i) Find the angle between the vectors sqrt2i+2j+2k\sqrt{2} \mathbf{i}+2 \mathbf{j}+2 \mathbf{k} and i+sqrt2j+sqrt2k\mathbf{i}+\sqrt{2} \mathbf{j}+\sqrt{2} \mathbf{k}.
Answer:
To find the angle between two vectors, we use the dot product formula. The dot product of two vectors A\mathbf{A} and B\mathbf{B} is given by A*B=|A||B|cos(theta)\mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos(\theta), where theta\theta is the angle between the vectors, and |A||\mathbf{A}| and |B||\mathbf{B}| are the magnitudes of the vectors.
The vectors given are A=sqrt2i+2j+2k\mathbf{A} = \sqrt{2} \mathbf{i} + 2 \mathbf{j} + 2 \mathbf{k} and B=i+sqrt2j+sqrt2k\mathbf{B} = \mathbf{i} + \sqrt{2} \mathbf{j} + \sqrt{2} \mathbf{k}.
Calculate the Dot Product A*B\mathbf{A} \cdot \mathbf{B}:
After calculating, we find that the angle theta\theta between the vectors is 00 radians, which is equivalent to 0^(@)0^\circ. This means the vectors are parallel to each other.
ii) Find the vector equation of the plane determined by the points (1,0,-1),(0,1,1)(1,0,-1),(0,1,1) and (-1,1,0)(-1,1,0).
Answer:
To find the vector equation of a plane determined by three points, we first need to find two direction vectors that lie in the plane. These can be obtained by subtracting the coordinates of the points. Let’s denote the points as A=(1,0,-1)A = (1,0,-1), B=(0,1,1)B = (0,1,1), and C=(-1,1,0)C = (-1,1,0). We can find two direction vectors, say vec(AB)\vec{AB} and vec(AC)\vec{AC}, by subtracting the coordinates of these points.
The vector vec(AB)\vec{AB} is given by:
vec(AB)=B-A=(0,1,1)-(1,0,-1)\vec{AB} = B – A = (0,1,1) – (1,0,-1)
Similarly, the vector vec(AC)\vec{AC} is:
vec(AC)=C-A=(-1,1,0)-(1,0,-1)\vec{AC} = C – A = (-1,1,0) – (1,0,-1)
Next, we find a normal vector to the plane by taking the cross product of vec(AB)\vec{AB} and vec(AC)\vec{AC}. The cross product of two vectors is perpendicular to both, and hence, will be normal to the plane.
The cross product vec(n)\vec{n} of vec(AB)\vec{AB} and vec(AC)\vec{AC} is given by:
(-1,-3,1)*(x-1,y,z+1)=0(-1, -3, 1) \cdot (x – 1, y, z + 1) = 0
Expanding this, we have:
-1(x-1)-3y+1(z+1)=0-1(x – 1) – 3y + 1(z + 1) = 0
-x+1-3y+z+1=0-x + 1 – 3y + z + 1 = 0
-x-3y+z+2=0-x – 3y + z + 2 = 0
Therefore, the vector equation of the plane is:
-x-3y+z+2=0-x – 3y + z + 2 = 0
iii) Check whether W={(x,y,z)inR^(3)∣x+y-z=0}W=\left\{(x, y, z) \in \mathbb{R}^3 \mid x+y-z=0\right\} is a subspace of R^(3)\mathbb{R}^3.
Answer:
To determine whether a set WW is a subspace of R^(3)\mathbb{R}^3, it must satisfy three conditions:
Non-emptiness: WW must contain the zero vector of R^(3)\mathbb{R}^3.
Closed under addition: For any two vectors vec(u)\vec{u} and vec(v)\vec{v} in WW, the sum vec(u)+ vec(v)\vec{u} + \vec{v} must also be in WW.
Closed under scalar multiplication: For any vector vec(u)\vec{u} in WW and any scalar cc, the product c vec(u)c\vec{u} must also be in WW.
Let’s check these conditions for W={(x,y,z)inR^(3)∣x+y-z=0}W = \left\{ (x, y, z) \in \mathbb{R}^3 \mid x + y – z = 0 \right\}.
Non-emptiness:
The zero vector in R^(3)\mathbb{R}^3 is (0,0,0)(0, 0, 0). For this vector, x=0x = 0, y=0y = 0, and z=0z = 0, so x+y-z=0+0-0=0x + y – z = 0 + 0 – 0 = 0. Therefore, the zero vector is in WW, satisfying the non-emptiness condition.
Closed under addition:
Let vec(u)=(x_(1),y_(1),z_(1))\vec{u} = (x_1, y_1, z_1) and vec(v)=(x_(2),y_(2),z_(2))\vec{v} = (x_2, y_2, z_2) be any two vectors in WW. This means x_(1)+y_(1)-z_(1)=0x_1 + y_1 – z_1 = 0 and x_(2)+y_(2)-z_(2)=0x_2 + y_2 – z_2 = 0. We need to check if their sum vec(u)+ vec(v)=(x_(1)+x_(2),y_(1)+y_(2),z_(1)+z_(2))\vec{u} + \vec{v} = (x_1 + x_2, y_1 + y_2, z_1 + z_2) also satisfies the condition of WW.
Therefore, vec(u)+ vec(v)\vec{u} + \vec{v} is in WW, satisfying the closure under addition.
Closed under scalar multiplication:
Let vec(u)=(x,y,z)\vec{u} = (x, y, z) be any vector in WW and cc be any scalar. Since vec(u)\vec{u} is in WW, x+y-z=0x + y – z = 0. We need to check if c vec(u)=(cx,cy,cz)c\vec{u} = (cx, cy, cz) also satisfies the condition of WW.
For c vec(u)c\vec{u}, we have:
cx+cy-cz=c(x+y-z)=c*0=0cx + cy – cz = c(x + y – z) = c \cdot 0 = 0
Therefore, c vec(u)c\vec{u} is in WW, satisfying the closure under scalar multiplication.
Since WW satisfies all three conditions, it is a subspace of R^(3)\mathbb{R}^3.
iv) Check whether the set of vectors {1+x,x+x^(2),1+x^(3)}\left\{1+x, x+x^2, 1+x^3\right\} is a linearly independent set of vectors in P_(3)\mathbf{P}_3, the vector space of polynomials of degree <= 3\leq 3.
Answer:
To determine whether the set of vectors {1+x,x+x^(2),1+x^(3)}\{1+x, x+x^2, 1+x^3\} is linearly independent in the vector space P_(3)\mathbf{P}_3 (the space of polynomials of degree <= 3\leq 3), we need to check if the only solution to the linear combination equation
For this polynomial to be identically zero, each coefficient must be zero. Therefore, we have the system of equations:
c_(1)+c_(3)=0c_1 + c_3 = 0
c_(1)+c_(2)=0c_1 + c_2 = 0
c_(2)=0c_2 = 0
c_(3)=0c_3 = 0
From equation 3, we immediately have c_(2)=0c_2 = 0. Substituting this into equation 2 gives c_(1)=0c_1 = 0. Finally, substituting c_(1)=0c_1 = 0 into equation 1 gives c_(3)=0c_3 = 0.
Since the only solution to the system is c_(1)=c_(2)=c_(3)=0c_1 = c_2 = c_3 = 0, the set of vectors {1+x,x+x^(2),1+x^(3)}\{1+x, x+x^2, 1+x^3\} is linearly independent in P_(3)\mathbf{P}_3.
v) Check whether T:R^(2)rarrR^(2)T: \mathbb{R}^2 \rightarrow \mathbb{R}^2, defined by T(x,y)=(-y,x)T(x, y)=(-y, x) is a linear transformation.
Answer:
To determine whether a function T:R^(2)rarrR^(2)T: \mathbb{R}^2 \rightarrow \mathbb{R}^2 is a linear transformation, it must satisfy two properties for all vectors vec(u), vec(v)inR^(2)\vec{u}, \vec{v} \in \mathbb{R}^2 and any scalar cc:
Let’s check these properties for TT defined by T(x,y)=(-y,x)T(x, y) = (-y, x).
Additivity:
Let vec(u)=(x_(1),y_(1))\vec{u} = (x_1, y_1) and vec(v)=(x_(2),y_(2))\vec{v} = (x_2, y_2) be any vectors in R^(2)\mathbb{R}^2. We need to check if T( vec(u)+ vec(v))=T( vec(u))+T( vec(v))T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v}).
Since T( vec(u)+ vec(v))=T( vec(u))+T( vec(v))T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v}), the additivity property is satisfied.
Homogeneity:
Let vec(u)=(x,y)\vec{u} = (x, y) be any vector in R^(2)\mathbb{R}^2 and cc be any scalar. We need to check if T(c vec(u))=cT( vec(u))T(c\vec{u}) = cT(\vec{u}).
First, calculate c vec(u)c\vec{u}:
c vec(u)=c(x,y)=(cx,cy)c\vec{u} = c(x, y) = (cx, cy)