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3D Geometry question

2025 · 8 Apr · Shift 2 · Q49
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3D Geometry question

2025 · 8 Apr · Shift 2 · Q49

JEE MainMathematics3D GeometryNumerical+4 / −1
Let the area of the triangle formed by the lines x+2=y−1=z,x−35=y−1=z−11x+2=y-1=z, \frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}x+2=y−1=z,5x−3​=−1y​=1z−1​ and x−3=y−33=z−21\frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1}−3x​=3y−3​=1z−2​ be AAA. Then A2A^2A2 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 56

We need the area of the triangle formed by the three given lines in 3D. This means the lines intersect pairwise and form a triangle whose vertices are the pairwise points of intersection.

1. Write each line in parametric form

Line L1L_1L1​

Given: x+2=y−1=zx+2=y-1=zx+2=y−1=z Let the common value be ttt. Then x=t−2,y=t+1,z=tx=t-2,\quad y=t+1,\quad z=tx=t−2,y=t+1,z=t So L1:(x,y,z)=(−2,1,0)+t(1,1,1).L_1:(x,y,z)=(-2,1,0)+t(1,1,1).L1​:(x,y,z)=(−2,1,0)+t(1,1,1).

Line L2L_2L2​

Given: x−35=y−1=z−11\frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}5x−3​=−1y​=1z−1​ Let the common value be sss. Then x=3+5s,y=−s,z=1+sx=3+5s,\quad y=-s,\quad z=1+sx=3+5s,y=−s,z=1+s So L2:(x,y,z)=(3,0,1)+s(5,−1,1).L_2:(x,y,z)=(3,0,1)+s(5,-1,1).L2​:(x,y,z)=(3,0,1)+s(5,−1,1).

Line L3L_3L3​

Given: x−3=y−33=z−21\frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1}−3x​=3y−3​=1z−2​ Let the common value be uuu. Then x=−3u,y=3+3u,z=2+ux=-3u,\quad y=3+3u,\quad z=2+ux=−3u,y=3+3u,z=2+u So L3:(x,y,z)=(0,3,2)+u(−3,3,1).L_3:(x,y,z)=(0,3,2)+u(-3,3,1).L3​:(x,y,z)=(0,3,2)+u(−3,3,1).

2. Find pairwise intersections

Intersection of L1L_1L1​ and L2L_2L2​

We solve t−2=3+5s,t+1=−s,t=1+s.t-2=3+5s,\quad t+1=-s,\quad t=1+s.t−2=3+5s,t+1=−s,t=1+s. From t=1+st=1+st=1+s, substitute into t+1=−st+1=-st+1=−s: (1+s)+1=−s  ⟹  2+2s=0  ⟹  s=−1.(1+s)+1=-s\implies 2+2s=0\implies s=-1.(1+s)+1=−s⟹2+2s=0⟹s=−1. Then t=1+s=0.t=1+s=0.t=1+s=0. So intersection point is P=L1∩L2=(−2,1,0).P=L_1\cap L_2=(-2,1,0).P=L1​∩L2​=(−2,1,0).

Intersection of L2L_2L2​ and L3L_3L3​

Solve 3+5s=−3u,−s=3+3u,1+s=2+u.3+5s=-3u,\quad -s=3+3u,\quad 1+s=2+u.3+5s=−3u,−s=3+3u,1+s=2+u. From the third equation, s−u=1.s-u=1.s−u=1. From the second equation, −s=3+3u  ⟹  s=−3−3u.-s=3+3u\implies s=-3-3u.−s=3+3u⟹s=−3−3u. Substitute into s−u=1s-u=1s−u=1: (−3−3u)−u=1  ⟹  −3−4u=1  ⟹  u=−1.(-3-3u)-u=1\implies -3-4u=1\implies u=-1.(−3−3u)−u=1⟹−3−4u=1⟹u=−1. Then s=−3−3(−1)=0.s=-3-3(-1)=0.s=−3−3(−1)=0. So intersection point is Q=L2∩L3=(3,0,1).Q=L_2\cap L_3=(3,0,1).Q=L2​∩L3​=(3,0,1).

Intersection of L3L_3L3​ and L1L_1L1​

Solve −3u=t−2,3+3u=t+1,2+u=t.-3u=t-2,\quad 3+3u=t+1,\quad 2+u=t.−3u=t−2,3+3u=t+1,2+u=t. From t=2+ut=2+ut=2+u, substitute into the second equation: 3+3u=(2+u)+1=3+u  ⟹  2u=0  ⟹  u=0.3+3u=(2+u)+1=3+u\implies 2u=0\implies u=0.3+3u=(2+u)+1=3+u⟹2u=0⟹u=0. Hence t=2.t=2.t=2. So intersection point is R=L3∩L1=(0,3,2).R=L_3\cap L_1=(0,3,2).R=L3​∩L1​=(0,3,2).

Thus the triangle has vertices P(−2,1,0),Q(3,0,1),R(0,3,2).P(-2,1,0),\quad Q(3,0,1),\quad R(0,3,2).P(−2,1,0),Q(3,0,1),R(0,3,2).

3. Use vector formula for area of triangle

Area of triangle PQRPQRPQR is A=12∥PQ→×PR→∥.A=\frac12\left\|\overrightarrow{PQ}\times\overrightarrow{PR}\right\|.A=21​​PQ​×PR​.

Compute the vectors: PQ→=Q−P=(3−(−2),0−1,1−0)=(5,−1,1),\overrightarrow{PQ}=Q-P=(3-(-2),0-1,1-0)=(5,-1,1),PQ​=Q−P=(3−(−2),0−1,1−0)=(5,−1,1), PR→=R−P=(0−(−2),3−1,2−0)=(2,2,2).\overrightarrow{PR}=R-P=(0-(-2),3-1,2-0)=(2,2,2).PR=R−P=(0−(−2),3−1,2−0)=(2,2,2).

Now,

\begin{vmatrix} \mathbf i & \mathbf j & \mathbf k\\ 5 & -1 & 1\\ 2 & 2 & 2 \end{vmatrix}.$$ So $$\overrightarrow{PQ}\times\overrightarrow{PR} =\mathbf i((-1)(2)-1(2)) -\mathbf j(5\cdot 2-1\cdot 2) +\mathbf k(5\cdot 2-(-1)\cdot 2).$$ $$= -4\mathbf i-8\mathbf j+12\mathbf k.$$ Its magnitude is $$\sqrt{(-4)^2+(-8)^2+12^2}=\sqrt{16+64+144}=\sqrt{224}=4\sqrt{14}.$$ Hence $$A=\frac12(4\sqrt{14})=2\sqrt{14}.$$ Therefore, $$A^2=(2\sqrt{14})^2=56.$$ --- ## 4. Compare with stored answer Derived answer: $$A^2=56$$ The stored correct answer is also $56$, so they agree.
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