Questions test
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= c
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=c (∞)
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= a (3)
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C
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B
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C
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b
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a
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false
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false
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True
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true
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false
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r is the position of any point on line
the equation of vector line passing through (25, 17, 42) and (-16, 12, 25) can be given as
r = (25, 17, 42) + t ((-16, 12, 25) – (27, 17, 42))
r = (25, 17, 42) + t(-41, -5, -17)
therefore, the equation of vector line passing through (25, 17, 42) and (-16, 12, 25) is
= r = (25, 17, 42) + t(-41, -5, -17)
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To determine if the given lines are parallel, we need to check if their direction vectors are parallel.
The direction vector of the first line is given by the ratios of the coefficients of x, y, and z in the equation. So, direction vector of first line
= <3, 1, -5>
The direction vector of the second line is obtained by writing the equation in vector form and taking the coefficients of x, y, and z. So, direction vector of second line
= <-6, -2, 10>
If the angles between the two vectors are zero , then one is a constant multiple of the second. Since our objective is to examine if a vector is a scalar multiple or not, then we need to find out if one vector is a scalar multiple of the other.
Let’s check if we can write the second vector as a scalar multiple of the first vector:
<-6, -2, 10>
k<3, 1, -5> is an example of a vector for some scalar k.
Equating the corresponding components, we get:Equating the corresponding components, we get:
-6 = 3k
-2 = k
10 = -5k
Here Eq(1) givs k = -2, but the third Eq(3) presents k = -2, too, which obviously is an impossibility.
Accordingly, there is no scalar multiple that the two vectors may be multiplied in order to get the same lines parallel, therefore, the two lines are non-parallel.
Question
The Cartesian equation of a plane with a normal vector n = (a, b, c) ) and passing through the point ( P(x0, y0, z0) \) is given by:
a(x – x0) + b(y – y0) + c(z – z0) = 0
Substituting the given values, the equation of the plane is:
-3(x – 2) – 7(y – 6) + 8(z – 1) = 0
Simplifying:
-3x + 6 – 7y + 42 + 8z – 8 = 0
Which simplifies to:
-3x – 7y + 8z + 40 = 0
So, the Cartesian equation of the plane is:
-3x – 7y + 8z + 40 = 0
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In order to locate the intercepts of a plane, 2x+6y-3z-12 = 0, we will need to follow these steps below.
we can take a double zero between two variables and solve the third one.
Setting x=0 and y=0,
we get -3z-12=0,
so the z-intercept is (-4, 0, 0).
Setting x=0 and z=0,
we get 6y-12=0,
so the y-intercept is (0, 2, 0).
Setting \(y=0\) and \(z=0\),
we get 2x-12=0,
so the x-intercept is (6, 0, 0).
Therefore, the intercepts of the plane are (6, 0, 0), (0, 2, 0), and (-4, 0, 0).
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Since the normal vector of two planes to be perpendicular must itself be perpendicular, meaning it needs both normal vectors to be perpendicula.
The normal vector of plane π1 is
n1 = 2,-k,3 and the
normal vector of plane π2 is
n2 = 2k+3,-2,-2
Therefore, we need to find the value of \( k \) such that the dot product of these two vectors is zero:
n1.n2 = 2(2k+3) – 2(-k) – 6 = 4k + 12 + 2k – 6 = 6k + 6 = 0\]
Solving for k , we get:
k = -1
Therefore, the value of k that will make the planes π1and π2 perpendicular is
k = -1.
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To be sure if they intersect in a line, or if they are parallel and different, or parallel and coincident, we must use the normals of the planes.
The normal of a plane is the polynomial coefficients in the equation of the plane.
For plane x₁: π + 2y + 3x – 3 = 0,
the normal vector is [3, 2, 1].
For plane x₂: 2x – y + z – 7 = 0,
the normal vector is [2, -1, 1].
By the book definition, if the two planes are parallel and their normal vector are the scalar multiple of each other.
In this example, the scalar plot [3, 2, 1] and [2, -1, 1] are not scalar multiples.
as I slide down the side of the plane, meaning I will not be parallel.
Provided that the planes are compatible, the normal vectors will be identical.
The problem comes in this example because it is determined that 3, 2, 1 and 2, -1, 1 aren’t equal.
in which the lines are not coincident so.
Consequently, the line is insinuated ( showed) as the point where the two planes meet.
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The angle θ that a vector a = (x,y,z) makes with the y -axis can be found using the formula:The angle θ that a vector a = (x,y,z) makes with the y -axis can be found using the formula:
cos(θ) = a .j/ ||a|| ||j||
in which k=j is the unit vector in the positive \( y \)-direction.denotes the dot product.
In this case,
j = (0,1,0)
and ||a|| = \sqrt{12^2 + (-3)^2 + 4^2} = 13.
So, vec{a. vec{j} = 12(0) + (-3)(1) + 4(0) = -3.
Thus, cos(θ) = {-3}/{13.1 = – {3}/{13}.
Therefore;
101.5°
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In order to determine a vector which is perpendicular to both vec{a} = [1, 2, 3] and vec{b} = [4, 2, -3] the following steps should be taken.
cross product of vec{a} and vec {b} can be represented by the dot product.
The cross product of two vectors vec{a} and vec{b} is given by:
vec{a} * vec{b} = matrix (i ,1, 4), (j,2,2), (k,3-3)
Calculating the determinant, we get:
vec{a} * vec{b} = (2*-3 – 3*2)i – (1*-3 – 3*4)j + (1*2 – 2*4)k
vec{a} * vec{b} = (-12 – 6)i – (-3 – 12)j + (2 – 8)k
vec{a} * vec{b} = -18i + 15j – 6k
So, a vector perpendicular to both vec{a} and vec{b} is vec{v} = [-18, 15, -6].
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To determine if the point (1,2,3) is on the plane, we can substitute the coordinates of the point into the equation of the plane and check if it satisfies the equation.
The equation of the plane is given by:
vec{r}=(-3,4,0)+t(6,4,-2)+s(1,-4,4)
Substituting the coordinates of the point (1,2,3) into the equation:
(1,2,3)=(-3,4,0)+t(6,4,-2)+s(1,-4,4)
This gives us a system of equations:
1 = -3 + 6t + s
2 = 4 + 4t – 4s
3 = 0 – 2t + 4s
Solving this system of equations, we find that t = 1 and s = 1.
Substituting these values back into the equation,
we find that the point (1,2,3) satisfies the equation of the plane.
Therefore, the point (1,2,3) is on the plane.
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We can find the point of intersection between two lines by setting their position vectors equal to each other and solving for the parameters $s$ and $t$. That is,
\begin{align*}
(0,1,3)+s(5,-1,2) = (-1,2,9) + t(1,0,2)
{5s = t-1
-1s =2
2s +3 = 2t + 9
Solving the second equation gives s=-2.
Substituting this into the first equation gives t=5s+1=-9.
Substituting both s and t into the third equation gives
-1=1, which is a contradiction.
Therefore, the two lines do not intersect and are parallel.
Question
3x + 20y – 8z = 15
20x + 15y – 7z = 28
6x + 5y -4z = 7
We are going to substitute these equations by multiplying by 4, 5 and 20 respectively.
Therefore, we get
15x + 100y – 40z = 75
80x + 60y – 28z = 112
120x + 100y – 80z = 140
We will subtract the first equation from the second; first from third, and now we substitute the answer we get.
Therefore, the solution of the system is;
x=1, y=-1, z= 2
