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Vector Algebra question

2023 · 1 Feb · Shift 1 · Q42
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  5. /2023 · 1 Feb · Shift 1 · Q42

Vector Algebra question

2023 · 1 Feb · Shift 1 · Q42

JEE MainMathematicsVector AlgebraNumerical+4 / −1
A(2,6,2),B(−4,0,λ),C(2,3,−1)A(2,6,2), B(-4,0, \lambda), C(2,3,-1)A(2,6,2),B(−4,0,λ),C(2,3,−1) and D(4,5,0),∣λ∣≤5D(4,5,0),|\lambda| \leq 5D(4,5,0),∣λ∣≤5 are the vertices of a quadrilateral ABCDA B C DABCD. If its area is 18 square units, then 5−6λ5-6 \lambda5−6λ is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 11

  1. Use the area formula for a quadrilateral in space

For quadrilateral ABCDABCDABCD, if we take diagonal ACACAC, then

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

Each triangle area is half the magnitude of a cross product.

So,

Area(△ABC)=12 ∣AB→×AC→∣,\text{Area}(\triangle ABC)=\frac12\,|\overrightarrow{AB}\times \overrightarrow{AC}|,Area(△ABC)=21​∣AB×AC∣, Area(△ACD)=12 ∣AC→×AD→∣.\text{Area}(\triangle ACD)=\frac12\,|\overrightarrow{AC}\times \overrightarrow{AD}|.Area(△ACD)=21​∣AC×AD∣.

But since ABCDABCDABCD is a quadrilateral, the two triangles formed by diagonal ACACAC lie in the same plane, so the total area is

Area(ABCD)=12 ∣AB→×AC→∣+12 ∣AC→×AD→∣.\text{Area}(ABCD)=\frac12\,|\overrightarrow{AB}\times \overrightarrow{AC}|+\frac12\,|\overrightarrow{AC}\times \overrightarrow{AD}|.Area(ABCD)=21​∣AB×AC∣+21​∣AC×AD∣.

A simpler standard approach for a planar quadrilateral is to use

Area(ABCD)=12 ∣AC→×BD→∣.\text{Area}(ABCD)=\frac12\,|\overrightarrow{AC}\times \overrightarrow{BD}|.Area(ABCD)=21​∣AC×BD∣.

We will use this.

  1. Find the diagonal vectors

Given:

A(2,6,2),B(−4,0,λ),C(2,3,−1),D(4,5,0).A(2,6,2),\quad B(-4,0,\lambda),\quad C(2,3,-1),\quad D(4,5,0).A(2,6,2),B(−4,0,λ),C(2,3,−1),D(4,5,0).

Compute

AC→=C−A=(2−2, 3−6, −1−2)=(0,−3,−3),\overrightarrow{AC}=C-A=(2-2,\,3-6,\,-1-2)=(0,-3,-3),AC=C−A=(2−2,3−6,−1−2)=(0,−3,−3), BD→=D−B=(4−(−4), 5−0, 0−λ)=(8,5,−λ).\overrightarrow{BD}=D-B=(4-(-4),\,5-0,\,0-\lambda)=(8,5,-\lambda).BD=D−B=(4−(−4),5−0,0−λ)=(8,5,−λ).
  1. Compute the cross product
AC→×BD→=∣i^j^k^0−3−385−λ∣.\overrightarrow{AC}\times \overrightarrow{BD} = \begin{vmatrix} \hat i & \hat j & \hat k\\ 0 & -3 & -3\\ 8 & 5 & -\lambda \end{vmatrix}.AC×BD=​i^08​j^​−35​k^−3−λ​​.

Expanding,

AC→×BD→=i^[(−3)(−λ)−(−3)(5)]−j^[0(−λ)−(−3)(8)]+k^[0⋅5−(−3)(8)].\overrightarrow{AC}\times \overrightarrow{BD} =\hat i\big[(-3)(-\lambda)-(-3)(5)\big] -\hat j\big[0(-\lambda)-(-3)(8)\big] +\hat k\big[0\cdot 5-(-3)(8)\big].AC×BD=i^[(−3)(−λ)−(−3)(5)]−j^​[0(−λ)−(−3)(8)]+k^[0⋅5−(−3)(8)].

Thus,

AC→×BD→=i^(3λ+15)−j^(24)+k^(24).\overrightarrow{AC}\times \overrightarrow{BD} =\hat i(3\lambda+15)-\hat j(24)+\hat k(24).AC×BD=i^(3λ+15)−j^​(24)+k^(24).

So,

AC→×BD→=(3λ+15,−24,24).\overrightarrow{AC}\times \overrightarrow{BD}=(3\lambda+15,-24,24).AC×BD=(3λ+15,−24,24).

Its magnitude is

∣AC→×BD→∣=(3λ+15)2+(−24)2+242|\overrightarrow{AC}\times \overrightarrow{BD}| =\sqrt{(3\lambda+15)^2+(-24)^2+24^2}∣AC×BD∣=(3λ+15)2+(−24)2+242​ =9(λ+5)2+576+576=9(λ+5)2+1152.=\sqrt{9(\lambda+5)^2+576+576} =\sqrt{9(\lambda+5)^2+1152}.=9(λ+5)2+576+576​=9(λ+5)2+1152​.
  1. Use the given area

Given area of quadrilateral is 181818:

12 ∣AC→×BD→∣=18.\frac12\,|\overrightarrow{AC}\times \overrightarrow{BD}|=18.21​∣AC×BD∣=18.

So,

∣AC→×BD→∣=36.|\overrightarrow{AC}\times \overrightarrow{BD}|=36.∣AC×BD∣=36.

Hence,

9(λ+5)2+1152=36.\sqrt{9(\lambda+5)^2+1152}=36.9(λ+5)2+1152​=36.

Squaring both sides,

9(λ+5)2+1152=1296.9(\lambda+5)^2+1152=1296.9(λ+5)2+1152=1296. 9(λ+5)2=144.9(\lambda+5)^2=144.9(λ+5)2=144. (λ+5)2=16.(\lambda+5)^2=16.(λ+5)2=16. λ+5=±4.\lambda+5=\pm 4.λ+5=±4.

Thus,

λ=−1orλ=−9.\lambda=-1 \quad \text{or} \quad \lambda=-9.λ=−1orλ=−9.

Given ∣λ∣≤5|\lambda|\le 5∣λ∣≤5, only

λ=−1\lambda=-1λ=−1

is valid.

  1. Find 5−6λ5-6\lambda5−6λ
5−6λ=5−6(−1)=5+6=11.5-6\lambda=5-6(-1)=5+6=11.5−6λ=5−6(−1)=5+6=11.
  1. Compare with stored answer

Derived answer is 111111, which matches the stored correct answer.

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