation using the quadratic formula
λ=−b±b2−4ac2a\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}
{2a}λ=2a−b±b2−4ac:
λ=−4±16+202\lambda = \frac{-4 \pm \sqrt{16 + 20}}
{2}λ=2−4±16+20 λ=−4±362\lambda = \frac{-4 \pm \sqrt{36}}
{2}λ=2−4±36 λ=−4±62\lambda = \frac{-4 \pm 6}{2}λ=2−4±6
λ=1 or λ=−5\lambda = 1 \text{ or } \lambda = -5λ=1 or λ=−5
Thus, the planes UUU and VVV are orthogonal for λ=1\lambda =
1λ=1 and λ=−5\lambda = -5λ=−5.
1.1 (b) Parallel Planes
, Two planes are parallel if their normal vectors are parallel. This
means that one normal vector is a scalar multiple of the other.
For planes UUU and VVV:
nU=(λ,5,−2λ)\mathbf{n}_U = (\lambda, 5, -2\lambda)nU
=(λ,5,−2λ) nV=(−λ,1,2)\mathbf{n}_V = (-\lambda, 1, 2)nV
=(−λ,1,2)
We need to find λ\lambdaλ such that:
(λ,5,−2λ)=k(−λ,1,2)(\lambda, 5, -2\lambda) = k(-\lambda, 1, 2)
(λ,5,−2λ)=k(−λ,1,2)
Equate the components:
λ=−kλ\lambda = -k\lambdaλ=−kλ 5=k5 = k5=k −2λ=2k-2\lambda
= 2k−2λ=2k
From 5=k5 = k5=k, substitute kkk into the third equation:
−2λ=2×5-2\lambda = 2 \times 5−2λ=2×5 −2λ=10-2\lambda =
10−2λ=10 λ=−5\lambda = -5λ=−5
Substitute λ=−5\lambda = -5λ=−5 into λ=−kλ\lambda = -k\
lambdaλ=−kλ:
−5=−5k-5 = -5k−5=−5k k=1k = 1k=1
So, the planes UUU and VVV are parallel when λ=−5\lambda = -
5λ=−5.
1.2 Equation of a Plane Parallel to a Given Plane
2024 - DUE 10 September 2024 ; 100% TRUSTED
Complete, trusted solutions and explanations.
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 break down each part of the question step-by-step:
1.1 (a) Orthogonal Planes
To determine for which value(s) of λ\lambdaλ the planes UUU
and VVV are orthogonal, we need to check the dot product of
their normal vectors.
The planes are given by: 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
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).
Two planes are orthogonal if their normal vectors are
orthogonal. This means their dot product should be zero:
,nU⋅nV=(λ,5,−2λ)⋅(−λ,1,2)\mathbf{n}_U \cdot \mathbf{n}_V = (\
lambda, 5, -2\lambda) \cdot (-\lambda, 1, 2)nU⋅nV
=(λ,5,−2λ)⋅(−λ,1,2)
Calculate the dot product:
nU⋅nV=λ(−λ)+5⋅1+(−2λ)⋅2\mathbf{n}_U \cdot \mathbf{n}_V = \
lambda(-\lambda) + 5 \cdot 1 + (-2\lambda) \cdot 2nU⋅nV
=λ(−λ)+5⋅1+(−2λ)⋅2 =−λ2+5−4λ= -\lambda^2 + 5 - 4\
lambda=−λ2+5−4λ
Set the dot product to zero:
−λ2+5−4λ=0-\lambda^2 + 5 - 4\lambda = 0−λ2+5−4λ=0
Rearrange into standard quadratic form:
λ2+4λ−5=0\lambda^2 + 4\lambda - 5 = 0λ2+4λ−5=0
Solve this quadratic equation using the quadratic formula
λ=−b±b2−4ac2a\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}
{2a}λ=2a−b±b2−4ac:
λ=−4±16+202\lambda = \frac{-4 \pm \sqrt{16 + 20}}
{2}λ=2−4±16+20 λ=−4±362\lambda = \frac{-4 \pm \sqrt{36}}
{2}λ=2−4±36 λ=−4±62\lambda = \frac{-4 \pm 6}{2}λ=2−4±6
λ=1 or λ=−5\lambda = 1 \text{ or } \lambda = -5λ=1 or λ=−5
Thus, the planes UUU and VVV are orthogonal for λ=1\lambda =
1λ=1 and λ=−5\lambda = -5λ=−5.
1.1 (b) Parallel Planes
, Two planes are parallel if their normal vectors are parallel. This
means that one normal vector is a scalar multiple of the other.
For planes UUU and VVV:
nU=(λ,5,−2λ)\mathbf{n}_U = (\lambda, 5, -2\lambda)nU
=(λ,5,−2λ) nV=(−λ,1,2)\mathbf{n}_V = (-\lambda, 1, 2)nV
=(−λ,1,2)
We need to find λ\lambdaλ such that:
(λ,5,−2λ)=k(−λ,1,2)(\lambda, 5, -2\lambda) = k(-\lambda, 1, 2)
(λ,5,−2λ)=k(−λ,1,2)
Equate the components:
λ=−kλ\lambda = -k\lambdaλ=−kλ 5=k5 = k5=k −2λ=2k-2\lambda
= 2k−2λ=2k
From 5=k5 = k5=k, substitute kkk into the third equation:
−2λ=2×5-2\lambda = 2 \times 5−2λ=2×5 −2λ=10-2\lambda =
10−2λ=10 λ=−5\lambda = -5λ=−5
Substitute λ=−5\lambda = -5λ=−5 into λ=−kλ\lambda = -k\
lambdaλ=−kλ:
−5=−5k-5 = -5k−5=−5k k=1k = 1k=1
So, the planes UUU and VVV are parallel when λ=−5\lambda = -
5λ=−5.
1.2 Equation of a Plane Parallel to a Given Plane
