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MAT1503 Assignment 5 (COMPLETE ANSWERS) 2024 - DUE 10 September 2024
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MAT1503 Assignment 5 (COMPLETE ANSWERS) 2024 - DUE 10 September 2024 ; 100% TRUSTED Complete, trusted solutions and explanations. Ensure your success with us ....
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MAT1503 Assignment 5 (COMPLETE ANSWERS) 2024 - DUE 10 September 2024 ; 100% TRUSTED Complete, trusted solutions and explanations. Ensure your success with us ....
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MAT1503 Assignment 5 (COMPLETE ANSWERS) 2024 - DUE 10 September 2024 ; 100% TRUSTED Complete, trusted solutions and explanations. Ensure your success with us ....
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,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)
1.1 Planes U and V
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
To determine when two planes are orthogonal, we need to
check when their normal vectors are orthogonal.
Normal Vector of Plane U: nU=(λ,5,−2λ)\mathbf{n_U} = (\
lambda, 5, -2\lambda)nU=(λ,5,−2λ)
Normal Vector of Plane V: nV=(−λ,1,2)\mathbf{n_V} = (-\
lambda, 1, 2)nV=(−λ,1,2)
,Two vectors are orthogonal if their dot product is zero. Thus, we
compute the dot product of nU\mathbf{n_U}nU and nV\
mathbf{n_V}nV:
nU⋅nV=(λ)(−λ)+(5)(1)+(−2λ)(2)\mathbf{n_U} \cdot \mathbf{n_V}
= (\lambda)(-\lambda) + (5)(1) + (-2\lambda)(2)nU⋅nV=(λ)(−λ)
+(5)(1)+(−2λ)(2) nU⋅nV=−λ2+5−4λ\mathbf{n_U} \cdot \
mathbf{n_V} = -\lambda^2 + 5 - 4\lambdanU⋅nV=−λ2+5−4λ
Set the dot product to zero for orthogonality:
−λ2+5−4λ=0-\lambda^2 + 5 - 4\lambda = 0−λ2+5−4λ=0
Rearrange:
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
Here a=1a = 1a=1, b=4b = 4b=4, and c=−5c = -5c=−5:
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
Thus:
λ=22=1orλ=−102=−5\lambda = \frac{2}{2} = 1 \quad \text{or} \
quad \lambda = \frac{-10}{2} = -5λ=22=1orλ=2−10=−5
b) Parallel Planes
, For the planes to be parallel, their normal vectors must be
scalar multiples of each other:
(λ,5,−2λ) and (−λ,1,2)(\lambda, 5, -2\lambda) \text{ and } (-\
lambda, 1, 2)(λ,5,−2λ) and (−λ,1,2)
We need to find if there exists a scalar kkk such that:
(λ,5,−2λ)=k(−λ,1,2)(\lambda, 5, -2\lambda) = k(-\lambda, 1, 2)
(λ,5,−2λ)=k(−λ,1,2)
Equating components, we get:
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 −2λ=2k-2\lambda =
2k−2λ=2k:
−2λ=2⋅5-2\lambda = 2 \cdot 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=−k(−5)-5 = -k(-5)−5=−k(−5) −5=5k-5 = 5k−5=5k k=−1k = -
1k=−1
The value λ=−5\lambda = -5λ=−5 satisfies the parallel condition
with k=−1k = -1k=−1.
1.2 Equation for Plane Parallel to Given Plane
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