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+1 vote
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in Mathematics by (30.8k points)
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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 |λ| is equal to _______. 

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1 Answer

+1 vote
by (32.1k points)

Correct answer is 43

\(\bar{a}_{1}=\lambda \hat{i}+2 \hat{j}+\hat{k}\)

\(\bar{a}_{2}=-2 \hat{i}-5 \hat{j}+4 \hat{k}\)

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

\(\vec{q}-=-3 \hat{i}+2 \hat{j}+4 \hat{k}\)

\((\lambda+2) \hat{i}+7 \hat{j}-3 \hat{k}=\bar{a}_{1}-\bar{a}_{2}\)

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

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

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

\(132=|6 \lambda+126|\)

\(\lambda=1, \lambda=-43\)

\(| \lambda|=43\)

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