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3D Geometry question

2024 · 5 Apr · Shift 2 · Q54
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  5. /2024 · 5 Apr · Shift 2 · Q54

3D Geometry question

2024 · 5 Apr · Shift 2 · Q54

JEE MainMathematics3D GeometryNumerical+4 / −1
Let the point (−1,α,β)(-1, \alpha, \beta)(−1,α,β) lie on the line of the shortest distance between the lines x+2−3=y−24=z−52\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}−3x+2​=4y−2​=2z−5​ and x+2−1=y+62=z−10\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}−1x+2​=2y+6​=0z−1​. Then (α−β)2(\alpha-\beta)^2(α−β)2 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 729/25

  1. Write the two lines in vector form

For x+2−3=y−24=z−52=λ,\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}=\lambda,−3x+2​=4y−2​=2z−5​=λ, we get x=−2−3λ,y=2+4λ,z=5+2λ.x=-2-3\lambda,\quad y=2+4\lambda,\quad z=5+2\lambda.x=−2−3λ,y=2+4λ,z=5+2λ. So line L1L_1L1​ passes through A(−2,2,5)A(-2,2,5)A(−2,2,5) and has direction vector d⃗1=(−3,4,2).\vec d_1=(-3,4,2).d1​=(−3,4,2).

For x+2−1=y+62=z−10=μ,\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}=\mu,−1x+2​=2y+6​=0z−1​=μ, we get x=−2−μ,y=−6+2μ,z=1.x=-2-\mu,\quad y=-6+2\mu,\quad z=1.x=−2−μ,y=−6+2μ,z=1. So line L2L_2L2​ passes through B(−2,−6,1)B(-2,-6,1)B(−2,−6,1) and has direction vector d⃗2=(−1,2,0).\vec d_2=(-1,2,0).d2​=(−1,2,0).

  1. Let points on the two lines be

On L1L_1L1​: P=A+td⃗1=(−2−3t,  2+4t,  5+2t).P=A+t\vec d_1=(-2-3t,\;2+4t,\;5+2t).P=A+td1​=(−2−3t,2+4t,5+2t).

On L2L_2L2​: Q=B+sd⃗2=(−2−s,  −6+2s,  1).Q=B+s\vec d_2=(-2-s,\;-6+2s,\;1).Q=B+sd2​=(−2−s,−6+2s,1).

The line of shortest distance joins points PPP and QQQ such that PQ→⊥d⃗1andPQ→⊥d⃗2.\overrightarrow{PQ} \perp \vec d_1 \quad \text{and} \quad \overrightarrow{PQ} \perp \vec d_2.PQ​⊥d1​andPQ​⊥d2​.

Now, PQ→=Q−P=(2t−s,  −8+2s−4t,  −4−2t).\overrightarrow{PQ}=Q-P=(2t-s,\;-8+2s-4t,\;-4-2t).PQ​=Q−P=(2t−s,−8+2s−4t,−4−2t).

  1. Apply perpendicularity conditions

Condition 1:

PQ→⋅d⃗1=0\overrightarrow{PQ}\cdot \vec d_1=0PQ​⋅d1​=0 (2t−s)(−3)+(−8+2s−4t)(4)+(−4−2t)(2)=0.(2t-s)(-3)+(-8+2s-4t)(4)+(-4-2t)(2)=0.(2t−s)(−3)+(−8+2s−4t)(4)+(−4−2t)(2)=0.

Expanding: −6t+3s−32+8s−16t−8−4t=0-6t+3s-32+8s-16t-8-4t=0−6t+3s−32+8s−16t−8−4t=0 11s-26t-40=0. \tag{1}

Condition 2:

PQ→⋅d⃗2=0\overrightarrow{PQ}\cdot \vec d_2=0PQ​⋅d2​=0 (2t−s)(−1)+(−8+2s−4t)(2)+(−4−2t)(0)=0.(2t-s)(-1)+(-8+2s-4t)(2)+(-4-2t)(0)=0.(2t−s)(−1)+(−8+2s−4t)(2)+(−4−2t)(0)=0.

Expanding: −2t+s−16+4s−8t=0-2t+s-16+4s-8t=0−2t+s−16+4s−8t=0 5s-10t-16=0. \tag{2}

  1. Solve for ttt and sss

From (2): 5s=10t+16  ⟹  s=2t+165.5s=10t+16\implies s=2t+\frac{16}{5}. 5s=10t+16⟹s=2t+516​.

Substitute into (1): 11(2t+165)−26t−40=011\left(2t+\frac{16}{5}\right)-26t-40=011(2t+516​)−26t−40=0 22t+1765−26t−40=022t+\frac{176}{5}-26t-40=022t+5176​−26t−40=0 −4t+1765−2005=0-4t+\frac{176}{5}-\frac{200}{5}=0−4t+5176​−5200​=0 −4t−245=0-4t-\frac{24}{5}=0−4t−524​=0 t=−65.t=-\frac{6}{5}. t=−56​.

Then s=2(−65)+165=45.s=2\left(-\frac{6}{5}\right)+\frac{16}{5}=\frac{4}{5}. s=2(−56​)+516​=54​.

  1. Find the required point on the shortest distance line with x=−1x=-1x=−1

The points of shortest distance are

On L1L_1L1​:

=\left(\frac85,\; -\frac{14}{5},\; \frac{13}{5}\right).$$ On $L_2$: $$Q=\left(-2-\frac45,\;-6+2\cdot\frac45,\;1\right) =\left(-\frac{14}{5},\; -\frac{22}{5},\;1\right).$$ So the line of shortest distance is the line through $P$ and $Q$. Its direction vector is $$Q-P=\left(-\frac{22}{5},\; -\frac{8}{5},\; -\frac{8}{5}\right) \propto (11,4,4).$$ Hence equation of this line is $$ (x,y,z)=\left(\frac85,-\frac{14}{5},\frac{13}{5}\right)+\lambda(11,4,4). $$ Now given point $(-1,\alpha,\beta)$ lies on it. So set $$\frac85+11\lambda=-1.$$ Thus $$11\lambda=-\frac{13}{5}\implies \lambda=-\frac{13}{55}. $$ Then $$\alpha=-\frac{14}{5}+4\lambda=-\frac{14}{5}-\frac{52}{55} =-\frac{154+52}{55}=-\frac{206}{55},$$ $$\beta=\frac{13}{5}+4\lambda=\frac{13}{5}-\frac{52}{55} =\frac{143-52}{55}=\frac{91}{55}. $$ Therefore, $$\alpha-\beta=-\frac{206}{55}-\frac{91}{55}=-\frac{297}{55}=-\frac{27}{5}. $$ So $$ (\alpha-\beta)^2=\left(\frac{27}{5}\right)^2=\frac{729}{25}. $$ This is **not an integer**, so let us simplify the shortest-distance line more carefully using the segment direction itself. Since $$Q-P=\left(-\frac{22}{5},-\frac85,-\frac85\right),$$ we can parametrize from $Q$ as $$ (x,y,z)=\left(-\frac{14}{5},-\frac{22}{5},1\right)+u\left(\frac{22}{5},\frac85,\frac85\right). $$ Set $x=-1$: $$-\frac{14}{5}+\frac{22}{5}u=-1$$ $$-14+22u=-5$$ $$22u=9\implies u=\frac{9}{22}. $$ Then $$\alpha=-\frac{22}{5}+\frac85\cdot\frac{9}{22} =-\frac{22}{5}+\frac{36}{55} =-\frac{242-36}{55}=-\frac{206}{55},$$ $$\beta=1+\frac85\cdot\frac{9}{22}=1+\frac{36}{55}=\frac{91}{55},$$ which gives the same result. So the computation is consistent. Thus the mathematically derived value is $$\boxed{\frac{729}{25}}.$$ Since the question is of integer type, the stored answer $25$ appears inconsistent with the given data.
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