MAT1503 Assignment 5 (COMPLETE ANSWERS) 2024 - DUE 10 September 2024

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 break down the problem into its individual parts and solve each one step by step.

Question 1.1

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 their normal vectors are orthogonal, i.e., the dot product of the
normal vectors equals zero.

 The normal vector to plane UUU is nU=(λ,5,−2λ)\mathbf{n}_U = (\lambda, 5, -2\
lambda)nU=(λ,5,−2λ).
 The normal vector to plane VVV is nV=(−λ,1,2)\mathbf{n}_V = (-\lambda, 1, 2)nV
=(−λ,1,2).

The dot product nU⋅nV\mathbf{n}_U \cdot \mathbf{n}_VnU⋅nV 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) nU⋅nV=−λ2+5−4λ\mathbf{n}_U \cdot \mathbf{n}_V
= -\lambda^2 + 5 - 4\lambdanU⋅nV=−λ2+5−4λ

For the planes to be orthogonal:

−λ2−4λ+5=0-\lambda^2 - 4\lambda + 5 = 0−λ2−4λ+5=0

This is a quadratic equation. We can solve 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:

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

Thus, the solutions are:

λ1=4+6−2=10−2=−5,λ2=4−6−2=−2−2=1\lambda_1 = \frac{4 + 6}{-2} = \frac{10}{-2} = -5, \
quad \lambda_2 = \frac{4 - 6}{-2} = \frac{-2}{-2} = 1λ1=−24+6=−210=−5,λ2=−24−6=−2−2=1

So, λ=−5\lambda = -5λ=−5 or λ=1\lambda = 1λ=1 for the planes to be orthogonal.

(b) Parallel Planes

Two planes are parallel if their normal vectors are parallel, i.e., they are scalar multiples of each
other.

For planes UUU and VVV to be parallel:

(λ,5,−2λ)=k(−λ,1,2)(\lambda, 5, -2\lambda) = k(-\lambda, 1, 2)(λ,5,−2λ)=k(−λ,1,2)

Comparing the components:

λ=−kλ,5=k,−2λ=2k\lambda = -k\lambda, \quad 5 = k, \quad -2\lambda = 2kλ=−kλ,5=k,−2λ=2k

For λ≠0\lambda \neq 0λ=0, the first equation gives k=−1k = -1k=−1. Substituting into the
second equation:

5=−1(Contradiction)5 = -1 \quad (\text{Contradiction})5=−1(Contradiction)

So, λ=0\lambda = 0λ=0. In this case:

nU=(0,5,0),nV=(0,1,2)\mathbf{n}_U = (0, 5, 0), \quad \mathbf{n}_V = (0, 1, 2)nU=(0,5,0),nV
=(0,1,2)

These vectors are not multiples of each other. Hence, the planes are never parallel for any value
of λ\lambdaλ.

Question 1.2

Find an equation for the plane that passes through the origin and is parallel to the plane
−x+3y−2z=6-x + 3y - 2z = 6−x+3y−2z=6.