Math / Geometry

QuestionLet $P=(0,0,0), Q=(1,-1,-1), R=(-2,1,1)$ . Find the area of the triangle $P Q R$ . area $=$ $\square$ Let $T=(4,4,1), U=(9,7,7), V=(-6,7,1)$ . Find the area of the triangle $T U V$ . area = $\square$

Studdy Solution

STEP 1

What is this asking? We need to find the area of two triangles in 3D space, given the coordinates of their vertices. Watch out! Remember that the area of a triangle in 3D space is half the magnitude of the cross product of two of its sides.
Don't forget to take the magnitude after calculating the cross product!

STEP 2

1. Find the area of triangle PQR
2. Find the area of triangle TUV

STEP 3

Let's define the vectors that represent the sides of triangle $PQR$ .
We can do this by subtracting the coordinates of the vertices.
We'll call these vectors $\vec{PQ}$ and $\vec{PR}$ .

STEP 4

Calculate $\vec{PQ}$ : $\vec{PQ} = Q - P = (1-0, -1-0, -1-0) = (1, -1, -1)$

STEP 5

Calculate $\vec{PR}$ : $\vec{PR} = R - P = (-2-0, 1-0, 1-0) = (-2, 1, 1)$

STEP 6

Now, let's compute the cross product of $\vec{PQ}$ and $\vec{PR}$ , which we'll call $\vec{N}$ .
The cross product gives us a vector perpendicular to both $\vec{PQ}$ and $\vec{PR}$ , and its magnitude is twice the area of the triangle. $\vec{N} = \vec{PQ} \times \vec{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & -1 \\ -2 & 1 & 1 \end{vmatrix} = (0)\mathbf{i} - (3)\mathbf{j} + (-1)\mathbf{k} = (0, -3, -1)$

STEP 7

The magnitude of $\vec{N}$ is given by: $|\vec{N}| = \sqrt{0^2 + (-3)^2 + (-1)^2} = \sqrt{0 + 9 + 1} = \sqrt{10}$

STEP 8

Finally, the area of triangle $PQR$ is half the magnitude of the cross product: $\text{Area}(PQR) = \frac{1}{2} |\vec{N}| = \frac{1}{2} \sqrt{10}$

STEP 9

Let's define the vectors representing the sides of triangle $TUV$ , $\vec{TU}$ and $\vec{TV}$ .

STEP 10

Calculate $\vec{TU}$ : $\vec{TU} = U - T = (9-4, 7-4, 7-1) = (5, 3, 6)$

STEP 11

Calculate $\vec{TV}$ : $\vec{TV} = V - T = (-6-4, 7-4, 1-1) = (-10, 3, 0)$

STEP 12

Compute the cross product $\vec{M} = \vec{TU} \times \vec{TV}$ : $\vec{M} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5 & 3 & 6 \\ -10 & 3 & 0 \end{vmatrix} = (0-18)\mathbf{i} - (0+60)\mathbf{j} + (15+30)\mathbf{k} = (-18, -60, 45)$

STEP 13

Calculate the magnitude of $\vec{M}$ : $|\vec{M}| = \sqrt{(-18)^2 + (-60)^2 + 45^2} = \sqrt{324 + 3600 + 2025} = \sqrt{5949} = 3 \sqrt{661}$ Since $3 \cdot 25.7 = 77.1$ and $\sqrt{661} \approx 25.7$ , we have $|\vec{M}| = 77.1$ .

STEP 14

The area of triangle $TUV$ is: $\text{Area}(TUV) = \frac{1}{2} |\vec{M}| = \frac{1}{2} \cdot 3\sqrt{661} = \frac{3}{2}\sqrt{661}$

STEP 15

Area of triangle $PQR = \frac{\sqrt{10}}{2}$ . Area of triangle $TUV = \frac{3\sqrt{661}}{2}$ .

Was this helpful?

QuestionLet $P=(0,0,0), Q=(1,-1,-1), R=(-2,1,1)$ . Find the area of the triangle $P Q R$ . area $=$ $\square$ Let $T=(4,4,1), U=(9,7,7), V=(-6,7,1)$ . Find the area of the triangle $T U V$ . area = $\square$

Studdy Solution

STEP 1

STEP 2

1. Find the area of triangle PQR
2. Find the area of triangle TUV

STEP 3

Let's define the vectors that represent the sides of triangle $PQR$ .
We can do this by subtracting the coordinates of the vertices.
We'll call these vectors $\vec{PQ}$ and $\vec{PR}$ .

STEP 4

Calculate $\vec{PQ}$ : $\vec{PQ} = Q - P = (1-0, -1-0, -1-0) = (1, -1, -1)$

STEP 5

Calculate $\vec{PR}$ : $\vec{PR} = R - P = (-2-0, 1-0, 1-0) = (-2, 1, 1)$

STEP 6

STEP 7

The magnitude of $\vec{N}$ is given by: $|\vec{N}| = \sqrt{0^2 + (-3)^2 + (-1)^2} = \sqrt{0 + 9 + 1} = \sqrt{10}$

STEP 8

Finally, the area of triangle $PQR$ is half the magnitude of the cross product: $\text{Area}(PQR) = \frac{1}{2} |\vec{N}| = \frac{1}{2} \sqrt{10}$

STEP 9

Let's define the vectors representing the sides of triangle $TUV$ , $\vec{TU}$ and $\vec{TV}$ .

STEP 10

Calculate $\vec{TU}$ : $\vec{TU} = U - T = (9-4, 7-4, 7-1) = (5, 3, 6)$

STEP 11

Calculate $\vec{TV}$ : $\vec{TV} = V - T = (-6-4, 7-4, 1-1) = (-10, 3, 0)$

STEP 12

Compute the cross product $\vec{M} = \vec{TU} \times \vec{TV}$ : $\vec{M} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5 & 3 & 6 \\ -10 & 3 & 0 \end{vmatrix} = (0-18)\mathbf{i} - (0+60)\mathbf{j} + (15+30)\mathbf{k} = (-18, -60, 45)$

STEP 13

Calculate the magnitude of $\vec{M}$ : $|\vec{M}| = \sqrt{(-18)^2 + (-60)^2 + 45^2} = \sqrt{324 + 3600 + 2025} = \sqrt{5949} = 3 \sqrt{661}$ Since $3 \cdot 25.7 = 77.1$ and $\sqrt{661} \approx 25.7$ , we have $|\vec{M}| = 77.1$ .

STEP 14

The area of triangle $TUV$ is: $\text{Area}(TUV) = \frac{1}{2} |\vec{M}| = \frac{1}{2} \cdot 3\sqrt{661} = \frac{3}{2}\sqrt{661}$

STEP 15

Area of triangle $PQR = \frac{\sqrt{10}}{2}$ . Area of triangle $TUV = \frac{3\sqrt{661}}{2}$ .

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

STEP 2

1. Find the area of triangle PQR2. Find the area of triangle TUV

STEP 3

Let's **define** the vectors that represent the sides of triangle PQRPQRPQR.We can do this by subtracting the coordinates of the vertices.We'll call these vectors PQ⃗\vec{PQ}PQ​ and PR⃗\vec{PR}PR.

STEP 4

**Calculate** PQ⃗\vec{PQ}PQ​: PQ⃗=Q−P=(1−0,−1−0,−1−0)=(1,−1,−1) \vec{PQ} = Q - P = (1-0, -1-0, -1-0) = (1, -1, -1) PQ​=Q−P=(1−0,−1−0,−1−0)=(1,−1,−1)

STEP 5

**Calculate** PR⃗\vec{PR}PR: PR⃗=R−P=(−2−0,1−0,1−0)=(−2,1,1) \vec{PR} = R - P = (-2-0, 1-0, 1-0) = (-2, 1, 1) PR=R−P=(−2−0,1−0,1−0)=(−2,1,1)

STEP 6

STEP 7

The **magnitude** of N⃗\vec{N}N is given by: ∣N⃗∣=02+(−3)2+(−1)2=0+9+1=10 |\vec{N}| = \sqrt{0^2 + (-3)^2 + (-1)^2} = \sqrt{0 + 9 + 1} = \sqrt{10} ∣N∣=02+(−3)2+(−1)2​=0+9+1​=10​

STEP 8

Finally, the **area** of triangle PQRPQRPQR is half the magnitude of the cross product: Area(PQR)=12∣N⃗∣=1210 \text{Area}(PQR) = \frac{1}{2} |\vec{N}| = \frac{1}{2} \sqrt{10} Area(PQR)=21​∣N∣=21​10​

STEP 9

Let's **define** the vectors representing the sides of triangle TUVTUVTUV, TU⃗\vec{TU}TU and TV⃗\vec{TV}TV.

STEP 10

**Calculate** TU⃗\vec{TU}TU: TU⃗=U−T=(9−4,7−4,7−1)=(5,3,6) \vec{TU} = U - T = (9-4, 7-4, 7-1) = (5, 3, 6) TU=U−T=(9−4,7−4,7−1)=(5,3,6)

STEP 11

**Calculate** TV⃗\vec{TV}TV: TV⃗=V−T=(−6−4,7−4,1−1)=(−10,3,0) \vec{TV} = V - T = (-6-4, 7-4, 1-1) = (-10, 3, 0) TV=V−T=(−6−4,7−4,1−1)=(−10,3,0)

STEP 12

STEP 13

STEP 14

The **area** of triangle TUVTUVTUV is: Area(TUV)=12∣M⃗∣=12⋅3661=32661 \text{Area}(TUV) = \frac{1}{2} |\vec{M}| = \frac{1}{2} \cdot 3\sqrt{661} = \frac{3}{2}\sqrt{661} Area(TUV)=21​∣M∣=21​⋅3661​=23​661​

STEP 15

Area of triangle PQR=102PQR = \frac{\sqrt{10}}{2}PQR=210​​. Area of triangle TUV=36612TUV = \frac{3\sqrt{661}}{2}TUV=23661​​.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Find the area of triangle PQR
2. Find the area of triangle TUV

Let's define the vectors that represent the sides of triangle $PQR$ .
We can do this by subtracting the coordinates of the vertices.
We'll call these vectors $\vec{PQ}$ and $\vec{PR}$ .

Calculate $\vec{PQ}$ : $\vec{PQ} = Q - P = (1-0, -1-0, -1-0) = (1, -1, -1)$

Calculate $\vec{PR}$ : $\vec{PR} = R - P = (-2-0, 1-0, 1-0) = (-2, 1, 1)$

The magnitude of $\vec{N}$ is given by: $|\vec{N}| = \sqrt{0^2 + (-3)^2 + (-1)^2} = \sqrt{0 + 9 + 1} = \sqrt{10}$

Finally, the area of triangle $PQR$ is half the magnitude of the cross product: $\text{Area}(PQR) = \frac{1}{2} |\vec{N}| = \frac{1}{2} \sqrt{10}$

Let's define the vectors representing the sides of triangle $TUV$ , $\vec{TU}$ and $\vec{TV}$ .

Calculate $\vec{TU}$ : $\vec{TU} = U - T = (9-4, 7-4, 7-1) = (5, 3, 6)$

Calculate $\vec{TV}$ : $\vec{TV} = V - T = (-6-4, 7-4, 1-1) = (-10, 3, 0)$

The area of triangle $TUV$ is: $\text{Area}(TUV) = \frac{1}{2} |\vec{M}| = \frac{1}{2} \cdot 3\sqrt{661} = \frac{3}{2}\sqrt{661}$

Area of triangle $PQR = \frac{\sqrt{10}}{2}$ . Area of triangle $TUV = \frac{3\sqrt{661}}{2}$ .