JEE MAIN - Mathematics (2021 - 22th July Evening Shift - No. 7)
If the shortest distance between the straight lines $$3(x - 1) = 6(y - 2) = 2(z - 1)$$ and $$4(x - 2) = 2(y - \lambda ) = (z - 3),\lambda \in R$$ is $${1 \over {\sqrt {38} }}$$, then the integral value of $$\lambda$$ is equal to :
3
2
5
$$-$$1
Explanation
$${L_1}:{{(x - 1)} \over 2} = {{(y - 2)} \over 1} = {{(z - 1)} \over 3}\overrightarrow {{r_1}} = 2\widehat i + \widehat j + 3\widehat k$$
$${L_2}:{{(x - 2)} \over 1} = {{y - \lambda } \over 2} = {{z - 3} \over 4}\overrightarrow {{r_2}} = \widehat i + 2\widehat j + 4\widehat k$$
_22th_July_Evening_Shift_en_7_1.png)
Shortest distance = Projection of $${\overrightarrow a }$$ on $${\overrightarrow {{r_1}} \times \overrightarrow {{r_2}} }$$
$$ = {{\left| {\overrightarrow a .\left( {\overrightarrow {{r_1}} \times \overrightarrow {{r_2}} } \right)} \right|} \over {\left| {\overrightarrow {{r_1}} \times \overrightarrow {{r_2}} } \right|}}$$
$$\left| {\overrightarrow a .\left( {\overrightarrow {{r_1}} \times \overrightarrow {{r_2}} } \right)} \right| = \left| {\matrix{ 1 & {\lambda - 2} & 2 \cr 2 & 1 & 3 \cr 1 & 2 & 4 \cr } } \right| = \left| {14 - 5\lambda } \right|$$
$$\left| {\overrightarrow {{r_1}} \times \overrightarrow {{r_2}} } \right| = \sqrt {38} $$
$$\therefore$$ $${1 \over {\sqrt {38} }} = {{\left| {14 - 5\lambda } \right|} \over {\sqrt {38} }}$$
$$ \Rightarrow \left| {14 - 5\lambda } \right| = 1$$
$$ \Rightarrow 14 - 5\lambda = 1$$ or $$14 - 5\lambda = - 1$$
$$ \Rightarrow \lambda = {{13} \over 5}$$ or 3
$$\therefore$$ Integral value of $$\lambda$$ = 3.
$${L_2}:{{(x - 2)} \over 1} = {{y - \lambda } \over 2} = {{z - 3} \over 4}\overrightarrow {{r_2}} = \widehat i + 2\widehat j + 4\widehat k$$
Shortest distance = Projection of $${\overrightarrow a }$$ on $${\overrightarrow {{r_1}} \times \overrightarrow {{r_2}} }$$
$$ = {{\left| {\overrightarrow a .\left( {\overrightarrow {{r_1}} \times \overrightarrow {{r_2}} } \right)} \right|} \over {\left| {\overrightarrow {{r_1}} \times \overrightarrow {{r_2}} } \right|}}$$
$$\left| {\overrightarrow a .\left( {\overrightarrow {{r_1}} \times \overrightarrow {{r_2}} } \right)} \right| = \left| {\matrix{ 1 & {\lambda - 2} & 2 \cr 2 & 1 & 3 \cr 1 & 2 & 4 \cr } } \right| = \left| {14 - 5\lambda } \right|$$
$$\left| {\overrightarrow {{r_1}} \times \overrightarrow {{r_2}} } \right| = \sqrt {38} $$
$$\therefore$$ $${1 \over {\sqrt {38} }} = {{\left| {14 - 5\lambda } \right|} \over {\sqrt {38} }}$$
$$ \Rightarrow \left| {14 - 5\lambda } \right| = 1$$
$$ \Rightarrow 14 - 5\lambda = 1$$ or $$14 - 5\lambda = - 1$$
$$ \Rightarrow \lambda = {{13} \over 5}$$ or 3
$$\therefore$$ Integral value of $$\lambda$$ = 3.
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