The shortest distance between the lines x-44 = y+25 = z+33 and x-13 = y-34 = z-42 is:

Given: The lines are x-44 = y+25 = z+33 and x-13 = y-34 = z-42. Find: The shortest distance between these two lines. The solution contains incorrect intermediate working for different lines, but the hint correctly indicates the required method: find a vector perpendicular to both direction vectors and project the joining vector onto it. Take points on the two lines as A(4,-2,-3), B(1,3,4) and their direction vectors as d1 = 4,5,3, d2 = 3,4,2. A vector perpendicular to both lines is their cross product: d1 d2 = vmatrix i & j & k 4 & 5 & 3 3 & 4 & 2 vmatrix = -2, 1, 1. Its magnitude is | d1 d2 | = (-2)^2 + 1^2 + 1^2 = 6. The joining vector is AB = 1-4, 3-(-2), 4-(-3) = -3, 5, 7. The shortest distance between two skew lines is Distance = | AB ( d1 d2)|| d1 d2|. Now, AB ( d1 d2) = -3,5,7 -2,1,1 = 6 + 5 + 7 = 18. Therefore, Distance = 18 6 = 3 6. Therefore, the shortest distance is 3 6, so the correct option is B.

The shortest distance between the lines x-44 = y+25 = z+33 and x-… — Solution | JEE Mathematics

MCQMediumJEE 2023Skew Lines & Shortest Distance

The shortest distance between the lines $\frac{x-4}{4} = \frac{y+2}{5} = \frac{z+3}{3}$ and $\frac{x-1}{3} = \frac{y-3}{4} = \frac{z-4}{2}$ is:

A
$2\sqrt{6}$
B
$3\sqrt{6}$
C
$6\sqrt{3}$
D
$6\sqrt{2}$

Answer

Correct answer:B

Step-by-step solution

Solution and Explanation

Given: The lines are $\frac{x-4}{4} = \frac{y+2}{5} = \frac{z+3}{3}$ and $\frac{x-1}{3} = \frac{y-3}{4} = \frac{z-4}{2}$ .

Find: The shortest distance between these two lines.

The solution contains incorrect intermediate working for different lines, but the hint correctly indicates the required method: find a vector perpendicular to both direction vectors and project the joining vector onto it.

Take points on the two lines as

A(4,-2,-3), \qquad B(1,3,4)

and their direction vectors as

\vec{d_1} = \langle 4,5,3 \rangle, \qquad \vec{d_2} = \langle 3,4,2 \rangle.

A vector perpendicular to both lines is their cross product:

\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & 3 \\ 3 & 4 & 2 \end{vmatrix} = \langle -2, 1, 1 \rangle.

Its magnitude is

\left|\vec{d_1} \times \vec{d_2}\right| = \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{6}.

The joining vector is

\overrightarrow{AB} = \langle 1-4, 3-(-2), 4-(-3) \rangle = \langle -3, 5, 7 \rangle.

The shortest distance between two skew lines is

\text{Distance} = \frac{|\overrightarrow{AB} \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}.

Now,

\overrightarrow{AB} \cdot (\vec{d_1} \times \vec{d_2}) = \langle -3,5,7 \rangle \cdot \langle -2,1,1 \rangle = 6 + 5 + 7 = 18.

Therefore,

\text{Distance} = \frac{18}{\sqrt{6}} = 3\sqrt{6}.

Therefore, the shortest distance is $3\sqrt{6}$ , so the correct option is B.

Common mistakes

Using the direction ratios incorrectly. The direction vectors are $\langle 4,5,3 \rangle$ and $\langle 3,4,2 \rangle$ , taken directly from the symmetric forms. Reading signs from the point coordinates into the direction vector gives a wrong result.
Choosing the wrong points on the lines. For $\frac{x-4}{4} = \frac{y+2}{5} = \frac{z+3}{3}$ , the point is $A(4,-2,-3)$ , not $A(4,2,3)$ . The constants must be read with their correct signs.
Computing the cross product incorrectly. The shortest distance formula depends on $\vec{d_1} \times \vec{d_2}$ being perpendicular to both lines. Any sign error in the determinant changes the final distance.
Using the magnitude of the joining vector directly. The shortest distance is not $|\overrightarrow{AB}|$ . You must project the joining vector onto the common perpendicular direction using $\frac{|\overrightarrow{AB} \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}$ .

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