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3D Geometry question

2024 · 5 Apr · Shift 1 · Q34
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  5. /2024 · 5 Apr · Shift 1 · Q34

3D Geometry question

2024 · 5 Apr · Shift 1 · Q34

JEE MainMathematics3D GeometryMCQ+4 / −1
If the line 2−x3=3y−24λ+1=4−z\frac{2-x}{3}=\frac{3 y-2}{4 \lambda+1}=4-z32−x​=4λ+13y−2​=4−z makes a right angle with the line x+33μ=1−2y6=5−z7\frac{x+3}{3 \mu}=\frac{1-2 y}{6}=\frac{5-z}{7}3μx+3​=61−2y​=75−z​, then 4λ+9μ4 \lambda+9 \mu4λ+9μ is equal to :
  1. A
    4
  2. B
    13
  3. C
    5
  4. D
    6
View written solutionFree

Correct answer: D

  1. Write each line in symmetric/parametric form and extract direction ratios

The first line is

2−x3=3y−24λ+1=4−z.\frac{2-x}{3}=\frac{3y-2}{4\lambda+1}=4-z.32−x​=4λ+13y−2​=4−z.

Let the common value be ttt. Then

2−x3=t⇒x=2−3t,\frac{2-x}{3}=t \Rightarrow x=2-3t,32−x​=t⇒x=2−3t, 3y−24λ+1=t⇒3y−2=(4λ+1)t⇒y=2+(4λ+1)t3,\frac{3y-2}{4\lambda+1}=t \Rightarrow 3y-2=(4\lambda+1)t \Rightarrow y=\frac{2+(4\lambda+1)t}{3},4λ+13y−2​=t⇒3y−2=(4λ+1)t⇒y=32+(4λ+1)t​, 4−z=t⇒z=4−t.4-z=t \Rightarrow z=4-t.4−z=t⇒z=4−t.

So the direction ratios of the first line are

(−3, 4λ+13, −1).(-3,\ \frac{4\lambda+1}{3},\ -1).(−3, 34λ+1​, −1).

Multiplying by 333, an equivalent set is

(−9, 4λ+1, −3).(-9,\ 4\lambda+1,\ -3).(−9, 4λ+1, −3).
  1. For the second line

Given

x+33μ=1−2y6=5−z7.\frac{x+3}{3\mu}=\frac{1-2y}{6}=\frac{5-z}{7}.3μx+3​=61−2y​=75−z​.

Let the common value be sss. Then

x+33μ=s⇒x=−3+3μs,\frac{x+3}{3\mu}=s \Rightarrow x=-3+3\mu s,3μx+3​=s⇒x=−3+3μs, 1−2y6=s⇒1−2y=6s⇒y=1−6s2,\frac{1-2y}{6}=s \Rightarrow 1-2y=6s \Rightarrow y=\frac{1-6s}{2},61−2y​=s⇒1−2y=6s⇒y=21−6s​, 5−z7=s⇒z=5−7s.\frac{5-z}{7}=s \Rightarrow z=5-7s.75−z​=s⇒z=5−7s.

Hence direction ratios are

(3μ, −3, −7).(3\mu,\ -3,\ -7).(3μ, −3, −7).
  1. Use the condition that the lines are perpendicular

For two lines to make a right angle, the dot product of their direction ratios must be zero:

(−9)(3μ)+(4λ+1)(−3)+(−3)(−7)=0.(-9)(3\mu)+(4\lambda+1)(-3)+(-3)(-7)=0.(−9)(3μ)+(4λ+1)(−3)+(−3)(−7)=0.

Now simplify:

−27μ−3(4λ+1)+21=0-27\mu-3(4\lambda+1)+21=0−27μ−3(4λ+1)+21=0 −27μ−12λ−3+21=0-27\mu-12\lambda-3+21=0−27μ−12λ−3+21=0 −27μ−12λ+18=0-27\mu-12\lambda+18=0−27μ−12λ+18=0 12λ+27μ=18.12\lambda+27\mu=18.12λ+27μ=18.

Divide by 333:

4λ+9μ=6.4\lambda+9\mu=6.4λ+9μ=6.
  1. Match with the options

Thus,

4λ+9μ=6.4\lambda+9\mu=6.4λ+9μ=6.

So the correct option is D.

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