Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

3D Geometry question

2025 · Shift 1 · Q21
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Advanced
  3. /Mathematics
  4. /3D Geometry
  5. /2025 · Shift 1 · Q21

3D Geometry question

2025 · Shift 1 · Q21

JEE AdvancedMathematics3D GeometryMultiple correct+4 / −2
Let L1L_1L1​ be the line of intersection of the planes given by the equations 2x+3y+z=42x + 3y + z = 42x+3y+z=4 and x+2y+z=5x + 2y + z = 5x+2y+z=5. Let L2L_2L2​ be the line passing through the point P(2,−1,3)P(2, -1, 3)P(2,−1,3) and parallel to L1L_1L1​. Let MMM denote the plane given by the equation 2x+y−2z=62x + y - 2z = 62x+y−2z=6. Suppose that the line L2L_2L2​ meets the plane MMM at the point QQQ. Let RRR be the foot of the perpendicular drawn from PPP to the plane MMM. Then which of the following statements is (are) TRUE?
  1. A
    The length of the line segment PQPQPQ is 939\sqrt{3}93​
  2. B
    The length of the line segment QRQRQR is 151515
  3. C
    The area of △PQR\triangle PQR△PQR is 32234\dfrac{3}{2}\sqrt{234}23​234​
  4. D
    The acute angle between the line segments PQPQPQ and PRPRPR is cos⁡−1(123)\cos^{-1}\left(\dfrac{1}{2\sqrt{3}}\right)cos−1(23​1​)
View written solutionFree

Correct answer: A, C

  1. Find the direction vector of L1L_1L1​.

The line L1L_1L1​ is the intersection of the planes 2x+3y+z=42x+3y+z=42x+3y+z=4 and x+2y+z=5.x+2y+z=5.x+2y+z=5.

Their normal vectors are n⃗1=(2,3,1),n⃗2=(1,2,1).\vec n_1=(2,3,1), \qquad \vec n_2=(1,2,1).n1​=(2,3,1),n2​=(1,2,1).

So a direction vector of L1L_1L1​ is d⃗=n⃗1×n⃗2.\vec d=\vec n_1\times \vec n_2.d=n1​×n2​.

Compute:

d⃗=∣i^j^k^231121∣=i^(3⋅1−1⋅2)−j^(2⋅1−1⋅1)+k^(2⋅2−3⋅1).\vec d= \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 3 & 1 \\ 1 & 2 & 1 \end{vmatrix} =\hat i(3\cdot 1-1\cdot 2)-\hat j(2\cdot 1-1\cdot 1)+\hat k(2\cdot 2-3\cdot 1).d=​i^21​j^​32​k^11​​=i^(3⋅1−1⋅2)−j^​(2⋅1−1⋅1)+k^(2⋅2−3⋅1).

Thus, d⃗=(1,−1,1).\vec d=(1,-1,1).d=(1,−1,1).

Hence L2L_2L2​, being parallel to L1L_1L1​ and passing through P(2,−1,3)P(2,-1,3)P(2,−1,3), has parametric form x=2+t,y=−1−t,z=3+t.x=2+t,\quad y=-1-t,\quad z=3+t.x=2+t,y=−1−t,z=3+t.

  1. Find point Q=L2∩MQ=L_2\cap MQ=L2​∩M.

Plane MMM is 2x+y−2z=6.2x+y-2z=6.2x+y−2z=6.

Substitute the parametric coordinates of L2L_2L2​: 2(2+t)+(−1−t)−2(3+t)=6.2(2+t)+(-1-t)-2(3+t)=6.2(2+t)+(−1−t)−2(3+t)=6.

Simplify: 4+2t−1−t−6−2t=64+2t-1-t-6-2t=64+2t−1−t−6−2t=6 −3−t=6-3-t=6−3−t=6 t=−9.t=-9.t=−9.

So Q=(2−9,",−1−(−9)?Q=(2-9,",-1-(-9)?Q=(2−9,",−1−(−9)? Carefully: x=2+t=2−9=−7,x=2+t=2-9=-7,x=2+t=2−9=−7, y=−1−t=−1−(−9)=8,y=-1-t=-1-(-9)=8,y=−1−t=−1−(−9)=8, z=3+t=3−9=−6.z=3+t=3-9=-6.z=3+t=3−9=−6. Therefore, Q=(−7,8,−6).Q=(-7,8,-6).Q=(−7,8,−6).

Now, PQ→=Q−P=(−9,9,−9)=9(−1,1,−1).\overrightarrow{PQ}=Q-P=(-9,9,-9)=9(-1,1,-1).PQ​=Q−P=(−9,9,−9)=9(−1,1,−1). So PQ=(−9)2+92+(−9)2=93.PQ=\sqrt{(-9)^2+9^2+(-9)^2}=9\sqrt{3}.PQ=(−9)2+92+(−9)2​=93​.

Therefore, Option A is true.

  1. Find the foot of perpendicular RRR from PPP to plane MMM.

Plane MMM has normal vector n⃗=(2,1,−2).\vec n=(2,1,-2).n=(2,1,−2).

So the perpendicular from P(2,−1,3)P(2,-1,3)P(2,−1,3) to the plane is along direction (2,1,−2)(2,1,-2)(2,1,−2). Let R=P+λ(2,1,−2)=(2+2λ,−1+λ,3−2λ).R=P+\lambda(2,1,-2)=(2+2\lambda,-1+\lambda,3-2\lambda).R=P+λ(2,1,−2)=(2+2λ,−1+λ,3−2λ).

Since RRR lies on plane MMM, 2(2+2λ)+(−1+λ)−2(3−2λ)=6.2(2+2\lambda)+(-1+\lambda)-2(3-2\lambda)=6.2(2+2λ)+(−1+λ)−2(3−2λ)=6.

Simplify: 4+4λ−1+λ−6+4λ=64+4\lambda-1+\lambda-6+4\lambda=64+4λ−1+λ−6+4λ=6 −3+9λ=6-3+9\lambda=6−3+9λ=6 9λ=99\lambda=99λ=9 λ=1.\lambda=1.λ=1.

Hence, R=(4,0,1).R=(4,0,1).R=(4,0,1).

Then PR→=R−P=(2,1,−2).\overrightarrow{PR}=R-P=(2,1,-2).PR=R−P=(2,1,−2). So PR=22+12+(−2)2=3.PR=\sqrt{2^2+1^2+(-2)^2}=3.PR=22+12+(−2)2​=3.

Also, QR→=R−Q=(11,−8,7).\overrightarrow{QR}=R-Q=(11,-8,7).QR​=R−Q=(11,−8,7). Thus, QR=112+(−8)2+72=121+64+49=234=326.QR=\sqrt{11^2+(-8)^2+7^2}=\sqrt{121+64+49}=\sqrt{234}=3\sqrt{26}.QR=112+(−8)2+72​=121+64+49​=234​=326​.

This is not 151515.

Therefore, Option B is false.

  1. Find the area of △PQR\triangle PQR△PQR.

Since PRPRPR is perpendicular to plane MMM, and both QQQ and RRR lie in plane MMM, the line segment QRQRQR lies in plane MMM. Hence, PR⊥QR.PR\perp QR.PR⊥QR. So △PQR\triangle PQR△PQR is right-angled at RRR.

Therefore,

=\frac12\cdot 3\cdot \sqrt{234} =\frac{3}{2}\sqrt{234}.$$ Therefore, **Option C is true**. --- 5. **Find the angle between** $PQ$ and $PR$. We have $$\overrightarrow{PQ}=(-9,9,-9), \qquad \overrightarrow{PR}=(2,1,-2).$$ Their dot product is $$\overrightarrow{PQ}\cdot\overrightarrow{PR}=(-9)(2)+(9)(1)+(-9)(-2)=-18+9+18=9.$$ Now, $$|PQ|=9\sqrt{3}, \qquad |PR|=3.$$ So $$\cos\theta=\frac{\overrightarrow{PQ}\cdot\overrightarrow{PR}}{|PQ||PR|} =\frac{9}{(9\sqrt{3})(3)}=\frac{1}{3\sqrt{3}}.$$ Thus, $$\theta=\cos^{-1}\left(\frac{1}{3\sqrt{3}}\right),$$ not $$\cos^{-1}\left(\frac{1}{2\sqrt{3}}\right).$$ Therefore, **Option D is false**. --- 6. **Final conclusion** The true statements are: - **A** - **C**
Next

More from 3D Geometry

  • Let R3 denote the three-dimensional space. Take two points P=(1,2,3) and Q=(4,2,7). Let dist(X,Y) denote the distance between two points X and Y in R3. Let S={X∈R3:(dist(X,P))2−(dist(X,Q))2=50} and T={Y∈R3:(dist(Y,Q))2−(dist(Y,P))2=50}.​…2024 · Multiple correct
  • Let γ∈R be such that the lines L1​:1x+11​=2y+21​=3z+29​ and L2​:3x+16​=2y+11​=γz+4​ intersect. Let R1​ be the point of intersection of L1​ and L2​. Let O=(0,0,0)… Includes table2024 · MCQ
  • A straight line drawn from the point P(1,3,2), parallel to the line 1x−2​=2y−4​=1z−6​, intersects the plane L1​:x−y+3z=6 at the point Q. Another straight line which passes through Q and is perpendicular…2024 · Multiple correct
  • Let ℓ1​ and ℓ2​ be the lines r1​=λ(i^+j^​+k^) and r2​=(j^​−k^)+μ(i^+k^), respectively. Let X be the set of all the planes H that contain the line ℓ1​. For a plane H… Includes table2023 · MCQ
  • Let P1​ and P2​ be two planes given by ​P1​:10x+15y+12z−60=0P2​:−2x+5y+4z−20=0​ Which of the following straight lines can be an edge of some tetrahedron whose two faces lie…2022 · Multiple correct
  • Let S be the reflection of a point Q with respect to the plane given by r=−(t+p)^+t^​+(1+p)k^ where t,p are real parameters and ^,^​,k^ are the unit vectors along…2022 · Multiple correct
  • Let L1 and L2 be the following straight lines. L1​:1x−1​=−1y​=3z−1​ and L2​:−3x−1​=−1y​=1z−1​. Suppose the straight line L:lx−α​=my−1​=−2z−γ​…2020 · Multiple correct
  • Let α 2 + β 2 + γ 2 e 0 and α+γ= 1. Suppose the point (3, 2, − 1) is the mirror image of the point (1, 0, − 1) with respect to the plane α x + β y + γ z = δ. Then which of…2020 · Multiple correct