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MCQEasyJEE 2024Basics of Vectors

JEE Mathematics 2024 Question with Solution

The position vectors of the vertices A, B, and C of a triangle are A⃗=2i^−3j^+3k^\vec{A} = 2\hat{i} - 3\hat{j} + 3\hat{k}, B⃗=2i^+2j^+3k^\vec{B} = 2\hat{i} + 2\hat{j} + 3\hat{k}, and C⃗=−i^+j^+3k^\vec{C} = -\hat{i} + \hat{j} + 3\hat{k} respectively. If ℓ\ell is the length of the angle bisector ADAD of ∠BAC\angle BAC, then 2ℓ22\ell^2 equals:

  • A

    4949

  • B

    4242

  • C

    5050

  • D

    4545

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The vertices are A=(2,−3,3)A=(2,-3,3), B=(2,2,3)B=(2,2,3), and C=(−1,1,3)C=(-1,1,3).

Find: The value of 2ℓ22\ell^2, where ℓ\ell is the length of the angle bisector ADAD of ∠BAC\angle BAC.

First find the side lengths adjacent to angle AA:

AB⃗=B⃗−A⃗=(0,5,0)\vec{AB}=\vec{B}-\vec{A}=(0,5,0) ∣AB⃗∣=02+52+02=5|\vec{AB}|=\sqrt{0^2+5^2+0^2}=5 AC⃗=C⃗−A⃗=(−3,4,0)\vec{AC}=\vec{C}-\vec{A}=(-3,4,0) ∣AC⃗∣=(−3)2+42+02=5|\vec{AC}|=\sqrt{(-3)^2+4^2+0^2}=5

Since AB=ACAB=AC, triangle ABCABC is isosceles. Therefore, the angle bisector from AA meets BCBC at its midpoint. Hence,

D⃗=B⃗+C⃗2\vec{D}=\frac{\vec{B}+\vec{C}}{2} D⃗=(2,2,3)+(−1,1,3)2=(12,32,3)\vec{D}=\frac{(2,2,3)+(-1,1,3)}{2}=\left(\frac12,\frac32,3\right)

Now compute AD⃗\vec{AD}:

AD⃗=D⃗−A⃗=(12−2,32−(−3),3−3)=(−32,92,0)\vec{AD}=\vec{D}-\vec{A}=\left(\frac12-2,\frac32-(-3),3-3\right)=\left(-\frac32,\frac92,0\right)

Therefore,

\ell^2=|\vec{AD}|^2=\left(-\frac32\right)^2+\left(\frac92\right)^2= rac94+\frac{81}{4}= rac{90}{4}= rac{45}{2}

So,

2ℓ2=2×452=452\ell^2=2\times \frac{45}{2}=45

Therefore, the correct option is D.

Using the angle bisector idea

Given: A=(2,−3,3)A=(2,-3,3), B=(2,2,3)B=(2,2,3), C=(−1,1,3)C=(-1,1,3).

Find: 2ℓ22\ell^2 for the angle bisector ADAD of ∠BAC\angle BAC.

By the angle bisector theorem,

BDDC=ABAC\frac{BD}{DC}=\frac{AB}{AC}

From the side lengths obtained in the solution,

AB=5,AC=5AB=5,\quad AC=5

Hence,

BDDC=1\frac{BD}{DC}=1

so DD is the midpoint of BCBC.

Now use the midpoint formula:

D=(2+(−1)2,2+12,3+32)=(12,32,3)D=\left(\frac{2+(-1)}{2},\frac{2+1}{2},\frac{3+3}{2}\right)=\left(\frac12,\frac32,3\right)

Then,

AD2=(2−12)2+(−3−32)2+(3−3)2AD^2=\left(2-\frac12\right)^2+\left(-3-\frac32\right)^2+(3-3)^2 AD^2=\left(\frac32\right)^2+\left(-\frac92\right)^2= rac94+\frac{81}{4}= rac{45}{2}

Therefore,

2ℓ2=2AD2=452\ell^2=2AD^2=45

So the correct answer is 4545, that is, option D.

Common mistakes

  • Assuming the angle bisector theorem must be applied with unequal ratios without first checking whether AB=ACAB=AC. Here the triangle is isosceles at AA, so the bisector goes to the midpoint of BCBC. Always compute ABAB and ACAC first.

  • Using the wrong midpoint coordinates for DD by adding coordinates incorrectly. The midpoint of B(2,2,3)B(2,2,3) and C(−1,1,3)C(-1,1,3) is (12,32,3)\left(\frac12,\frac32,3\right), not any other point. Average each coordinate separately.

  • Making an error while forming AD⃗\vec{AD} from the coordinates of AA and DD. The vector is obtained by subtraction, and the squared length uses the squares of all components. Write AD⃗=D⃗−A⃗\vec{AD}=\vec{D}-\vec{A} carefully before taking magnitude.

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