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JEE Mathematics 2023 Question with Solution

If the foot of the perpendicular drawn from (1,9,7)(1, 9, 7) to the line passing through the point (3,2,1)(3, 2, 1) and parallel to the planes x+2y+z=0x + 2y + z = 0 and 3y−z=33y - z = 3 is (α,β,γ)(\alpha, \beta, \gamma), then α+β+γ\alpha + \beta + \gamma is equal to:

  • A

    −1-1

  • B

    33

  • C

    11

  • D

    55

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The line passes through (3,2,1)(3,2,1) and is parallel to the intersection of the planes x+2y+z=0x+2y+z=0 and 3y−z=33y-z=3. The point is P(1,9,7)P(1,9,7).

Find: α+β+γ\alpha+\beta+\gamma where (α,β,γ)(\alpha,\beta,\gamma) is the foot of the perpendicular from PP to the line.

Sketch showing point P(1,9,7), the line with direction ratios (-5,1,3), and foot of perpendicular M on the line.

Direction ratio of line

∣i^j^k^12103−1∣\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 0 & 3 & -1 \end{vmatrix} =i^(−5)−j^(−1)+k^(3)= \hat{i}(-5) - \hat{j}(-1) + \hat{k}(3) =−5i^+j^+3k^= -5\hat{i} + \hat{j} + 3\hat{k}

So a direction vector is (−5,1,3)(-5,1,3).

A general point on the line is

M(−5λ+3,  λ+2,  3λ+1)M(-5\lambda + 3,\; \lambda + 2,\; 3\lambda + 1)

Since PM→\overrightarrow{PM} is perpendicular to the line direction (−5,1,3)(-5,1,3),

PM→⊥(−5i^+j^+3k^)\overrightarrow{PM} \perp (-5\hat{i}+\hat{j}+3\hat{k})

Hence,

−5(−5λ+2)+(λ−7)+3(3λ−6)=0-5(-5\lambda + 2) + (\lambda - 7) + 3(3\lambda - 6) = 0 25λ+λ+9λ−10−7−18=025\lambda + \lambda + 9\lambda - 10 - 7 - 18 = 0 λ=1\lambda = 1

Therefore,

M=(−2,3,4)=(α,β,γ)M = (-2,3,4) = (\alpha,\beta,\gamma)

So,

α+β+γ=−2+3+4=5\alpha + \beta + \gamma = -2 + 3 + 4 = 5

Therefore, the correct option is D.

Parametric Solution

Given: The line passes through (3,2,1)(3,2,1) and is parallel to the intersection of the planes x+2y+z=0x + 2y + z = 0 and 3y−z=33y - z = 3.

Find: The value of α+β+γ\alpha+\beta+\gamma.

From 3y−z=33y-z=3, we get

z=3y−3z = 3y-3

Substitute in x+2y+z=0x+2y+z=0:

x+2y+(3y−3)=0x+2y+(3y-3)=0 x+5y−3=0x+5y-3=0 x=−5y+3x=-5y+3

Thus the direction ratios are (−5,1,3)(-5,1,3).

Parametric form of the line:

x=3−5t,x=3-5t, y=2+t,y=2+t, z=1+3tz=1+3t

So the foot of the perpendicular is (α,β,γ)=(3−5t,2+t,1+3t)(\alpha,\beta,\gamma)=(3-5t,2+t,1+3t).

Vector from (1,9,7)(1,9,7) to this point is

(2−5t,−7+t,−6+3t)(2-5t,-7+t,-6+3t)

This is perpendicular to (−5,1,3)(-5,1,3). Hence,

(2−5t)(−5)+(−7+t)(1)+(−6+3t)(3)=0(2-5t)(-5)+(-7+t)(1)+(-6+3t)(3)=0 −10+25t−7+t−18+9t=0-10+25t-7+t-18+9t=0 35t=3535t=35 t=1t=1

Now,

α=3−5=−2,\alpha=3-5=-2, β=2+1=3,\beta=2+1=3, γ=1+3=4\gamma=1+3=4

Therefore,

α+β+γ=5\alpha+\beta+\gamma=5

So the correct option is D.

Common mistakes

  • Using the normals of the two planes directly as the line direction. This is wrong because the required line is parallel to the intersection of the planes, so its direction vector is the cross product of the plane normals. Use (1,2,1)×(0,3,−1)(1,2,1) \times (0,3,-1).

  • Writing the perpendicular condition incorrectly. The vector from the given point to the foot must be perpendicular to the line direction, so its dot product with (−5,1,3)(-5,1,3) must be zero. Do not equate coordinates directly.

  • Making sign errors while forming PM→\overrightarrow{PM}. If M=(3−5t,2+t,1+3t)M=(3-5t,2+t,1+3t) and P=(1,9,7)P=(1,9,7), then PM→=M−P=(2−5t,−7+t,−6+3t)\overrightarrow{PM}=M-P=(2-5t,-7+t,-6+3t). A wrong sign changes the value of tt.

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