If the shortest distance between the line joining the points (1, 2, 3) and (2, 3, 4), and the line x - 1 \over 2 = y + 1 \over - 1 = z - 2 \over 0 is \alpha , then 28 \alpha^2 is equal to \underline\hspace2cm .

3D Geometry PYQ — JEE Main 2023 · 25 Jan · Shift 2 · Q45 with Solution

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Correct answer: 18

Write the two lines in vector form

The first line passes through the points $A(1,2,3)$ and $B(2,3,4)$ .

Its direction vector is $\vec d_1 = B-A = (2-1,3-2,4-3) = (1,1,1).$

So line $L_1$ can be written as $\vec r = (1,2,3) + \lambda(1,1,1).$

The second line is $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-2}{0}.$

Hence it passes through the point $C(1,-1,2)$ and has direction vector $\vec d_2 = (2,-1,0).$

So line $L_2$ is $\vec r = (1,-1,2) + \mu(2,-1,0).$
Use formula for shortest distance between two skew lines

The shortest distance is $\alpha = \frac{|(\vec{CA})\cdot (\vec d_1 \times \vec d_2)|}{|\vec d_1 \times \vec d_2|},$ where $\vec{CA} = A-C = (1,2,3)-(1,-1,2) = (0,3,1).$
Compute the cross product
$\begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 1 & 1 \\ 2 & -1 & 0 \end{vmatrix}.$$ Expanding, $$\vec d_1 \times \vec d_2 = \hat i(1\cdot 0 - 1\cdot (-1)) - \hat j(1\cdot 0 - 1\cdot 2) + \hat k(1\cdot (-1) - 1\cdot 2).$$ $$\vec d_1 \times \vec d_2 = (1,2,-3).$$ Therefore, $$|\vec d_1 \times \vec d_2| = \sqrt{1^2+2^2+(-3)^2} = \sqrt{14}.$$$
Compute the scalar triple product

$\vec{CA}\cdot (\vec d_1 \times \vec d_2) = (0,3,1)\cdot (1,2,-3).$

$= 0\cdot 1 + 3\cdot 2 + 1\cdot (-3) = 3.$

Hence, $\alpha = \frac{|3|}{\sqrt{14}} = \frac{3}{\sqrt{14}}.$
Find $28\alpha^2$

$\alpha^2 = \frac{9}{14}.$

Therefore, $28\alpha^2 = 28\cdot \frac{9}{14} = 18.$
Final answer

$\boxed{18}$
Comparison with stored answer

Stored correct answer = $18$ .

Our derived answer matches it.

View written solutionFree

Correct answer: 18

Write the two lines in vector form

The first line passes through the points $A(1,2,3)$ and $B(2,3,4)$ .

Its direction vector is $\vec d_1 = B-A = (2-1,3-2,4-3) = (1,1,1).$

So line $L_1$ can be written as $\vec r = (1,2,3) + \lambda(1,1,1).$

The second line is $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-2}{0}.$

Hence it passes through the point $C(1,-1,2)$ and has direction vector $\vec d_2 = (2,-1,0).$

So line $L_2$ is $\vec r = (1,-1,2) + \mu(2,-1,0).$
Use formula for shortest distance between two skew lines

The shortest distance is $\alpha = \frac{|(\vec{CA})\cdot (\vec d_1 \times \vec d_2)|}{|\vec d_1 \times \vec d_2|},$ where $\vec{CA} = A-C = (1,2,3)-(1,-1,2) = (0,3,1).$
Compute the cross product
$\begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 1 & 1 \\ 2 & -1 & 0 \end{vmatrix}.$$ Expanding, $$\vec d_1 \times \vec d_2 = \hat i(1\cdot 0 - 1\cdot (-1)) - \hat j(1\cdot 0 - 1\cdot 2) + \hat k(1\cdot (-1) - 1\cdot 2).$$ $$\vec d_1 \times \vec d_2 = (1,2,-3).$$ Therefore, $$|\vec d_1 \times \vec d_2| = \sqrt{1^2+2^2+(-3)^2} = \sqrt{14}.$$$
Compute the scalar triple product

$\vec{CA}\cdot (\vec d_1 \times \vec d_2) = (0,3,1)\cdot (1,2,-3).$

$= 0\cdot 1 + 3\cdot 2 + 1\cdot (-3) = 3.$

Hence, $\alpha = \frac{|3|}{\sqrt{14}} = \frac{3}{\sqrt{14}}.$
Find $28\alpha^2$

$\alpha^2 = \frac{9}{14}.$

Therefore, $28\alpha^2 = 28\cdot \frac{9}{14} = 18.$
Final answer

$\boxed{18}$
Comparison with stored answer

Stored correct answer = $18$ .

Our derived answer matches it.