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Vector Algebra question

2018 · 15 Apr · Shift 2 · Q30
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  5. /2018 · 15 Apr · Shift 2 · Q30

Vector Algebra question

2018 · 15 Apr · Shift 2 · Q30

JEE MainMathematicsVector AlgebraMCQ+4 / −1
If the position vectors of the vertices A, B and C of a Δ\DeltaΔ ABC are respectively 4i^+7j^+8k^,2i^+3j^+4k^,4\widehat i + 7\widehat j + 8\widehat k,2\widehat i + 3\widehat j + 4\widehat k,4i+7j​+8k,2i+3j​+4k, and 2i^+5j^+7k^,2\widehat i + 5\widehat j + 7\widehat k,2i+5j​+7k, then the position vectors of the point, where the bisector of ∠\angle∠ A meets BC is :
  1. A
    12(4i^+8j^+11k^){1 \over 2}\left( {4\widehat i + 8\widehat j + 11\widehat k} \right)21​(4i+8j​+11k)
  2. B
    13(6i^+11j^+15k^){1 \over 3}\left( {6\widehat i + 11\widehat j + 15\widehat k} \right)31​(6i+11j​+15k)
  3. C
    13(6i^+13j^+18k^){1 \over 3}\left( {6\widehat i + 13\widehat j + 18\widehat k} \right)31​(6i+13j​+18k)
  4. D
    14(8i^+14j^+19k^){1 \over 4}\left( {8\widehat i + 14\widehat j + 19\widehat k} \right)41​(8i+14j​+19k)
View written solutionFree

Correct answer: C

  1. Let the position vectors of the vertices be
A⃗=4i^+7j^+8k^,B⃗=2i^+3j^+4k^,C⃗=2i^+5j^+7k^.\vec A=4\hat i+7\hat j+8\hat k, \quad \vec B=2\hat i+3\hat j+4\hat k, \quad \vec C=2\hat i+5\hat j+7\hat k.A=4i^+7j^​+8k^,B=2i^+3j^​+4k^,C=2i^+5j^​+7k^.
  1. We need the point where the internal bisector of ∠A\angle A∠A meets the side BCBCBC.

By the Angle Bisector Theorem, if the bisector of ∠A\angle A∠A meets BCBCBC at PPP, then

BPPC=ABAC.\frac{BP}{PC}=\frac{AB}{AC}.PCBP​=ACAB​.

So first compute ABABAB and ACACAC.

  1. Compute ABABAB:
AB⃗=B⃗−A⃗=(2−4)i^+(3−7)j^+(4−8)k^=−2i^−4j^−4k^.\vec{AB}=\vec B-\vec A=(2-4)\hat i+(3-7)\hat j+(4-8)\hat k =-2\hat i-4\hat j-4\hat k.AB=B−A=(2−4)i^+(3−7)j^​+(4−8)k^=−2i^−4j^​−4k^.

Hence,

AB=(−2)2+(−4)2+(−4)2=4+16+16=36=6.AB=\sqrt{(-2)^2+(-4)^2+(-4)^2}=\sqrt{4+16+16}=\sqrt{36}=6.AB=(−2)2+(−4)2+(−4)2​=4+16+16​=36​=6.
  1. Compute ACACAC:
AC⃗=C⃗−A⃗=(2−4)i^+(5−7)j^+(7−8)k^=−2i^−2j^−k^.\vec{AC}=\vec C-\vec A=(2-4)\hat i+(5-7)\hat j+(7-8)\hat k =-2\hat i-2\hat j-\hat k.AC=C−A=(2−4)i^+(5−7)j^​+(7−8)k^=−2i^−2j^​−k^.

Hence,

AC=(−2)2+(−2)2+(−1)2=4+4+1=3.AC=\sqrt{(-2)^2+(-2)^2+(-1)^2}=\sqrt{4+4+1}=3.AC=(−2)2+(−2)2+(−1)2​=4+4+1​=3.
  1. Therefore,
BPPC=ABAC=63=2:1.\frac{BP}{PC}=\frac{AB}{AC}=\frac{6}{3}=2:1.PCBP​=ACAB​=36​=2:1.

So PPP divides BCBCBC internally in the ratio 2:12:12:1.

  1. Using the section formula: if a point divides the line joining position vectors B⃗\vec BB and C⃗\vec CC in the ratio 2:12:12:1, then
P⃗=2C⃗+1B⃗2+1.\vec P=\frac{2\vec C+1\vec B}{2+1}.P=2+12C+1B​.

Substitute B⃗\vec BB and C⃗\vec CC:

P⃗=2(2i^+5j^+7k^)+(2i^+3j^+4k^)3.\vec P=\frac{2(2\hat i+5\hat j+7\hat k)+(2\hat i+3\hat j+4\hat k)}{3}.P=32(2i^+5j^​+7k^)+(2i^+3j^​+4k^)​.
  1. Simplify:
2C⃗=4i^+10j^+14k^2\vec C=4\hat i+10\hat j+14\hat k2C=4i^+10j^​+14k^

and hence

2C⃗+B⃗=(4+2)i^+(10+3)j^+(14+4)k^=6i^+13j^+18k^.2\vec C+\vec B=(4+2)\hat i+(10+3)\hat j+(14+4)\hat k =6\hat i+13\hat j+18\hat k.2C+B=(4+2)i^+(10+3)j^​+(14+4)k^=6i^+13j^​+18k^.

Therefore,

P⃗=13(6i^+13j^+18k^).\vec P=\frac{1}{3}(6\hat i+13\hat j+18\hat k).P=31​(6i^+13j^​+18k^).
  1. This matches option C.
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