Find the area of triangle PQR, P(0,-2,0), Q(4,1,-2), R(5,3,1)

Question

May Dunn · Accepted Answer

Area of a vector with vertices at P,Q and R is

A r e a = \frac{1}{2} | \vec{P Q} \times \vec{P R} |

\vec{P Q} = < 4, 2, 0 ≻ < 0, 0, - 3 \geq < 4, 2, 3 >

\vec{P R} = < 3, 3, 1 ≻ < 0, 0, - 3 \geq < 3, 3, 4 >

A r e a = \frac{1}{2} | < 4, 2, 3 > \times < 3, 3, 4 > | = \frac{1}{2} | \begin{matrix} i & j & k \\ 4 & 2 & 3 \\ 3 & 3 & 4 \end{matrix} |

= \frac{| i (8 - 9) - j (16 - 9) + k (12 - 6) |}{2}

= \frac{∣ - i - 7 j + 6 k}{2}

= \frac{\sqrt{1^{2} + 7^{2} + 6^{2}}}{2}

= \frac{\sqrt{1 + 49 + 36}}{2} = \frac{\sqrt{86}}{2}

madeleinejames20 · Accepted Answer

Step 1:First, let&#039;s find the vectors PQ and PR. The vector PQ is given by subtracting the coordinates of point P from those of point Q:PQ=[4−01−(−2)−2−0]=[43−2]Similarly, the vector PR is given by subtracting the coordinates of point P from those of point R:PR=[5−03−(−2)1−0]=[551]Step 2:Next, we calculate the cross product of PQ and PR. The cross product is obtained by taking the determinants of the following matrix:PQ×PR=|𝐢𝐣𝐤43−2551|Expanding this determinant, we have:PQ×PR=(3·1−(−2)·5)𝐢−(4·1−(−2)·5)𝐣+(4·5−3·5)𝐤Simplifying the expression, we get:PQ×PR=[−13145]Step 3:Now, we can calculate the magnitude of this cross product vector:|PQ×PR|=(−13)2+142+52Simplifying further, we obtain:|PQ×PR|=169+196+25=390Finally, we can find the area of triangle PQR by substituting this value into the formula:{Area}=12(390)=3902Hence, the area of triangle PQR is 3902.

Eliza Beth13 · Accepted Answer

Answer:12390Explanation:First, we find the vectors 𝐯1 and 𝐯2 that are formed by the edges PQ and PR, respectively:𝐯1=𝐐−𝐏=(4,1,−2)−(0,−2,0)=(4,3,−2)𝐯2=𝐑−𝐏=(5,3,1)−(0,−2,0)=(5,5,1)Next, we calculate the cross product of 𝐯1 and 𝐯2:𝐧=𝐯1×𝐯2Using the determinant method, we have:𝐧=|𝐢𝐣𝐤43−2551|Expanding the determinant, we get:𝐧=𝐢|3−251|−𝐣|4−251|+𝐤|4355|Calculating the determinants, we find:𝐧=𝐢((3×1)−(−2×5))−𝐣((4×1)−(−2×5))+𝐤((4×5)−(3×5))𝐧=𝐢(3+10)−𝐣(4+10)+𝐤(20−15)𝐧=𝐢(13)−𝐣(14)+𝐤(5)Therefore, the normal vector to the triangle is 𝐧=13𝐢−14𝐣+5𝐤.Now, we can calculate the magnitude of the normal vector:|𝐧|=132+(−14)2+52=169+196+25=390Finally, we can find the area of triangle PQR using the formula:A=12|𝐧|Substituting the values, we have:A=12390Therefore, the area of triangle PQR is 12390 square units.

Mr Solver · Accepted Answer

To find the area of triangle PQR with vertices P(0,-2,0), Q(4,1,-2), and R(5,3,1), we can use the formula for the area of a triangle in 3D space. Let's denote the position vectors of the vertices as

𝐏

,

𝐐

, and

𝐑

, respectively. The area of the triangle is given by:

Area = \frac{1}{2} ‖ (𝐐 - 𝐏) \times (𝐑 - 𝐏) ‖

where

\times

represents the cross product and

‖ \cdot ‖

represents the magnitude of a vector.
First, let's calculate the position vectors of

𝐐 - 𝐏

and

𝐑 - 𝐏

:

𝐐 - 𝐏 = [\begin{matrix} 4 - 0 \\ 1 - (- 2) \\ (- 2) - 0 \end{matrix}] = [\begin{matrix} 4 \\ 3 \\ - 2 \end{matrix}]

𝐑 - 𝐏 = [\begin{matrix} 5 - 0 \\ 3 - (- 2) \\ 1 - 0 \end{matrix}] = [\begin{matrix} 5 \\ 5 \\ 1 \end{matrix}]

Next, let's calculate the cross product of

𝐐 - 𝐏

and

𝐑 - 𝐏

:

(𝐐 - 𝐏) \times (𝐑 - 𝐏) = [\begin{matrix} 4 \\ 3 \\ - 2 \end{matrix}] \times [\begin{matrix} 5 \\ 5 \\ 1 \end{matrix}] = [\begin{matrix} (- 2 \times 5) - (3 \times 1) \\ (4 \times 1) - (5 \times - 2) \\ (4 \times 5) - (3 \times 5) \end{matrix}] = [\begin{matrix} - 13 \\ 14 \\ 5 \end{matrix}]

Finally, let's calculate the magnitude of

(𝐐 - 𝐏) \times (𝐑 - 𝐏)

and divide it by 2 to obtain the area:

‖ (𝐐 - 𝐏) \times (𝐑 - 𝐏) ‖ = \sqrt{(- 13)^{2} + 14^{2} + 5^{2}} = \sqrt{169 + 196 + 25} = \sqrt{390} = 3 \sqrt{10}

Therefore, the area of triangle PQR is

\frac{1}{2} \times 3 \sqrt{10} = \frac{3}{2} \sqrt{10}

.

Find the area of triangle PQR, P(0,-2,0), Q(4,1,-2), R(5,3,1)

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