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MAT1503 Assignment 5 (COMPLETE ANSWERS) 2024 - DUE 10 September 2024
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- Linear Algebra I
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- University Of South Africa
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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 questions and solve each part.
Question 1.1
Given the planes UUU and VVV:
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) Determine the value(s) of λ\lambdaλ for which the planes UUU and VVV are
orthogonal.
The planes are orthogonal if the dot product of their normal vectors is zero.
The normal vector for UUU is nU=(λ,5,−2λ)\mathbf{n}_U = (\lambda, 5, -2\lambda)nU
=(λ,5,−2λ).
The normal vector for 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) =−λ2+5−4λ= -\lambda^2 + 5 - 4\lambda=−λ2+5−4λ
Setting the dot product to zero for orthogonality:
−λ2−4λ+5=0-\lambda^2 - 4\lambda + 5 = 0−λ2−4λ+5=0
This is a quadratic equation. Solving for λ\lambdaλ:
,λ=−(−4)±(−4)2−4(−1)(5)2(−1)=4±16+20−2=4±36−2=4±6−2\lambda = \frac{-(-4) \pm \sqrt{(-
4)^2 - 4(-1)(5)}}{2(-1)} = \frac{4 \pm \sqrt{16 + 20}}{-2} = \frac{4 \pm \sqrt{36}}{-2} = \
frac{4 \pm 6}{-2}λ=2(−1)−(−4)±(−4)2−4(−1)(5)=−24±16+20=−24±36=−24±6
So,
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
The planes UUU and VVV are orthogonal when λ=1\lambda = 1λ=1 or λ=−5\lambda = -5λ=−5.
(b) Determine the value(s) of λ\lambdaλ for which the planes UUU and VVV are parallel.
The planes are parallel if their normal vectors are scalar multiples of each other. This implies:
(λ,5,−2λ)=k(−λ,1,2)(\lambda, 5, -2\lambda) = k(-\lambda, 1, 2)(λ,5,−2λ)=k(−λ,1,2)
This gives us the following system of equations:
λ=−kλ(1)\lambda = -k\lambda \quad \text{(1)}λ=−kλ(1) 5=k(2)5 = k \quad \text{(2)}5=k(2)
−2λ=2k(3)-2\lambda = 2k \quad \text{(3)}−2λ=2k(3)
From equation (2), k=5k = 5k=5. Substituting this into equation (1):
λ=−5λ\lambda = -5\lambdaλ=−5λ λ+5λ=0 ⟹ λ=0\lambda + 5\lambda = 0 \implies \lambda =
0λ+5λ=0⟹λ=0
So, the planes UUU and VVV are parallel when λ=0\lambda = 0λ=0.
Question 1.2
Find an equation for the plane that passes through the origin (0,0,0)(0, 0, 0)(0,0,0) and is parallel
to the plane −x+3y−2z=6-x + 3y - 2z = 6−x+3y−2z=6.
If the plane passes through the origin and is parallel to the given plane, its equation will have the
same normal vector as the given plane but no constant term. Thus, the equation is:
−x+3y−2z=0-x + 3y - 2z = 0−x+3y−2z=0
Question 1.3
Find the distance between the point (−1,−2,0)(-1, -2, 0)(−1,−2,0) and the plane 3x−y+4z=−23x -
y + 4z = -23x−y+4z=−2.
The distance ddd from a point (x1,y1,z1)(x_1, y_1, z_1)(x1,y1,z1) to a plane ax+by+cz+d=0ax +
by + cz + d = 0ax+by+cz+d=0 is given by:
, d=∣ax1+by1+cz1+d∣a2+b2+c2d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 +
c^2}}d=a2+b2+c2∣ax1+by1+cz1+d∣
Substituting the values:
a=3, b=−1, c=4, d=2 (Note: shift −2 to the left side)a = 3, \, b = -1, \, c = 4, \, d = 2 \, (\text{Note:
shift } -2 \text{ to the left side})a=3,b=−1,c=4,d=2(Note: shift −2 to the left side) x1=−1, y1=−2,
z1=0x_1 = -1, \, y_1 = -2, \, z_1 = 0x1=−1,y1=−2,z1=0 d=∣3(−1)−1(−2)+4(0)+2∣32+
(−1)2+42=∣−3+2+2∣9+1+16=∣1∣26=126d = \frac{|3(-1) - 1(-2) + 4(0) + 2|}{\sqrt{3^2 + (-1)^2 +
4^2}} = \frac{|-3 + 2 + 2|}{\sqrt{9 + 1 + 16}} = \frac{|1|}{\sqrt{26}} = \frac{1}{\
sqrt{26}}d=32+(−1)2+42∣3(−1)−1(−2)+4(0)+2∣=9+1+16∣−3+2+2∣=26∣1∣=261
Thus, the distance between the point (−1,−2,0)(-1, -2, 0)(−1,−2,0) and the plane 3x−y+4z=−23x -
y + 4z = -23x−y+4z=−2 is 126\frac{1}{\sqrt{26}}261 units.
Question 2: 11 Marks (2.1) Find the angle between the
two vectors ⃗v = ⟨−1, 1, 0,−1⟩ ⃗v = ⟨1,−1, 3,−2⟩. Determine
(3) whether both vectors are perpendicular, parallel or
neither. (2.2) Find the direction cosines and the direction
angles for the vector ⃗r = ⟨0,−1,−2, 3 (3) 4 ⟩. (2.3)
HMW:Additional Exercises. Let ⃗r (t) = ⟨t,−1t , t2 − 2⟩.
Evaluate the derivative of ⃗r (t)|t=1 . Calculate the
derivative of V(t) · ⃗r (t) whenever V(1) = ⟨−1, 1,−3⟩ and V
′(1) = ⟨1,−2, 5⟩. (2.4) HMW:Additional Exercises. Assume
that a wagon is pulled horizontally by an exercising force
of 5 lb on the handle at an angle of 45 with the
horizontal. (a) Illustrate the problem using a rough sketch.
(b) Determine the amount of work done in moving the
wagon 30 lb. 36 MAT1503/101/0/2024 (2.5)
HMW:Additional Exercises. Let the vector ⃗v = ⟨3500,
4250⟩ gives the number of units of two models of solar
lamps fabricated by electronics company. Assume that
the vector ⃗a = ⟨1008.00, 699, 99⟩ gives the prices (in
Rand-ZA) of the two models of solar lamps, respectively.
(a) Calculate the dot product of the two vectors ⃗a and ⃗v.
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