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Vector Algebra question

2024 · 5 Apr · Shift 1 · Q49
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  5. /2024 · 5 Apr · Shift 1 · Q49

Vector Algebra question

2024 · 5 Apr · Shift 1 · Q49

JEE MainMathematicsVector AlgebraMCQ+4 / −1
If A(1,−1,2),B(5,7,−6),C(3,4,−10)\mathrm{A}(1,-1,2), \mathrm{B}(5,7,-6), \mathrm{C}(3,4,-10)A(1,−1,2),B(5,7,−6),C(3,4,−10) and D(−1,−4,−2)\mathrm{D}(-1,-4,-2)D(−1,−4,−2) are the vertices of a quadrilateral ABCD, then its area is :
  1. A
    24724 \sqrt{7}247​
  2. B
    48748 \sqrt{7}487​
  3. C
    242924 \sqrt{29}2429​
  4. D
    122912 \sqrt{29}1229​
View written solutionFree

Correct answer: D

  1. Use diagonals to find area of quadrilateral

For a quadrilateral with vertices A,B,C,DA,B,C,DA,B,C,D taken in order, its area is

Area=12∣AC→×BD→∣\text{Area} = \frac{1}{2}\left|\overrightarrow{AC} \times \overrightarrow{BD}\right|Area=21​​AC×BD​

So first compute the diagonals.

  1. Find vectors AC→\overrightarrow{AC}AC and BD→\overrightarrow{BD}BD

Given:

A(1,−1,2),B(5,7,−6),C(3,4,−10),D(−1,−4,−2)A(1,-1,2),\quad B(5,7,-6),\quad C(3,4,-10),\quad D(-1,-4,-2)A(1,−1,2),B(5,7,−6),C(3,4,−10),D(−1,−4,−2)

We get

AC→=C−A=(3−1, 4−(−1), −10−2)=(2,5,−12)\overrightarrow{AC}=C-A=(3-1,\,4-(-1),\,-10-2)=(2,5,-12)AC=C−A=(3−1,4−(−1),−10−2)=(2,5,−12)

and

BD→=D−B=(−1−5, −4−7, −2−(−6))=(−6,−11,4)\overrightarrow{BD}=D-B=(-1-5,\,-4-7,\,-2-(-6))=(-6,-11,4)BD=D−B=(−1−5,−4−7,−2−(−6))=(−6,−11,4)
  1. Compute the cross product
AC→×BD→=∣i^j^k^25−12−6−114∣\overrightarrow{AC}\times\overrightarrow{BD} = \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 5 & -12 \\ -6 & -11 & 4 \end{vmatrix}AC×BD=​i^2−6​j^​5−11​k^−124​​

Now expand:

=i^ (5⋅4−(−12)(−11))−j^ (2⋅4−(−12)(−6))+k^ (2⋅(−11)−5⋅(−6))= \hat i\,(5\cdot 4-(-12)(-11)) - \hat j\,(2\cdot 4-(-12)(-6)) + \hat k\,(2\cdot(-11)-5\cdot(-6))=i^(5⋅4−(−12)(−11))−j^​(2⋅4−(−12)(−6))+k^(2⋅(−11)−5⋅(−6)) =i^ (20−132)−j^ (8−72)+k^ (−22+30)= \hat i\,(20-132)-\hat j\,(8-72)+\hat k\,(-22+30)=i^(20−132)−j^​(8−72)+k^(−22+30) =−112i^+64j^+8k^= -112\hat i +64\hat j +8\hat k=−112i^+64j^​+8k^
  1. Find its magnitude
∣AC→×BD→∣=(−112)2+642+82\left|\overrightarrow{AC}\times\overrightarrow{BD}\right| =\sqrt{(-112)^2+64^2+8^2}​AC×BD​=(−112)2+642+82​ =12544+4096+64=16704=\sqrt{12544+4096+64}= \sqrt{16704}=12544+4096+64​=16704​

Factorize:

16704=16⋅1044=16⋅36⋅2916704=16\cdot 1044=16\cdot 36\cdot 2916704=16⋅1044=16⋅36⋅29

So,

16704=4⋅629=2429\sqrt{16704}=4\cdot 6\sqrt{29}=24\sqrt{29}16704​=4⋅629​=2429​
  1. Area of the quadrilateral
Area=12⋅2429=1229\text{Area}=\frac{1}{2}\cdot 24\sqrt{29}=12\sqrt{29}Area=21​⋅2429​=1229​
  1. Match with options
1229\boxed{12\sqrt{29}}1229​​

Hence the correct option is D.

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