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Vector Algebra question

2021 · 20 Jul · Shift 1 · Q40
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Vector Algebra question

2021 · 20 Jul · Shift 1 · Q40

JEE MainMathematicsVector AlgebraNumerical+4 / −1
If the shortest distance between the lines r1→=αi^+2j^+2k^+λ(i^−2j^+2k^)\overrightarrow {{r_1}} = \alpha \widehat i + 2\widehat j + 2\widehat k + \lambda (\widehat i - 2\widehat j + 2\widehat k)r1​​=αi+2j​+2k+λ(i−2j​+2k), λ∈\lambda\inλ∈ R, α\alphaα> 0 and r2→=−4i^−k^+μ(3i^−2j^−2k^)\overrightarrow {{r_2}} = - 4\widehat i - \widehat k + \mu (3\widehat i - 2\widehat j - 2\widehat k)r2​​=−4i−k+μ(3i−2j​−2k), μ∈\mu\inμ∈ R is 9, then α\alphaα is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 6

  1. Write the two lines in point-direction form

The given lines are

L1:r⃗=(α,2,2)+λ(1,−2,2)L_1: \vec r= (\alpha,2,2)+\lambda(1,-2,2)L1​:r=(α,2,2)+λ(1,−2,2)

and

L2:r⃗=(−4,0,−1)+μ(3,−2,−2).L_2: \vec r= (-4,0,-1)+\mu(3,-2,-2).L2​:r=(−4,0,−1)+μ(3,−2,−2).

So,

  • a point on L1L_1L1​ is A(α,2,2)A(\alpha,2,2)A(α,2,2),
  • direction vector of L1L_1L1​ is d⃗1=(1,−2,2)\vec d_1=(1,-2,2)d1​=(1,−2,2),
  • a point on L2L_2L2​ is B(−4,0,−1)B(-4,0,-1)B(−4,0,−1),
  • direction vector of L2L_2L2​ is d⃗2=(3,−2,−2)\vec d_2=(3,-2,-2)d2​=(3,−2,−2).
  1. Use formula for shortest distance between two skew lines

For two lines

r⃗=a⃗+λd⃗1,r⃗=b⃗+μd⃗2,\vec r=\vec a+\lambda \vec d_1, \qquad \vec r=\vec b+\mu \vec d_2,r=a+λd1​,r=b+μd2​,

the shortest distance is

D=∣(b⃗−a⃗)⋅(d⃗1×d⃗2)∣∣d⃗1×d⃗2∣.D=\frac{|(\vec b-\vec a)\cdot(\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}.D=∣d1​×d2​∣∣(b−a)⋅(d1​×d2​)∣​.

Given shortest distance D=9D=9D=9.

  1. Compute d⃗1×d⃗2\vec d_1\times \vec d_2d1​×d2​
d⃗1×d⃗2=∣i^j^k^1−223−2−2∣\vec d_1\times \vec d_2= \begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & -2 & 2\\ 3 & -2 & -2 \end{vmatrix}d1​×d2​=​i^13​j^​−2−2​k^2−2​​ =i^[(−2)(−2)−2(−2)]−j^[1(−2)−2(3)]+k^[1(−2)−(−2)(3)]=\hat i\big[(-2)(-2)-2(-2)\big] -\hat j\big[1(-2)-2(3)\big] +\hat k\big[1(-2)-(-2)(3)\big]=i^[(−2)(−2)−2(−2)]−j^​[1(−2)−2(3)]+k^[1(−2)−(−2)(3)] =i^(4+4)−j^(−2−6)+k^(−2+6)=8i^+8j^+4k^.=\hat i(4+4)-\hat j(-2-6)+\hat k(-2+6) =8\hat i+8\hat j+4\hat k.=i^(4+4)−j^​(−2−6)+k^(−2+6)=8i^+8j^​+4k^.

Thus,

d⃗1×d⃗2=(8,8,4)=4(2,2,1).\vec d_1\times \vec d_2=(8,8,4)=4(2,2,1).d1​×d2​=(8,8,4)=4(2,2,1).

Its magnitude is

∣d⃗1×d⃗2∣=82+82+42=64+64+16=144=12.|\vec d_1\times \vec d_2|=\sqrt{8^2+8^2+4^2}= \sqrt{64+64+16}=\sqrt{144}=12.∣d1​×d2​∣=82+82+42​=64+64+16​=144​=12.
  1. Compute b⃗−a⃗\vec b-\vec ab−a
b⃗−a⃗=(−4,0,−1)−(α,2,2)=(−4−α,−2,−3).\vec b-\vec a = (-4,0,-1)-(\alpha,2,2)=(-4-\alpha,-2,-3).b−a=(−4,0,−1)−(α,2,2)=(−4−α,−2,−3).

Now,

(b⃗−a⃗)⋅(d⃗1×d⃗2)=(−4−α,−2,−3)⋅(8,8,4).(\vec b-\vec a)\cdot(\vec d_1\times \vec d_2) =(-4-\alpha,-2,-3)\cdot(8,8,4).(b−a)⋅(d1​×d2​)=(−4−α,−2,−3)⋅(8,8,4). =8(−4−α)+8(−2)+4(−3)=−32−8α−16−12=−60−8α.=8(-4-\alpha)+8(-2)+4(-3) =-32-8\alpha-16-12 =-60-8\alpha.=8(−4−α)+8(−2)+4(−3)=−32−8α−16−12=−60−8α.

So,

D=∣−60−8α∣12=9.D=\frac{|{-60-8\alpha}|}{12}=9.D=12∣−60−8α∣​=9.
  1. Solve for α\alphaα
∣−60−8α∣12=9  ⟹  ∣−60−8α∣=108.\frac{|{-60-8\alpha}|}{12}=9 \implies |{-60-8\alpha}|=108.12∣−60−8α∣​=9⟹∣−60−8α∣=108. ∣8α+60∣=108.|8\alpha+60|=108.∣8α+60∣=108.

So,

8α+60=108or8α+60=−108.8\alpha+60=108 \quad \text{or} \quad 8\alpha+60=-108.8α+60=108or8α+60=−108.

From the first,

8α=48  ⟹  α=6.8\alpha=48 \implies \alpha=6.8α=48⟹α=6.

From the second,

8α=−168  ⟹  α=−21.8\alpha=-168 \implies \alpha=-21.8α=−168⟹α=−21.

But given α>0\alpha>0α>0, therefore

α=6.\boxed{\alpha=6}.α=6​.
  1. Compare with stored answer

Stored correct answer: 666

Our derived answer is also 666, so the answer agrees.

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