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STD 12 - 11. three dimension geometry — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 11. three dimension geometry1 Mark
MCQ
If the shortest distance between the lines $\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}$ and $\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}$ is $\frac{44}{\sqrt{30}}$, then the largest possible value of $|\lambda|$ is equal to ..........
  • A
    $45$
  • B
    $49$
  • ✓
    $43$
  • D
    $40$
✓

Answer

Correct option: C.
$43$
c
$ \overline{\mathrm{a}}_1=\lambda \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} $

$ \overline{\mathrm{a}}_2=-2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} $

$ \overrightarrow{\mathrm{p}}-=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}} $

$ \overrightarrow{\mathrm{q}}-=-3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} $

$(\lambda+2) \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}=\overline{\mathrm{a}}_1-\overline{\mathrm{a}}_2$

$ \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}-=-6 \hat{\mathrm{i}}-15 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

$ \frac{44}{\sqrt{30}}=\frac{|-6 \lambda-12-105-9|}{\sqrt{(-6)^2+(-15)^2+3^2}} $

$ \frac{44}{\sqrt{30}}=\frac{|6 \lambda+126|}{3 \sqrt{30}} $

$ 132=|6 \lambda+126| $

$ \lambda=1, \lambda=-43 $

$ |\lambda|=43$

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