MAT1503 Assignment 5
(COMPLETE ANSWERS) 2024
- DUE 10 September 2024
100% GUARANTEED
,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
Let the planes UUU and VVV be given by the equations:
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 vectors to these planes are:
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)
(a) For the planes to be orthogonal:
Two planes are orthogonal if their normal vectors are perpendicular, i.e., their dot product is
zero.
The dot product of nU\mathbf{n}_UnU and nV\mathbf{n}_VnV is:
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) =−λ2+5−4λ= -\lambda^2 + 5 - 4\lambda=−λ2+5−4λ
For the planes to be orthogonal, set the dot product to zero:
−λ2−4λ+5=0-\lambda^2 - 4\lambda + 5 = 0−λ2−4λ+5=0
This is a quadratic equation in λ\lambdaλ. Solving it 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, so:
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\lambda = \frac{4 \pm \sqrt{36}}{-2}λ=−24±36 λ=4±6−2\lambda = \frac{4 \pm 6}{-
2}λ=−24±6 λ1=10−2=−5andλ2=−2−2=1\lambda_1 = \frac{10}{-2} = -5 \quad \text{and} \quad \
lambda_2 = \frac{-2}{-2} = 1λ1=−210=−5andλ2=−2−2=1
Thus, the planes are orthogonal for λ=−5\lambda = -5λ=−5 and λ=1\lambda = 1λ=1.
(b) For the planes to be parallel:
Two planes are parallel if their normal vectors are parallel, i.e., proportional.
This means there should exist 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)
From the first components:
λ=−kλ⇒k=−1\lambda = -k\lambda \Rightarrow k = -1λ=−kλ⇒k=−1
From the second components:
5=1k=k5 = 1k = k5=1k=k
So k=5k = 5k=5, which contradicts k=−1k = -1k=−1 from the first equation. Hence, the planes
UUU and VVV are never parallel for any value of λ\lambdaλ.
Question 1.2
We need to find the equation of a plane passing through the origin (0, 0, 0) and parallel to the
plane given by:
−x+3y−2z=6-x + 3y - 2z = 6−x+3y−2z=6
A plane parallel to the given plane will have the same normal vector, so the equation will be:
−x+3y−2z=d-x + 3y - 2z = d−x+3y−2z=d
Since it passes through the origin (0, 0, 0), substitute these coordinates into the equation to find
ddd:
−0+3(0)−2(0)=d-0 + 3(0) - 2(0) = d−0+3(0)−2(0)=d d=0d = 0d=0
(COMPLETE ANSWERS) 2024
- DUE 10 September 2024
100% GUARANTEED
,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
Let the planes UUU and VVV be given by the equations:
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 vectors to these planes are:
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)
(a) For the planes to be orthogonal:
Two planes are orthogonal if their normal vectors are perpendicular, i.e., their dot product is
zero.
The dot product of nU\mathbf{n}_UnU and nV\mathbf{n}_VnV is:
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) =−λ2+5−4λ= -\lambda^2 + 5 - 4\lambda=−λ2+5−4λ
For the planes to be orthogonal, set the dot product to zero:
−λ2−4λ+5=0-\lambda^2 - 4\lambda + 5 = 0−λ2−4λ+5=0
This is a quadratic equation in λ\lambdaλ. Solving it 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, so:
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\lambda = \frac{4 \pm \sqrt{36}}{-2}λ=−24±36 λ=4±6−2\lambda = \frac{4 \pm 6}{-
2}λ=−24±6 λ1=10−2=−5andλ2=−2−2=1\lambda_1 = \frac{10}{-2} = -5 \quad \text{and} \quad \
lambda_2 = \frac{-2}{-2} = 1λ1=−210=−5andλ2=−2−2=1
Thus, the planes are orthogonal for λ=−5\lambda = -5λ=−5 and λ=1\lambda = 1λ=1.
(b) For the planes to be parallel:
Two planes are parallel if their normal vectors are parallel, i.e., proportional.
This means there should exist 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)
From the first components:
λ=−kλ⇒k=−1\lambda = -k\lambda \Rightarrow k = -1λ=−kλ⇒k=−1
From the second components:
5=1k=k5 = 1k = k5=1k=k
So k=5k = 5k=5, which contradicts k=−1k = -1k=−1 from the first equation. Hence, the planes
UUU and VVV are never parallel for any value of λ\lambdaλ.
Question 1.2
We need to find the equation of a plane passing through the origin (0, 0, 0) and parallel to the
plane given by:
−x+3y−2z=6-x + 3y - 2z = 6−x+3y−2z=6
A plane parallel to the given plane will have the same normal vector, so the equation will be:
−x+3y−2z=d-x + 3y - 2z = d−x+3y−2z=d
Since it passes through the origin (0, 0, 0), substitute these coordinates into the equation to find
ddd:
−0+3(0)−2(0)=d-0 + 3(0) - 2(0) = d−0+3(0)−2(0)=d d=0d = 0d=0
