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Vector Algebra question

2023 · 8 Apr · Shift 2 · Q29
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  5. /2023 · 8 Apr · Shift 2 · Q29

Vector Algebra question

2023 · 8 Apr · Shift 2 · Q29

JEE MainMathematicsVector AlgebraMCQ+4 / −1
The area of the quadrilateral ABCD\mathrm{ABCD}ABCD with vertices A(2,1,1),B(1,2,5),C(−2,−3,5)\mathrm{A}(2,1,1), \mathrm{B}(1,2,5), \mathrm{C}(-2,-3,5)A(2,1,1),B(1,2,5),C(−2,−3,5) and D(1,−6,−7)\mathrm{D}(1,-6,-7)D(1,−6,−7) is equal to :
  1. A
    48
  2. B
    8388 \sqrt{38}838​
  3. C
    54
  4. D
    9389 \sqrt{38}938​
View written solutionFree

Correct answer: B

  1. Check whether the four points are coplanar

For a quadrilateral in space, the four vertices must be coplanar. Let us verify this first.

Take

AB→=B−A=(1−2, 2−1, 5−1)=(−1,1,4)\overrightarrow{AB}=B-A=(1-2,\,2-1,\,5-1)=(-1,1,4)AB=B−A=(1−2,2−1,5−1)=(−1,1,4) AC→=C−A=(−2−2, −3−1, 5−1)=(−4,−4,4)\overrightarrow{AC}=C-A=(-2-2,\,-3-1,\,5-1)=(-4,-4,4)AC=C−A=(−2−2,−3−1,5−1)=(−4,−4,4) AD→=D−A=(1−2, −6−1, −7−1)=(−1,−7,−8)\overrightarrow{AD}=D-A=(1-2,\,-6-1,\,-7-1)=(-1,-7,-8)AD=D−A=(1−2,−6−1,−7−1)=(−1,−7,−8)

Now compute the scalar triple product:

AB→⋅(AC→×AD→)\overrightarrow{AB}\cdot(\overrightarrow{AC}\times \overrightarrow{AD})AB⋅(AC×AD)

First,

AC→×AD→=∣i^j^k^−4−44−1−7−8∣\overrightarrow{AC}\times \overrightarrow{AD} =\begin{vmatrix} \hat i & \hat j & \hat k\\ -4 & -4 & 4\\ -1 & -7 & -8 \end{vmatrix}AC×AD=​i^−4−1​j^​−4−7​k^4−8​​ =i^[(−4)(−8)−4(−7)]−j^[(−4)(−8)−4(−1)]+k^[(−4)(−7)−(−4)(−1)]=\hat i\big[(-4)(-8)-4(-7)\big] -\hat j\big[(-4)(-8)-4(-1)\big] +\hat k\big[(-4)(-7)-(-4)(-1)\big]=i^[(−4)(−8)−4(−7)]−j^​[(−4)(−8)−4(−1)]+k^[(−4)(−7)−(−4)(−1)] =i^(32+28)−j^(32+4)+k^(28−4)=(60,−36,24)=\hat i(32+28)-\hat j(32+4)+\hat k(28-4) =(60,-36,24)=i^(32+28)−j^​(32+4)+k^(28−4)=(60,−36,24)

Then,

AB→⋅(60,−36,24)=(−1)(60)+(1)(−36)+(4)(24)=−60−36+96=0\overrightarrow{AB}\cdot(60,-36,24)=(-1)(60)+(1)(-36)+(4)(24) =-60-36+96=0AB⋅(60,−36,24)=(−1)(60)+(1)(−36)+(4)(24)=−60−36+96=0

Since the scalar triple product is zero, the points are coplanar, so the quadrilateral is well-defined.

  1. Split the quadrilateral into two triangles

We can find the area of quadrilateral ABCDABCDABCD as

Area(ABCD)=Area(△ABC)+Area(△ACD)\text{Area}(ABCD)=\text{Area}(\triangle ABC)+\text{Area}(\triangle ACD)Area(ABCD)=Area(△ABC)+Area(△ACD)

provided the diagonal ACACAC lies inside the quadrilateral, which it does for the planar ordering here.

The area of a triangle formed by vectors u⃗,v⃗\vec u,\vec vu,v is

12 ∣u⃗×v⃗∣\frac12\,|\vec u\times \vec v|21​∣u×v∣
  1. Area of △ABC\triangle ABC△ABC

We already have

AB→=(−1,1,4),AC→=(−4,−4,4)\overrightarrow{AB}=(-1,1,4),\qquad \overrightarrow{AC}=(-4,-4,4)AB=(−1,1,4),AC=(−4,−4,4)

Now,

AB→×AC→=∣i^j^k^−114−4−44∣\overrightarrow{AB}\times \overrightarrow{AC} =\begin{vmatrix} \hat i & \hat j & \hat k\\ -1 & 1 & 4\\ -4 & -4 & 4 \end{vmatrix}AB×AC=​i^−1−4​j^​1−4​k^44​​ =i^(1⋅4−4⋅(−4))−j^((−1)⋅4−4⋅(−4))+k^((−1)(−4)−1(−4))=\hat i(1\cdot 4-4\cdot(-4)) -\hat j((-1)\cdot 4-4\cdot(-4)) +\hat k((-1)(-4)-1(-4))=i^(1⋅4−4⋅(−4))−j^​((−1)⋅4−4⋅(−4))+k^((−1)(−4)−1(−4)) =i^(4+16)−j^(−4+16)+k^(4+4)=(20,−12,8)=\hat i(4+16)-\hat j(-4+16)+\hat k(4+4) =(20,-12,8)=i^(4+16)−j^​(−4+16)+k^(4+4)=(20,−12,8)

Its magnitude is

|(20,-12,8)|=\sqrt{20^2+(-12)^2+8^2} =\sqrt{400+144+64}=sqrt{608}=4\sqrt{38}

Therefore,

Area(△ABC)=12(438)=238\text{Area}(\triangle ABC)=\frac12(4\sqrt{38})=2\sqrt{38}Area(△ABC)=21​(438​)=238​
  1. Area of △ACD\triangle ACD△ACD

Use

AC→=(−4,−4,4),AD→=(−1,−7,−8)\overrightarrow{AC}=(-4,-4,4),\qquad \overrightarrow{AD}=(-1,-7,-8)AC=(−4,−4,4),AD=(−1,−7,−8)

From above,

AC→×AD→=(60,−36,24)\overrightarrow{AC}\times \overrightarrow{AD}=(60,-36,24)AC×AD=(60,−36,24)

Its magnitude is

∣(60,−36,24)∣=602+(−36)2+242=3600+1296+576=5472|(60,-36,24)|=\sqrt{60^2+(-36)^2+24^2} =\sqrt{3600+1296+576} =\sqrt{5472}∣(60,−36,24)∣=602+(−36)2+242​=3600+1296+576​=5472​ 5472=144⋅38=1238\sqrt{5472}=\sqrt{144\cdot 38}=12\sqrt{38}5472​=144⋅38​=1238​

Hence,

Area(△ACD)=12(1238)=638\text{Area}(\triangle ACD)=\frac12(12\sqrt{38})=6\sqrt{38}Area(△ACD)=21​(1238​)=638​
  1. Total area of quadrilateral
Area(ABCD)=238+638=838\text{Area}(ABCD)=2\sqrt{38}+6\sqrt{38}=8\sqrt{38}Area(ABCD)=238​+638​=838​

So the correct option is:

838\boxed{8\sqrt{38}}838​​

This corresponds to Option B.

  1. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

Hence, they agree.

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