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MCQMediumJEE 2023Equation of Plane

JEE Mathematics 2023 Question with Solution

Let two vertices of a triangle ABC be (2,4,6)(2, 4, 6) and (0,−2,−5)(0,-2,-5), and its centroid be (2,1,−1)(2, 1,-1). If the image of the third vertex in the plane x+2y+4z=11x + 2y + 4z = 11 is (α,β,γ)(\alpha, \beta, \gamma), then αβ+βγ+γα\alpha\beta + \beta\gamma + \gamma\alpha is equal to:

  • A

    7676

  • B

    7474

  • C

    7070

  • D

    7272

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Two vertices are A(2,4,6)A(2,4,6) and B(0,−2,−5)B(0,-2,-5), and the centroid is G(2,1,−1)G(2,1,-1). Let the third vertex be C(x,y,z)C(x,y,z).

Find: The value of αβ+βγ+γα\alpha\beta + \beta\gamma + \gamma\alpha, where (α,β,γ)(\alpha,\beta,\gamma) is the image of CC in the plane x+2y+4z=11x+2y+4z=11.

Using the centroid formula,

2+0+x3=2,4−2+y3=1,6−5+z3=−1\frac{2+0+x}{3}=2, \quad \frac{4-2+y}{3}=1, \quad \frac{6-5+z}{3}=-1

So,

x=4,y=1,z=−4x=4, \quad y=1, \quad z=-4

Hence, the third vertex is C(4,1,−4)C(4,1,-4).

Now reflect C(4,1,−4)C(4,1,-4) in the plane x+2y+4z−11=0x+2y+4z-11=0. Using the reflection relation shown in the solution,

α−41=β−12=γ+44=−4+2−16−111+4+16=−2521\frac{\alpha-4}{1}=\frac{\beta-1}{2}=\frac{\gamma+4}{4}=\frac{-4+2-16-11}{1+4+16}=\frac{-25}{21}

From this, the solution gives

α=6,β=5,γ=4\alpha=6, \quad \beta=5, \quad \gamma=4

Therefore,

αβ+βγ+γα=(6⋅5)+(5⋅4)+(4⋅6)=30+20+24=74\alpha\beta+\beta\gamma+\gamma\alpha=(6\cdot 5)+(5\cdot 4)+(4\cdot 6)=30+20+24=74

Therefore, the correct option is B and the required value is 7474. The provided the solution marks option C, but its own working and final value clearly give 7474, which corresponds to option B.

Coordinate Computation

Given: A(2,4,6)A(2,4,6), B(0,−2,−5)B(0,-2,-5), centroid G(2,1,−1)G(2,1,-1).

Find: The reflected image coordinates (α,β,γ)(\alpha,\beta,\gamma) and then compute αβ+βγ+γα\alpha\beta+\beta\gamma+\gamma\alpha.

For a triangle, centroid coordinates are the averages of the three vertices. So,

G=(xA+xB+xC3,yA+yB+yC3,zA+zB+zC3)G=\left(\frac{x_A+x_B+x_C}{3},\frac{y_A+y_B+y_C}{3},\frac{z_A+z_B+z_C}{3}\right)

Substitute the given coordinates:

(2+0+x3,4+(−2)+y3,6+(−5)+z3)=(2,1,−1)\left(\frac{2+0+x}{3},\frac{4+(-2)+y}{3},\frac{6+(-5)+z}{3}\right)=(2,1,-1)

Equating corresponding coordinates,

2+x3=2,2+y3=1,1+z3=−1\frac{2+x}{3}=2, \quad \frac{2+y}{3}=1, \quad \frac{1+z}{3}=-1

which gives

x=4,y=1,z=−4x=4, \quad y=1, \quad z=-4

Thus the third vertex is C(4,1,−4)C(4,1,-4). Using the reflection computation provided in the solution for the plane x+2y+4z−11=0x+2y+4z-11=0, we obtain the image point

(α,β,γ)=(6,5,4)(\alpha,\beta,\gamma)=(6,5,4)

Now compute the required expression:

αβ+βγ+γα=6⋅5+5⋅4+4⋅6\alpha\beta+\beta\gamma+\gamma\alpha=6\cdot 5+5\cdot 4+4\cdot 6 =30+20+24=74=30+20+24=74

Hence, the required value is 7474, so the correct option is B.

Common mistakes

  • Students may use the centroid formula incorrectly by averaging only two vertices. The centroid of a triangle is the average of all three vertices. Always write coordinate-wise equations involving AA, BB, and CC together.

  • A common error is mishandling signs in the yy- and zz-coordinates, especially with −2-2 and −5-5. This gives a wrong third vertex. Keep the negative coordinates intact while forming the centroid equations.

  • Some students may trust the listed option label from the solution's instead of the numerical working. Here the solution heading says option C, but the working gives 7474. Always verify the final value against the options.

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