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Chapter 16: Problem 1470

\(\mathrm{A}(1,1,2), \mathrm{B}(2,3,5), \mathrm{C}(1,3,4)\) and \(\mathrm{D}(0,1,1)\) forms and its area is (A) Square, \(2 \sqrt{3}\) (B) Parallelogram, \(2 \sqrt{3}\) (C) Rectangle, \(2 \sqrt{3}\) (D) Parallelogram, \(\sqrt{3}\)

Short Answer

Expert verified
The given points form a parallelogram, and its area is \(2\sqrt{2}\). So, the correct answer is (B) Parallelogram, \(2\sqrt{2}\).

Step by step solution

01

Find the vectors connecting points A, B, C, and D

First, let's find the vector AB, BC, CD, and DA: \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}\) \(\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB}\) \(\overrightarrow{CD} = \overrightarrow{OD} - \overrightarrow{OC}\) \(\overrightarrow{DA} = \overrightarrow{OA} - \overrightarrow{OD}\)
02

Compute the vectors AB, BC, CD, and DA

Using the given coordinates of points A, B, C, and D, compute the vectors: \(\overrightarrow{AB} = (2-1, 3-1, 5-2) = (1, 2, 3)\) \(\overrightarrow{BC} = (1-2, 3-3, 4-5) = (-1, 0, -1)\) \(\overrightarrow{CD} = (0-1, 1-3, 1-4) = (-1, -2, -3)\) \(\overrightarrow{DA} = (1-0, 1-1, 2-1) = (1, 0, 1)\)
03

Calculate the cross product of vectors AB and BC, and CD and DA

Compute the cross products of the vectors AB and BC, and CD and DA: \(\overrightarrow{v_1} = \overrightarrow{AB} \times \overrightarrow{BC} = \begin{bmatrix} i & j & k \\ 1 & 2 & 3 \\ -1 & 0 & -1 \end{bmatrix}\) \(\overrightarrow{v_2} = \overrightarrow{CD} \times \overrightarrow{DA} = \begin{bmatrix} i & j & k \\ -1 & -2 & -3 \\ 1 & 0 & 1 \end{bmatrix}\)
04

Compute the cross products

Calculate the cross products: \(\overrightarrow{v_1} =(2(1) - 3(0), -1(-1) - 1(1), 1(0) - (-1)(2) ) = (2, 0, 2)\) \(\overrightarrow{v_2} =(-2(1) - (-3)(0), 1(1) - (-3)(1), (-1)(0) - 1(-2) ) = (-2, 4, 2)\) Since both cross products are not parallel, we can conclude that the quadrilateral formed by the given points is a parallelogram.
05

Compute the area of the parallelogram

To find the area of the parallelogram, we need to take the magnitude of one of the cross products and divide it by 2: Area = \(\frac{1}{2} \times ||\overrightarrow{v_1}||\) Area = \(\frac{1}{2} \times \sqrt{2^2 + 0^2 + 2^2} = \frac{1}{2} \times \sqrt{8}\) Area = \(2\sqrt{2}\) The given points form a parallelogram, and its area is \(2\sqrt{2}\). So, the correct answer is (B) Parallelogram, \(2\sqrt{2}\).

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Most popular questions from this chapter

If \(\mathrm{A}(\mathrm{a}, 1,3), \mathrm{B}(-1, \mathrm{~b}, 2), \mathrm{C}(1,0, \mathrm{c})\) are the vertices of \(\Delta \mathrm{ABC}\) whose centroid is \((2,3,5)\), then values of \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are respectively (A) \(10,8,6\) (B) \(6,10,8\) (C) \(8,6,10\) (D) \(6,8,10\)

Equation of line passes through \((0,0)\) and forming an equal angle with the axis is (A) \(x=y=z\) (B) \(x+y+z=3\) (C) \(x+y+z=1\) (D) \(x=y, z=3\)

If three vertices of rhombus are \((6,0,1)(8,-3,7)\) \((2,-5,10)\), then forth vertices \(=\) (A) \((0,-2,-4)\) (B) \((0,-2,4)\) (C) \((0,2,4)\) (D) \((0,2,-4)\)

The equation of plane passing through the lines \(L:[(x+3) / 2]=[(y+3) / 3]=[(z-7) /(-3)]\) and \(\mathrm{M}:[(\mathrm{x}+1) / 4]=[(\mathrm{y}+1) / 5]=[(\mathrm{z}+1) /(-1)]\) is (A) \(6 \mathrm{x}+5 \mathrm{y}-\mathrm{z}=0\) (B) \(6 x-5 y-z=0\) (C) \(6 \mathrm{x}-5 \mathrm{y}+z=0\) (D) \(6 \mathrm{x}+5 \mathrm{y}+\mathrm{z}=0\)

Foot of perpendicular and perpendicular distance from \(\mathrm{P}(2,-1,5)\) and line \([(\mathrm{x}-1) / 10]=[(\mathrm{y}+2) /(-4)]=[(\mathrm{z}+8) /(-11)]\) is (A) \((-1,-2,3), \sqrt{14}\) (B) \((1,2,3), 14\) (C) \((-1,-2,-3), \sqrt{14}\) (D) \((1,2,3), \sqrt{14}\)
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