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NVAMediumJEE 2023Equation of Line in 3D

JEE Mathematics 2023 Question with Solution

Let the line L:x−12=y+1−1=z−31L: \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{1} intersect the plane 2x+y+3z=162x + y + 3z = 16 at the point PP. Let the point QQ be the foot of perpendicular from the point R(1,−1,−3)R(1, -1, -3) on the line LL. If α\alpha is the area of triangle PQRPQR, then α2\alpha^2 is equal to:

Answer

Correct answer:180

Step-by-step solution

Standard Method

Given: The line is

x−12=y+1−1=z−31\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{1}

and the plane is

2x+y+3z=162x + y + 3z = 16

Point R=(1,−1,−3)R = (1,-1,-3).

Find: α2\alpha^2, where α\alpha is the area of triangle PQRPQR.

Any point on LL can be written as

(x,y,z)=(2λ+1,−λ−1,λ+3)(x,y,z) = (2\lambda+1, -\lambda-1, \lambda+3)

Substitute in the plane equation:

2(2λ+1)+(−λ−1)+3(λ+3)=162(2\lambda+1)+(-\lambda-1)+3(\lambda+3)=16 6λ+10=166\lambda+10=16 λ=1\lambda=1

Hence,

P=(3,−2,4)P=(3,-2,4)

Let QQ be the foot of the perpendicular from RR to line LL. Then

Q=(2λ+1,−λ−1,λ+3)Q=(2\lambda+1,-\lambda-1,\lambda+3)

So the direction ratios of QRQR are

⟨2λ,−λ,λ+6⟩\langle 2\lambda, -\lambda, \lambda+6 \rangle

The direction ratios of line LL are

⟨2,−1,1⟩\langle 2,-1,1 \rangle

Since QR⊥LQR \perp L,

⟨2λ,−λ,λ+6⟩⋅⟨2,−1,1⟩=0\langle 2\lambda, -\lambda, \lambda+6 \rangle \cdot \langle 2,-1,1 \rangle = 0 4λ+λ+λ+6=04\lambda+\lambda+\lambda+6=0 6λ+6=06\lambda+6=0 λ=−1\lambda=-1

Therefore,

Q=(−1,0,2)Q=(-1,0,2)
A sketch showing point R(1, -1, -3) above point Q(-1, 0, 2) with a right angle at Q, and triangle with vertices P(3, -2, 4), Q(-1, 0, 2), and R(1, -1, -3).

Now,

QR→=(1−(−1),−1−0,−3−2)=2i^−j^−5k^\overrightarrow{QR} = (1-(-1), -1-0, -3-2) = 2\hat{i} - \hat{j} - 5\hat{k} QP→=(3−(−1),−2−0,4−2)=4i^−2j^+2k^\overrightarrow{QP} = (3-(-1), -2-0, 4-2) = 4\hat{i} - 2\hat{j} + 2\hat{k}

Using the area formula

α=12∣QR→×QP→∣\alpha = \frac{1}{2}\left| \overrightarrow{QR} \times \overrightarrow{QP} \right|

The solution gives

QR→×QP→=−12i^−24j^\overrightarrow{QR} \times \overrightarrow{QP} = -12\hat{i} - 24\hat{j}

So,

α=12144+576\alpha = \frac{1}{2}\sqrt{144+576}

Hence,

α2=7204=180\alpha^2 = \frac{720}{4} = 180

Therefore, the value of α2\alpha^2 is 180180.

Vector Area Interpretation

Given: Points PP and QQ lie on the line, with PP also on the plane, and QQ is the perpendicular projection of RR on the line.

Find: Area of triangle PQRPQR using vectors.

The hint uses the vector formula for area of a triangle in space:

Area of △PQR=12∣QR→×QP→∣\text{Area of } \triangle PQR = \frac{1}{2}\left|\overrightarrow{QR} \times \overrightarrow{QP}\right|

Once PP and QQ are found from the line conditions, form the two side vectors from the common vertex QQ and evaluate the cross product magnitude. Squaring the final area removes the square root and gives the required result 180180.

Common mistakes

  • Using the line intersection point incorrectly. The point PP must satisfy both the line and the plane. Taking any arbitrary point on the line is wrong; substitute the parametric coordinates into the plane equation first.

  • Finding the foot of perpendicular incorrectly. For point QQ, the vector QR→\overrightarrow{QR} must be perpendicular to the direction vector of line LL. Missing this dot-product condition leads to a wrong value of λ\lambda.

  • Using PR→\overrightarrow{PR} and PQ→\overrightarrow{PQ} without a common vertex in the stated formula. The cross-product area formula should use two sides of the triangle from the same point, such as QR→\overrightarrow{QR} and QP→\overrightarrow{QP}.

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