JEE MAIN - Mathematics (2021 - 16th March Evening Shift - No. 4)
If the foot of the perpendicular from point (4, 3, 8) on the line $${L_1}:{{x - a} \over l} = {{y - 2} \over 3} = {{z - b} \over 4}$$, l $$\ne$$ 0 is (3, 5, 7), then the shortest distance between the line L1 and line $${L_2}:{{x - 2} \over 3} = {{y - 4} \over 4} = {{z - 5} \over 5}$$ is equal to :
$${1 \over {\sqrt 6 }}$$
$${1 \over 2}$$
$${1 \over {\sqrt 3 }}$$
$$\sqrt {{2 \over 3}} $$
Explanation
(3, 5, 7) lie on given line L1
$${{3 - a} \over l} = {3 \over 3} = {{7 - b} \over 4}$$
$${{7 - b} \over 4} = 1 \Rightarrow b = 3$$
$${{3 - a} \over l} = 1 \Rightarrow 3 - a = l$$
M(4, 3, 8)
N(3, 5, 7)
DR'S of MN = (1, $$-$$2, 1)
MN $$ \bot $$ line L1
(1)(l) + ($$-$$2)(3) + 4(1) = 0
$$ \Rightarrow $$ l = 2
a = 1
a = 1, b = 3, l = 2
$${{x - 1} \over 2} = {{y - 2} \over 3} = {{z - 3} \over 4}$$
$${{x - 2} \over 3} = {{y - 4} \over 4} = {{z - 5} \over 5}$$
$$A = \, < 1,2,3 > $$
$$B = \, < 2,4,5 > $$
$$\overrightarrow {AB} = \, < 1,2,2 > $$
$$\overrightarrow p = 2\widehat i + 3\widehat j + 4\widehat k$$
$$\overrightarrow q = 3\widehat i + 4\widehat j + 5\widehat k$$
$$\overrightarrow p \, \times \overrightarrow q = - \widehat i + 2\widehat j - \widehat k$$
Shortest distance = $$\left| {{{\overrightarrow {AB} \,.\,(\overrightarrow p \times \overrightarrow q )} \over {|\overrightarrow p \times \overrightarrow q |}}} \right|\, = {1 \over {\sqrt 6 }}$$
$${{3 - a} \over l} = {3 \over 3} = {{7 - b} \over 4}$$
$${{7 - b} \over 4} = 1 \Rightarrow b = 3$$
$${{3 - a} \over l} = 1 \Rightarrow 3 - a = l$$
M(4, 3, 8)
N(3, 5, 7)
DR'S of MN = (1, $$-$$2, 1)
MN $$ \bot $$ line L1
(1)(l) + ($$-$$2)(3) + 4(1) = 0
$$ \Rightarrow $$ l = 2
a = 1
a = 1, b = 3, l = 2
$${{x - 1} \over 2} = {{y - 2} \over 3} = {{z - 3} \over 4}$$
$${{x - 2} \over 3} = {{y - 4} \over 4} = {{z - 5} \over 5}$$
$$A = \, < 1,2,3 > $$
$$B = \, < 2,4,5 > $$
$$\overrightarrow {AB} = \, < 1,2,2 > $$
$$\overrightarrow p = 2\widehat i + 3\widehat j + 4\widehat k$$
$$\overrightarrow q = 3\widehat i + 4\widehat j + 5\widehat k$$
$$\overrightarrow p \, \times \overrightarrow q = - \widehat i + 2\widehat j - \widehat k$$
Shortest distance = $$\left| {{{\overrightarrow {AB} \,.\,(\overrightarrow p \times \overrightarrow q )} \over {|\overrightarrow p \times \overrightarrow q |}}} \right|\, = {1 \over {\sqrt 6 }}$$
Comments (0)
