4. find the area of the parallelogram with sides $\vec{u}$ and $\vec{v}$:
(a) $\vec{u} = \langle 3, 1 \
angle$, $\vec{v} = \langle 2, 4 \
angle$.
(b) $\vec{u} = \langle 1, -1 \
angle$, $\vec{v} = \langle 2, 3 \
angle$.
(c) $\vec{u} = \langle 6, 5 \
angle$, $\vec{v} = \langle -3, -4 \
angle$.
(d) $\vec{u} = \langle 1, 2, 3 \
angle$, $\vec{v} = \langle 4, -1, -2 \
angle$.
(e) $\vec{u} = \langle 1, 1, 0 \
angle$, $\vec{v} = \langle 0, -1, 2 \
angle$.
(f) $\vec{u} = \langle 2, 0, 0 \
angle$, $\vec{v} = \langle 4, 5, 6 \
angle$.

Question

Accepted Answer

### Part (a)
# Explanation:
## Step1: Recall the formula for the area of a parallelogram with vectors $\vec{u}=\langle u_1, u_2
angle$ and $\vec{v}=\langle v_1, v_2
angle$ in 2D. The area is the absolute value of the determinant of the matrix formed by these vectors as columns (or rows), i.e., $|\det\begin{pmatrix}u_1&v_1\u_2&v_2\end{pmatrix}| = |u_1v_2 - u_2v_1|$.
For $\vec{u}=\langle 3,1
angle$ and $\vec{v}=\langle 2,4
angle$, the determinant is $3\times4 - 1\times2$.
## Step2: Calculate the determinant.
$3\times4 - 1\times2 = 12 - 2 = 10$. The absolute value of 10 is 10.
# Answer:
10

### Part (b)
# Explanation:
## Step1: Use the 2D area formula $|\det\begin{pmatrix}u_1&v_1\u_2&v_2\end{pmatrix}| = |u_1v_2 - u_2v_1|$ for $\vec{u}=\langle 1,-1
angle$ and $\vec{v}=\langle 2,3
angle$.
The determinant is $1\times3 - (-1)\times2$.
## Step2: Compute the determinant.
$1\times3 - (-1)\times2 = 3 + 2 = 5$. The absolute value is 5.
# Answer:
5

### Part (c)
# Explanation:
## Step1: Apply the 2D area formula $|\det\begin{pmatrix}u_1&v_1\u_2&v_2\end{pmatrix}| = |u_1v_2 - u_2v_1|$ for $\vec{u}=\langle 6,5
angle$ and $\vec{v}=\langle -3,-4
angle$.
The determinant is $6\times(-4) - 5\times(-3)$.
## Step2: Calculate the determinant.
$6\times(-4) - 5\times(-3)= -24 + 15 = -9$. The absolute value is 9.
# Answer:
9

### Part (d)
# Explanation:
## Step1: For 3D vectors $\vec{u}=\langle u_1, u_2, u_3
angle$ and $\vec{v}=\langle v_1, v_2, v_3
angle$, the area of the parallelogram is the magnitude of the cross product $\vec{u}\times\vec{v}$. The cross product formula is $\vec{u}\times\vec{v}=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\u_1&u_2&u_3\v_1&v_2&v_3\end{vmatrix}=\vec{i}(u_2v_3 - u_3v_2)-\vec{j}(u_1v_3 - u_3v_1)+\vec{k}(u_1v_2 - u_2v_1)$. Then the magnitude is $\sqrt{(u_2v_3 - u_3v_2)^2+(u_3v_1 - u_1v_3)^2+(u_1v_2 - u_2v_1)^2}$.
For $\vec{u}=\langle 1,2,3
angle$ and $\vec{v}=\langle 4,-1,-2
angle$, compute the cross product components:
- $i$-component: $2\times(-2)-3\times(-1)= -4 + 3 = -1$
- $j$-component: $-(1\times(-2)-3\times4)= -(-2 - 12)=14$
- $k$-component: $1\times(-1)-2\times4= -1 - 8 = -9$
## Step2: Calculate the magnitude of the cross product.
The magnitude is $\sqrt{(-1)^2 + 14^2 + (-9)^2}=\sqrt{1 + 196 + 81}=\sqrt{278}$.
# Answer:
$\sqrt{278}$

### Part (e)
# Explanation:
## Step1: Use the 3D cross product magnitude formula. For $\vec{u}=\langle 1,1,0
angle$ and $\vec{v}=\langle 0,-1,2
angle$, compute the cross product components:
- $i$-component: $1\times2 - 0\times(-1)= 2 - 0 = 2$
- $j$-component: $-(1\times2 - 0\times0)= -2$
- $k$-component: $1\times(-1)-1\times0= -1 - 0 = -1$
## Step2: Find the magnitude of the cross product.
The magnitude is $\sqrt{2^2 + (-2)^2 + (-1)^2}=\sqrt{4 + 4 + 1}=\sqrt{9}=3$.
# Answer:
3

### Part (f)
# Explanation:
## Step1: Apply the 3D cross product magnitude formula. For $\vec{u}=\langle 2,0,0
angle$ and $\vec{v}=\langle 4,5,6
angle$, compute the cross product components:
- $i$-component: $0\times6 - 0\times5= 0 - 0 = 0$
- $j$-component: $-(2\times6 - 0\times4)= -12$
- $k$-component: $2\times5 - 0\times4= 10 - 0 = 10$
## Step2: Calculate the magnitude of the cross product.
The magnitude is $\sqrt{0^2 + (-12)^2 + 10^2}=\sqrt{0 + 144 + 100}=\sqrt{244}=2\sqrt{61}$.
# Answer:
$2\sqrt{61}$

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QUESTION IMAGE

Question

Explanation:

Part (a)

Step1: Recall the formula for the area of a parallelogram with vectors \(\vec{u}=\langle u_1, u_2

Step2: Calculate the determinant.

Step1: Use the 2D area formula \(|\det

Step2: Compute the determinant.

Step1: Apply the 2D area formula \(|\det

Step2: Calculate the determinant.

Answer:

Part (b)