Find the area of the parallelogram.
-$\mathrm { P } _ { 1 } ( - 1,0,2 ) , \mathrm { P } _ { 2 } ( 2,1 , - 2 ) , \mathrm { P } _ { 3 } ( 2 , - 1,4 )$
A) $\sqrt { 26 }$
B) $4 \sqrt { 5 }$
C) $2 \sqrt { 91 }$
D) $\sqrt { 14 }$

Question

Quizplus · Accepted Answer

The area of the parallelogram can be found by calculating the cross product of two vectors that represent adjacent sides of the parallelogram and then finding the magnitude of this cross product. Let's use $P_1P_2$ and $P_1P_3$ as the vectors.First, find the vectors:- $P_1P_2 = (2 - (-1), 1 - 0, -2 - 2) = (3, 1, -4)$- $P_1P_3 = (2 - (-1), -1 - 0, 4 - 2) = (3, -1, 2)$Next, calculate the cross product of $P_1P_2$ and $P_1P_3$:- $P_1P_2 \times P_1P_3 = \begin{vmatrix} i & j & k \ 3 & 1 & -4 \ 3 & -1 & 2 \end{vmatrix} = i(1*2 - (-4)(-1)) - j(3*2 - (-4)*3) + k(3*(-1) - 1*3)$- $= i(2 - 4) - j(6 - 12) + k(-3 - 3)$- $= -2i + 6j - 6k$Finally, find the magnitude of the cross product to get the area:- $|\vec{A}| = \sqrt{(-2)^2 + 6^2 + (-6)^2} = \sqrt{4 + 36 + 36} = \sqrt{76} = 2\sqrt{19}$However, none of the provided options match $2\sqrt{19}$, which suggests a miscalculation in the explanation. Let's correct the calculation:- Correct cross product calculation: $P_1P_2 \times P_1P_3 = \begin{vmatrix} i & j & k \ 3 & 1 & -4 \ 3 & -1 & 2 \end{vmatrix} = i(1*2 - (-4)(-1)) - j(3*2 - 3*(-4)) + k(3*(-1) - 1*3)$- Correctly calculating the determinant gives: $i(2 + 4) - j(6 + 12) + k(-3 - 3)$- Which simplifies to: $6i - 18j - 6k$- The magnitude is: $\sqrt{6^2 + (-18)^2 + (-6)^2} = \sqrt{36 + 324 + 36} = \sqrt{396} = 2\sqrt{99}$Given the discrepancy in the calculation and the options provided, the correct approach to finding the area involves correctly calculating the cross product and its magnitude. The correct magnitude calculation based on the vectors from $P_1$ to $P_2$ and $P_1$ to $P_3$ should yield the area of the parallelogram, but the provided options and calculations have led to confusion. The correct answer should be based on the magnitude of the cross product of the vectors formed by the given points, which typically involves squaring the components of the cross product vector, summing them, and taking the square root. The mistake in the explanation led to an incorrect conclusion; the correct process involves accurately calculating the cross product and its magnitude.

Find the Area of the Parallelogram P1(−1,0,2),P2(2,1,−2),P3(2,−1,4)\mathrm { P } _ { 1 } ( - 1,0,2 ) , \mathrm { P } _ { 2 } ( 2,1 , - 2 ) , \mathrm { P } _ { 3 } ( 2 , - 1,4 )P1​(−1,0,2),P2​(2,1,−2),P3​(2,−1,4)

Find the Area of the Parallelogram $\mathrm { P } _ { 1 } ( - 1,0,2 ) , \mathrm { P } _ { 2 } ( 2,1 , - 2 ) , \mathrm { P } _ { 3 } ( 2 , - 1,4 )$