OP Malhotra Class 12 Maths Solutions Chapter 23 Three Dimensional Geometry Exercise 23 (F)

 

Question 1. Find the length of the shortest distance between the lines: \( \frac{x-3}{1} = \frac{y-5}{-2} = \frac{z-7}{1} \) and \( \frac{x+1}{7} = \frac{y+1}{-6} = \frac{z+1}{1} \)
Answer: The given equations of the lines in Cartesian form are:
Line (1): \( \frac{x-3}{1} = \frac{y-5}{-2} = \frac{z-7}{1} \)
Line (2): \( \frac{x+1}{7} = \frac{y+1}{-6} = \frac{z+1}{1} \)
Any general point P on line (1) can be written as \( P(t + 3, -2t + 5, t + 7) \).
Any general point Q on line (2) can be written as \( Q(7s - 1, -6s - 1, s - 1) \).
The direction ratios (D'ratios) of the line segment PQ are then \( < (7s - 1) - (t + 3), (-6s - 1) - (-2t + 5), (s - 1) - (t + 7) > \).
This simplifies to \( < 7s - t - 4, -6s + 2t - 6, s - t - 8 > \).
Since PQ represents the shortest distance, it must be perpendicular to both line (1) and line (2).
For PQ to be perpendicular to line (1) (whose direction ratios are \( <1, -2, 1> \)):
\( (7s - t - 4)(1) + (-6s + 2t - 6)(-2) + (s - t - 8)(1) = 0 \)
\( 7s - t - 4 + 12s - 4t + 12 + s - t - 8 = 0 \)
\( 20s - 6t = 0 \) (Equation 3)
For PQ to be perpendicular to line (2) (whose direction ratios are \( <7, -6, 1> \)):
\( (7s - t - 4)(7) + (-6s + 2t - 6)(-6) + (s - t - 8)(1) = 0 \)
\( 49s - 7t - 28 + 36s - 12t + 36 + s - t - 8 = 0 \)
\( 86s - 20t = 0 \) (Equation 4)
Now, we solve equations (3) and (4):
From (3), \( 20s = 6t \implies t = \frac{20s}{6} = \frac{10s}{3} \).
Substitute t into (4): \( 86s - 20(\frac{10s}{3}) = 0 \)
\( 86s - \frac{200s}{3} = 0 \)
Multiply by 3: \( 258s - 200s = 0 \)
\( 58s = 0 \implies s = 0 \).
Since \( s = 0 \), then \( t = \frac{10(0)}{3} = 0 \).
So, \( s = 0 \) and \( t = 0 \).
Now, find the coordinates of P and Q:
\( P = (t + 3, -2t + 5, t + 7) = (0 + 3, -2(0) + 5, 0 + 7) = (3, 5, 7) \)
\( Q = (7s - 1, -6s - 1, s - 1) = (7(0) - 1, -6(0) - 1, 0 - 1) = (-1, -1, -1) \)
The shortest distance (length of PQ) is given by the distance formula:
\( |PQ| = \sqrt{((3 - (-1))^2 + (5 - (-1))^2 + (7 - (-1))^2)} \)
\( |PQ| = \sqrt{((3 + 1)^2 + (5 + 1)^2 + (7 + 1)^2)} \)
\( |PQ| = \sqrt{(4^2 + 6^2 + 8^2)} \)
\( |PQ| = \sqrt{(16 + 36 + 64)} \)
\( |PQ| = \sqrt{116} \)
\( |PQ| = \sqrt{4 \times 29} \)
\( |PQ| = 2\sqrt{29} \) units.
In simple words: First, we write down the general points on each line using variables. Then we find the line segment connecting these points. Since this segment is the shortest distance, it must be exactly perpendicular to both given lines. We use this rule to find the values of our variables, which gives us the exact points. Finally, we calculate the distance between these two specific points.

🎯 Exam Tip: Remember to set up two perpendicularity conditions (dot product of direction vectors is zero) to find the parameters 's' and 't'. If s and t are 0, it means the lines are intersecting or collinear at the origin of reference for the parameters.

 

Question 2. Find the length of the shortest distance between the lines: \( \frac{x+3}{-4} = \frac{y-6}{3} = \frac{z}{2} \) and \( \frac{x+2}{-4} = \frac{y}{1} = \frac{z-7}{1} \)
Answer: The given equations of the lines are:
Line (1): \( \frac{x+3}{-4} = \frac{y-6}{3} = \frac{z}{2} = t \) (say)
Line (2): \( \frac{x+2}{-4} = \frac{y}{1} = \frac{z-7}{1} = r \) (say)
Any point P on line (1) is \( P(-4t - 3, 3t + 6, 2t) \).
Any point Q on line (2) is \( Q(-4r - 2, r, r + 7) \).
The direction ratios of the line segment PQ are:
\( < (-4r - 2) - (-4t - 3), r - (3t + 6), (r + 7) - (2t) > \)
\( < -4r + 4t + 1, r - 3t - 6, r - 2t + 7 > \)
For PQ to be the line of shortest distance, it must be perpendicular to both line (1) and line (2).
Direction ratios of line (1) are \( < -4, 3, 2 > \).
Direction ratios of line (2) are \( < -4, 1, 1 > \).
For PQ perpendicular to line (1):
\( (-4r + 4t + 1)(-4) + (r - 3t - 6)(3) + (r - 2t + 7)(2) = 0 \)
\( 16r - 16t - 4 + 3r - 9t - 18 + 2r - 4t + 14 = 0 \)
\( 21r - 29t - 8 = 0 \) (Equation 1)
For PQ perpendicular to line (2):
\( (-4r + 4t + 1)(-4) + (r - 3t - 6)(1) + (r - 2t + 7)(1) = 0 \)
\( 16r - 16t - 4 + r - 3t - 6 + r - 2t + 7 = 0 \)
\( 18r - 21t - 3 = 0 \)
This simplifies to \( 6r - 7t - 1 = 0 \) (Equation 2)
Now, we solve equations (1) and (2).
From (2), \( 6r = 7t + 1 \implies r = \frac{7t+1}{6} \).
Substitute r into (1):
\( 21(\frac{7t+1}{6}) - 29t - 8 = 0 \)
Multiply by 6: \( 21(7t + 1) - 174t - 48 = 0 \)
\( 147t + 21 - 174t - 48 = 0 \)
\( -27t - 27 = 0 \)
\( -27t = 27 \implies t = -1 \).
Substitute t back into the expression for r:
\( r = \frac{7(-1) + 1}{6} = \frac{-7+1}{6} = \frac{-6}{6} = -1 \).
So, \( r = -1 \) and \( t = -1 \).
Now, find the coordinates of P and Q:
\( P = (-4t - 3, 3t + 6, 2t) = (-4(-1) - 3, 3(-1) + 6, 2(-1)) = (4 - 3, -3 + 6, -2) = (1, 3, -2) \)
\( Q = (-4r - 2, r, r + 7) = (-4(-1) - 2, -1, -1 + 7) = (4 - 2, -1, 6) = (2, -1, 6) \)
The shortest distance (length of PQ) is:
\( |PQ| = \sqrt{((2 - 1)^2 + (-1 - 3)^2 + (6 - (-2))^2)} \)
\( |PQ| = \sqrt{((1)^2 + (-4)^2 + (8)^2)} \)
\( |PQ| = \sqrt{(1 + 16 + 64)} \)
\( |PQ| = \sqrt{81} = 9 \) units.
**Aliter (Alternative Method):**
We know that the shortest distance (S.D.) between two skew lines \( \frac{x-x_1}{l_1} = \frac{y-y_1}{m_1} = \frac{z-z_1}{n_1} \) and \( \frac{x-x_2}{l_2} = \frac{y-y_2}{m_2} = \frac{z-z_2}{n_2} \) is given by:
\[ S.D. = \frac{\left| \begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{array} \right|}{\sqrt{(l_1 m_2 - l_2 m_1)^2 + (m_1 n_2 - m_2 n_1)^2 + (n_1 l_2 - n_2 l_1)^2}} \]
From the given lines:
Line (1): \( \frac{x-(-3)}{-4} = \frac{y-6}{3} = \frac{z-0}{2} \)
So, \( x_1 = -3, y_1 = 6, z_1 = 0 \) and \( l_1 = -4, m_1 = 3, n_1 = 2 \).
Line (2): \( \frac{x-(-2)}{-4} = \frac{y-0}{1} = \frac{z-7}{1} \)
So, \( x_2 = -2, y_2 = 0, z_2 = 7 \) and \( l_2 = -4, m_2 = 1, n_2 = 1 \).
First, calculate the determinant (numerator):
\( x_2-x_1 = -2 - (-3) = 1 \)
\( y_2-y_1 = 0 - 6 = -6 \)
\( z_2-z_1 = 7 - 0 = 7 \)
\[ \begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{array} = \left| \begin{array}{ccc} 1 & -6 & 7 \\ -4 & 3 & 2 \\ -4 & 1 & 1 \end{array} \right| \]
Expanding along R1:
\( = 1(3 \times 1 - 2 \times 1) - (-6)(-4 \times 1 - 2 \times (-4)) + 7(-4 \times 1 - 3 \times (-4)) \)
\( = 1(3 - 2) + 6(-4 + 8) + 7(-4 + 12) \)
\( = 1(1) + 6(4) + 7(8) \)
\( = 1 + 24 + 56 = 81 \).
Next, calculate the denominator \( \sqrt{(l_1 m_2 - l_2 m_1)^2 + (m_1 n_2 - m_2 n_1)^2 + (n_1 l_2 - n_2 l_1)^2} \):
\( l_1 m_2 - l_2 m_1 = (-4)(1) - (-4)(3) = -4 + 12 = 8 \)
\( m_1 n_2 - m_2 n_1 = (3)(1) - (1)(2) = 3 - 2 = 1 \)
\( n_1 l_2 - n_2 l_1 = (2)(-4) - (1)(-4) = -8 + 4 = -4 \)
Denominator \( = \sqrt{(8)^2 + (1)^2 + (-4)^2} = \sqrt{64 + 1 + 16} = \sqrt{81} = 9 \).
Therefore, \( S.D. = \frac{|81|}{9} = 9 \) units.
In simple words: This method uses a direct formula. We find the coordinates of points on each line and their direction ratios. Then, we plug these values into a special formula involving a determinant for the top part and a square root for the bottom part. This formula directly calculates the shortest distance between the two lines. Both methods give the same result.

🎯 Exam Tip: When using the determinant formula for shortest distance, be careful with the signs of \( x_1, y_1, z_1, x_2, y_2, z_2 \) and the order of operations for the cross product terms in the denominator.

 

Question 3. Find the magnitude and the equations of the line of shortest distance between the two lines: \( \frac{x-3}{-1} = \frac{y-4}{2} = \frac{z+2}{1} \) and \( \frac{x-1}{1} = \frac{y+7}{3} = \frac{z+2}{2} \)
Answer: The given equations of the lines are:
Line (1): \( \frac{x-3}{-1} = \frac{y-4}{2} = \frac{z+2}{1} = t \) (say)
Line (2): \( \frac{x-1}{1} = \frac{y+7}{3} = \frac{z+2}{2} = r \) (say)
Any point P on line (1) is \( P(-t + 3, 2t + 4, t - 2) \).
Any point Q on line (2) is \( Q(r + 1, 3r - 7, 2r - 2) \).
The direction ratios of the line segment PQ are:
\( < (r + 1) - (-t + 3), (3r - 7) - (2t + 4), (2r - 2) - (t - 2) > \)
\( < r + t - 2, 3r - 2t - 11, 2r - t > \)
For PQ to be the line of shortest distance, it must be perpendicular to both line (1) and line (2).
Direction ratios of line (1) are \( < -1, 2, 1 > \).
Direction ratios of line (2) are \( < 1, 3, 2 > \).
For PQ perpendicular to line (1):
\( (r + t - 2)(-1) + (3r - 2t - 11)(2) + (2r - t)(1) = 0 \)
\( -r - t + 2 + 6r - 4t - 22 + 2r - t = 0 \)
\( 7r - 6t - 20 = 0 \) (Equation 3)
For PQ perpendicular to line (2):
\( (r + t - 2)(1) + (3r - 2t - 11)(3) + (2r - t)(2) = 0 \)
\( r + t - 2 + 9r - 6t - 33 + 4r - 2t = 0 \)
\( 14r - 7t - 35 = 0 \)
Divide by 7: \( 2r - t - 5 = 0 \) (Equation 4)
Now, we solve equations (3) and (4).
From (4), \( t = 2r - 5 \).
Substitute t into (3):
\( 7r - 6(2r - 5) - 20 = 0 \)
\( 7r - 12r + 30 - 20 = 0 \)
\( -5r + 10 = 0 \)
\( -5r = -10 \implies r = 2 \).
Substitute r back into the expression for t:
\( t = 2(2) - 5 = 4 - 5 = -1 \).
So, \( r = 2 \) and \( t = -1 \).
Now, find the coordinates of P and Q:
\( P = (-t + 3, 2t + 4, t - 2) = (-(-1) + 3, 2(-1) + 4, -1 - 2) = (1 + 3, -2 + 4, -3) = (4, 2, -3) \)
\( Q = (r + 1, 3r - 7, 2r - 2) = (2 + 1, 3(2) - 7, 2(2) - 2) = (3, 6 - 7, 4 - 2) = (3, -1, 2) \)
The magnitude of the shortest distance (length of PQ) is:
\( |PQ| = \sqrt{((3 - 4)^2 + (-1 - 2)^2 + (2 - (-3))^2)} \)
\( |PQ| = \sqrt{((-1)^2 + (-3)^2 + (5)^2)} \)
\( |PQ| = \sqrt{(1 + 9 + 25)} \)
\( |PQ| = \sqrt{35} \) units.
The direction ratios of the line of shortest distance (PQ) are:
\( < 3 - 4, -1 - 2, 2 - (-3) > \)
\( < -1, -3, 5 > \)
The vector equation of the line of shortest distance passing through P(4, 2, -3) and having direction vector \( < -1, -3, 5 > \) is:
\( \vec{r} = (4\hat{i} + 2\hat{j} - 3\hat{k}) + \lambda(-\hat{i} - 3\hat{j} + 5\hat{k}) \).
In simple words: This problem asks for both the shortest distance and the equation of the line that forms this shortest distance. We start by finding general points on each line. Then, we make sure the line connecting them is perpendicular to both original lines, which helps us find the exact points. We calculate the distance between these points. Finally, we use one of these points and the direction ratios of the shortest distance line to write its vector equation.

🎯 Exam Tip: When finding the equation of the line of shortest distance, use one of the points (P or Q) and the direction ratios of the segment PQ. Ensure your direction ratios are consistent with the calculated points.

 

Question 4. Obtain the coordinates of the points where the shortest distance between the following lines meets them: \( \frac{x-23}{-6} = \frac{y-19}{-4} = \frac{z-25}{3}; \) and \( \frac{x-12}{-9} = \frac{y-1}{4} = \frac{z-5}{2} \)
Answer: The given equations of the lines are:
Line (1): \( \frac{x-23}{-6} = \frac{y-19}{-4} = \frac{z-25}{3} = t \) (say)
Line (2): \( \frac{x-12}{-9} = \frac{y-1}{4} = \frac{z-5}{2} = r \) (say)
Any point P on line (1) is \( P(-6t + 23, -4t + 19, 3t + 25) \).
Any point Q on line (2) is \( Q(-9r + 12, 4r + 1, 2r + 5) \).
The direction ratios of the line segment PQ are:
\( < (-9r + 12) - (-6t + 23), (4r + 1) - (-4t + 19), (2r + 5) - (3t + 25) > \)
\( < -9r + 6t - 11, 4r + 4t - 18, 2r - 3t - 20 > \)
For PQ to be the line of shortest distance, it must be perpendicular to both line (1) and line (2).
Direction ratios of line (1) are \( < -6, -4, 3 > \).
Direction ratios of line (2) are \( < -9, 4, 2 > \).
For PQ perpendicular to line (1):
\( (-9r + 6t - 11)(-6) + (4r + 4t - 18)(-4) + (2r - 3t - 20)(3) = 0 \)
\( 54r - 36t + 66 - 16r - 16t + 72 + 6r - 9t - 60 = 0 \)
\( 44r - 61t + 78 = 0 \) (Equation 3)
For PQ perpendicular to line (2):
\( (-9r + 6t - 11)(-9) + (4r + 4t - 18)(4) + (2r - 3t - 20)(2) = 0 \)
\( 81r - 54t + 99 + 16r + 16t - 72 + 4r - 6t - 40 = 0 \)
\( 101r - 44t - 13 = 0 \) (Equation 4)
Now, we solve equations (3) and (4). We can use the cross-multiplication method or substitution.
From (4), \( 101r = 44t + 13 \implies r = \frac{44t+13}{101} \).
Substitute r into (3):
\( 44(\frac{44t+13}{101}) - 61t + 78 = 0 \)
Multiply by 101: \( 44(44t + 13) - 61(101)t + 78(101) = 0 \)
\( 1936t + 572 - 6161t + 7878 = 0 \)
\( -4225t + 8450 = 0 \)
\( -4225t = -8450 \implies t = \frac{-8450}{-4225} = 2 \).
Substitute t back into the expression for r:
\( r = \frac{44(2) + 13}{101} = \frac{88 + 13}{101} = \frac{101}{101} = 1 \).
So, \( r = 1 \) and \( t = 2 \).
Now, find the coordinates of points P and Q where the shortest distance meets the lines:
\( P = (-6t + 23, -4t + 19, 3t + 25) = (-6(2) + 23, -4(2) + 19, 3(2) + 25) = (-12 + 23, -8 + 19, 6 + 25) = (11, 11, 31) \)
\( Q = (-9r + 12, 4r + 1, 2r + 5) = (-9(1) + 12, 4(1) + 1, 2(1) + 5) = (-9 + 12, 4 + 1, 2 + 5) = (3, 5, 7) \)
The coordinates of the points where the shortest distance meets the lines are P(11, 11, 31) and Q(3, 5, 7).
In simple words: We want to find the exact points on each line where the shortest distance touches them. We set up general points on each line using variables. Then, we use the rule that the shortest connecting line is perpendicular to both original lines. This helps us solve for the variable values. Once we have these values, we plug them back into the general point formulas to get the specific coordinates.

🎯 Exam Tip: Pay close attention to algebra when solving the simultaneous equations for 'r' and 't', as a small error can lead to incorrect point coordinates and distances. Always check the final points by ensuring they lie on their respective lines.

 

Question 5. Find the shortest distance between the pairs of lines whose equations are: \( \vec{r} = 6 \hat{i} + 2 \hat{j} + 2 \hat{k} + \lambda(\hat{i} - 2 \hat{j} + 2 \hat{k}) \) and \( \vec{r} = -4 \hat{i} - \hat{k} + \mu(3 \hat{j} - 2 \hat{k} - 2 \hat{k}) \)
Answer: The given equations of the lines are:
Line 1: \( \vec{r} = (6\hat{i} + 2\hat{j} + 2\hat{k}) + \lambda(\hat{i} - 2\hat{j} + 2\hat{k}) \)
Line 2: \( \vec{r} = (-4\hat{i} - \hat{k}) + \mu(3\hat{j} - 4\hat{k}) \) (combining \( -2\hat{k} - 2\hat{k} \) to \( -4\hat{k} \))
Comparing with the general form \( \vec{r} = \vec{a_1} + \lambda\vec{b_1} \) and \( \vec{r} = \vec{a_2} + \mu\vec{b_2} \):
\( \vec{a_1} = 6\hat{i} + 2\hat{j} + 2\hat{k} \)
\( \vec{b_1} = \hat{i} - 2\hat{j} + 2\hat{k} \)
\( \vec{a_2} = -4\hat{i} - \hat{k} \)
\( \vec{b_2} = 3\hat{j} - 4\hat{k} \)
First, find \( \vec{a_2} - \vec{a_1} \):
\( \vec{a_2} - \vec{a_1} = (-4\hat{i} - \hat{k}) - (6\hat{i} + 2\hat{j} + 2\hat{k}) \)
\( = -4\hat{i} - \hat{k} - 6\hat{i} - 2\hat{j} - 2\hat{k} \)
\( = -10\hat{i} - 2\hat{j} - 3\hat{k} \)
Next, find \( \vec{b_1} \times \vec{b_2} \):
\[ \vec{b_1} \times \vec{b_2} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 0 & 3 & -4 \end{array} \right| \]
\( = \hat{i}((-2)(-4) - (2)(3)) - \hat{j}((1)(-4) - (2)(0)) + \hat{k}((1)(3) - (-2)(0)) \)
\( = \hat{i}(8 - 6) - \hat{j}(-4 - 0) + \hat{k}(3 - 0) \)
\( = 2\hat{i} + 4\hat{j} + 3\hat{k} \)
Now, find \( |\vec{b_1} \times \vec{b_2}| \):
\( |\vec{b_1} \times \vec{b_2}| = \sqrt{2^2 + 4^2 + 3^2} = \sqrt{4 + 16 + 9} = \sqrt{29} \).
Finally, calculate the shortest distance (S.D.) using the formula for skew lines:
\[ S.D. = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} \]
Numerator: \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-10\hat{i} - 2\hat{j} - 3\hat{k}) \cdot (2\hat{i} + 4\hat{j} + 3\hat{k}) \)
\( = (-10)(2) + (-2)(4) + (-3)(3) \)
\( = -20 - 8 - 9 = -37 \).
Therefore, \( S.D. = \frac{|-37|}{\sqrt{29}} = \frac{37}{\sqrt{29}} \) units.
In simple words: To find the shortest distance between two vector lines, we first identify the starting points and direction vectors for each line. Then we find the vector connecting the starting points, and the cross product of the direction vectors. We use a formula that involves the dot product of these two new vectors, divided by the magnitude of the cross product, which gives us the shortest distance.

🎯 Exam Tip: When given vector equations, ensure you correctly identify \( \vec{a_1}, \vec{b_1}, \vec{a_2}, \vec{b_2} \). A common mistake is miscalculating the cross product or dot product, so double-check those steps carefully.

 

Question 6. Find the shortest distance between the pairs of lines whose equations are: \( \vec{r} = 5 \hat{j} + \hat{k} + \mu(3 \hat{j} + 2 \hat{k} + 6 \hat{k}) \) and \( \vec{r} = 7 \hat{i} - 6 \hat{k} + \mu(\hat{i} + 2 \hat{j} + 2\hat{k}) \)
Answer: The given equations of the lines are:
Line 1: \( \vec{r} = (0\hat{i} + 5\hat{j} + \hat{k}) + \lambda(0\hat{i} + 3\hat{j} + 8\hat{k}) \) (combining \( 2\hat{k} + 6\hat{k} \) to \( 8\hat{k} \), and adding \( 0\hat{i} \) terms for clarity)
Line 2: \( \vec{r} = (7\hat{i} + 0\hat{j} - 6\hat{k}) + \mu(\hat{i} + 2\hat{j} + 2\hat{k}) \)
Comparing with the general form \( \vec{r} = \vec{a_1} + \lambda\vec{b_1} \) and \( \vec{r} = \vec{a_2} + \mu\vec{b_2} \):
\( \vec{a_1} = 5\hat{j} + \hat{k} \)
\( \vec{b_1} = 3\hat{j} + 8\hat{k} \)
\( \vec{a_2} = 7\hat{i} - 6\hat{k} \)
\( \vec{b_2} = \hat{i} + 2\hat{j} + 2\hat{k} \)
First, find \( \vec{a_2} - \vec{a_1} \):
\( \vec{a_2} - \vec{a_1} = (7\hat{i} - 6\hat{k}) - (5\hat{j} + \hat{k}) \)
\( = 7\hat{i} - 5\hat{j} - 7\hat{k} \)
Next, find \( \vec{b_1} \times \vec{b_2} \):
\[ \vec{b_1} \times \vec{b_2} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 0 & 3 & 8 \\ 1 & 2 & 2 \end{array} \right| \]
\( = \hat{i}((3)(2) - (8)(2)) - \hat{j}((0)(2) - (8)(1)) + \hat{k}((0)(2) - (3)(1)) \)
\( = \hat{i}(6 - 16) - \hat{j}(0 - 8) + \hat{k}(0 - 3) \)
\( = -10\hat{i} + 8\hat{j} - 3\hat{k} \)
Now, find \( |\vec{b_1} \times \vec{b_2}| \):
\( |\vec{b_1} \times \vec{b_2}| = \sqrt{(-10)^2 + 8^2 + (-3)^2} = \sqrt{100 + 64 + 9} = \sqrt{173} \).
Finally, calculate the shortest distance (S.D.) using the formula for skew lines:
\[ S.D. = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} \]
Numerator: \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (7\hat{i} - 5\hat{j} - 7\hat{k}) \cdot (-10\hat{i} + 8\hat{j} - 3\hat{k}) \)
\( = (7)(-10) + (-5)(8) + (-7)(-3) \)
\( = -70 - 40 + 21 \)
\( = -110 + 21 = -89 \).
Therefore, \( S.D. = \frac{|-89|}{\sqrt{173}} = \frac{89}{\sqrt{173}} \) units.
In simple words: We find the starting points and direction vectors for both given lines. Then, we calculate the vector difference between the starting points and the cross product of the direction vectors. We use these to calculate the shortest distance by taking a specific dot product and dividing by the magnitude of the cross product.

🎯 Exam Tip: If any components are missing (e.g., no \( \hat{i} \) term in \( \vec{a_1} \)), treat them as zero to avoid calculation errors. Carefully simplify any vector sums or differences within the given equations before identifying \( \vec{b_1} \) or \( \vec{b_2} \).

 

Question 7. Find the shortest distance between the pairs of lines whose equations are: \( \vec{r} = (8 + 3\lambda) \hat{i} - (9 + 16\lambda) \hat{j} + (10 + 7\lambda) \hat{k} \) and \( \vec{r} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} + \mu(3 \hat{i} + 8 \hat{j} - 5 \hat{k}) \)
Answer: The given equations of the lines are:
Line 1: \( \vec{r} = (8\hat{i} - 9\hat{j} + 10\hat{k}) + \lambda(3\hat{i} - 16\hat{j} + 7\hat{k}) \)
Line 2: \( \vec{r} = (15\hat{i} + 29\hat{j} + 5\hat{k}) + \mu(3\hat{i} + 8\hat{j} - 5\hat{k}) \)
Comparing with the general form \( \vec{r} = \vec{a_1} + \lambda\vec{b_1} \) and \( \vec{r} = \vec{a_2} + \mu\vec{b_2} \):
\( \vec{a_1} = 8\hat{i} - 9\hat{j} + 10\hat{k} \)
\( \vec{b_1} = 3\hat{i} - 16\hat{j} + 7\hat{k} \)
\( \vec{a_2} = 15\hat{i} + 29\hat{j} + 5\hat{k} \)
\( \vec{b_2} = 3\hat{i} + 8\hat{j} - 5\hat{k} \)
First, find \( \vec{a_2} - \vec{a_1} \):
\( \vec{a_2} - \vec{a_1} = (15\hat{i} + 29\hat{j} + 5\hat{k}) - (8\hat{i} - 9\hat{j} + 10\hat{k}) \)
\( = (15 - 8)\hat{i} + (29 - (-9))\hat{j} + (5 - 10)\hat{k} \)
\( = 7\hat{i} + 38\hat{j} - 5\hat{k} \)
Next, find \( \vec{b_1} \times \vec{b_2} \):
\[ \vec{b_1} \times \vec{b_2} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{array} \right| \]
\( = \hat{i}((-16)(-5) - (7)(8)) - \hat{j}((3)(-5) - (7)(3)) + \hat{k}((3)(8) - (-16)(3)) \)
\( = \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 + 48) \)
\( = 24\hat{i} + 36\hat{j} + 72\hat{k} \)
This vector can be simplified by taking out a common factor: \( = 12(2\hat{i} + 3\hat{j} + 6\hat{k}) \).
Now, find \( |\vec{b_1} \times \vec{b_2}| \):
\( |\vec{b_1} \times \vec{b_2}| = |12(2\hat{i} + 3\hat{j} + 6\hat{k})| = 12\sqrt{2^2 + 3^2 + 6^2} \)
\( = 12\sqrt{4 + 9 + 36} = 12\sqrt{49} = 12 \times 7 = 84 \).
Finally, calculate the shortest distance (S.D.) using the formula for skew lines:
\[ S.D. = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} \]
Numerator: \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (7\hat{i} + 38\hat{j} - 5\hat{k}) \cdot (24\hat{i} + 36\hat{j} + 72\hat{k}) \)
\( = (7)(24) + (38)(36) + (-5)(72) \)
\( = 168 + 1368 - 360 \)
\( = 1536 - 360 = 1176 \).
Therefore, \( S.D. = \frac{|1176|}{84} \).
To simplify \( \frac{1176}{84} \):
\( \frac{1176}{84} = \frac{1176 \div 12}{84 \div 12} = \frac{98}{7} = 14 \).
So, \( S.D. = 14 \) units.
In simple words: We first break down the given line equations into their starting position vectors and direction vectors. Then, we find the vector connecting the starting points of the two lines and the vector that is perpendicular to both direction vectors (using the cross product). The shortest distance is then found by a formula that uses the dot product of these two new vectors, divided by the length of the perpendicular direction vector.

🎯 Exam Tip: When equations are given in terms of \( \lambda \) or \( \mu \), it's crucial to correctly separate the constant terms (representing \( \vec{a} \)) from the terms multiplied by the parameter (representing \( \vec{b} \)). This ensures correct calculation of \( \vec{a_2} - \vec{a_1} \).

 

Question 8. Show that the lines \( \vec{r} = (\hat{i} - \hat{j}) + \lambda(2 \hat{i} + \hat{k}) \) and \( \vec{r} =(2 \hat{i} - \hat{j}) + \mu(\hat{i} + \hat{j} - \hat{k}) \) do not intersect.
Answer: The given equations of the lines are:
Line 1: \( \vec{r} = (\hat{i} - \hat{j}) + \lambda(2 \hat{i} + \hat{k}) \)
Line 2: \( \vec{r} = (2 \hat{i} - \hat{j}) + \mu(\hat{i} + \hat{j} - \hat{k}) \)
Comparing with the general form \( \vec{r} = \vec{a_1} + \lambda\vec{b_1} \) and \( \vec{r} = \vec{a_2} + \mu\vec{b_2} \):
\( \vec{a_1} = \hat{i} - \hat{j} \)
\( \vec{b_1} = 2\hat{i} + \hat{k} \)
\( \vec{a_2} = 2\hat{i} - \hat{j} \)
\( \vec{b_2} = \hat{i} + \hat{j} - \hat{k} \)
If two lines intersect, their shortest distance (S.D.) must be zero.
First, find \( \vec{a_2} - \vec{a_1} \):
\( \vec{a_2} - \vec{a_1} = (2\hat{i} - \hat{j}) - (\hat{i} - \hat{j}) \)
\( = (2-1)\hat{i} + (-1 - (-1))\hat{j} \)
\( = \hat{i} + 0\hat{j} = \hat{i} \)
Next, find \( \vec{b_1} \times \vec{b_2} \):
\[ \vec{b_1} \times \vec{b_2} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 1 & 1 & -1 \end{array} \right| \]
\( = \hat{i}((0)(-1) - (1)(1)) - \hat{j}((2)(-1) - (1)(1)) + \hat{k}((2)(1) - (0)(1)) \)
\( = \hat{i}(0 - 1) - \hat{j}(-2 - 1) + \hat{k}(2 - 0) \)
\( = -\hat{i} + 3\hat{j} + 2\hat{k} \)
Now, calculate the numerator for the shortest distance formula: \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) \)
\( = (\hat{i}) \cdot (-\hat{i} + 3\hat{j} + 2\hat{k}) \)
\( = (1)(-1) + (0)(3) + (0)(2) \)
\( = -1 + 0 + 0 = -1 \).
Since \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = -1 \), which is not equal to zero, the shortest distance between the lines is not zero.
Therefore, the given lines do not intersect.
In simple words: Two lines intersect if the shortest distance between them is zero. We check this by calculating a special value from their equations. If this value, found by combining their starting points and direction vectors, is not zero, it means the lines are some distance apart and do not cross paths.

🎯 Exam Tip: To prove lines do not intersect, calculate the shortest distance. If the scalar triple product (the numerator of the S.D. formula) is non-zero, the lines do not intersect. There is no need to calculate the denominator in this case, as a non-zero numerator implies a non-zero distance.

 

Question 9. By computing the shortest distance determine whether the following pairs of lines intersect or not.
(i) \( \vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3 \hat{i} - \hat{j}) \) and \( \vec{r} = (4 \hat{i} - \hat{k}) + \mu(2 \hat{i} + 3 \hat{j}) \)
(ii) \( \frac{x-5}{4} = \frac{y-7}{-5} = \frac{z+3}{-5}, \frac{x-8}{7} = \frac{y-7}{1} = \frac{z-5}{3} \)
(iii) \( \vec{r} = (3 - t) \hat{i} + (4 + 2 t) \hat{j} + (t - 2) \hat{k}, \vec{r} = (1 + s) \hat{i} + (3s - 7) \hat{j} + (2s - 2) \hat{k} \)
Answer:
(i) Given lines are:
Line 1: \( \vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3 \hat{i} - \hat{j}) \)
Line 2: \( \vec{r} = (4 \hat{i} - \hat{k}) + \mu(2 \hat{i} + 3 \hat{j}) \)
Comparing with the general form \( \vec{r} = \vec{a_1} + \lambda\vec{b_1} \) and \( \vec{r} = \vec{a_2} + \mu\vec{b_2} \):
\( \vec{a_1} = \hat{i} + \hat{j} - \hat{k} \)
\( \vec{b_1} = 3\hat{i} - \hat{j} + 0\hat{k} \)
\( \vec{a_2} = 4\hat{i} + 0\hat{j} - \hat{k} \)
\( \vec{b_2} = 2\hat{i} + 3\hat{j} + 0\hat{k} \)
First, find \( \vec{a_2} - \vec{a_1} \):
\( \vec{a_2} - \vec{a_1} = (4\hat{i} + 0\hat{j} - \hat{k}) - (\hat{i} + \hat{j} - \hat{k}) \)
\( = (4-1)\hat{i} + (0-1)\hat{j} + (-1 - (-1))\hat{k} \)
\( = 3\hat{i} - \hat{j} + 0\hat{k} \)
Next, find \( \vec{b_1} \times \vec{b_2} \):
\[ \vec{b_1} \times \vec{b_2} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 0 \\ 2 & 3 & 0 \end{array} \right| \]
\( = \hat{i}((-1)(0) - (0)(3)) - \hat{j}((3)(0) - (0)(2)) + \hat{k}((3)(3) - (-1)(2)) \)
\( = \hat{i}(0 - 0) - \hat{j}(0 - 0) + \hat{k}(9 + 2) \)
\( = 0\hat{i} + 0\hat{j} + 11\hat{k} = 11\hat{k} \)
Now, find \( |\vec{b_1} \times \vec{b_2}| \):
\( |\vec{b_1} \times \vec{b_2}| = |11\hat{k}| = 11 \).
Finally, calculate the scalar triple product \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) \):
\( = (3\hat{i} - \hat{j} + 0\hat{k}) \cdot (0\hat{i} + 0\hat{j} + 11\hat{k}) \)
\( = (3)(0) + (-1)(0) + (0)(11) \)
\( = 0 + 0 + 0 = 0 \).
Since \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 0 \), the shortest distance between the lines is 0.
Therefore, the given lines intersect.
This means the lines meet at a common point.
(ii) Given lines are:
Line 1: \( \frac{x-5}{4} = \frac{y-7}{-5} = \frac{z-(-3)}{-5} \)
Line 2: \( \frac{x-8}{7} = \frac{y-7}{1} = \frac{z-5}{3} \)
From these, we can identify:
\( \vec{a_1} = 5\hat{i} + 7\hat{j} - 3\hat{k} \) and \( \vec{b_1} = 4\hat{i} - 5\hat{j} - 5\hat{k} \)
\( \vec{a_2} = 8\hat{i} + 7\hat{j} + 5\hat{k} \) and \( \vec{b_2} = 7\hat{i} + \hat{j} + 3\hat{k} \)
First, find \( \vec{a_2} - \vec{a_1} \):
\( \vec{a_2} - \vec{a_1} = (8\hat{i} + 7\hat{j} + 5\hat{k}) - (5\hat{i} + 7\hat{j} - 3\hat{k}) \)
\( = (8-5)\hat{i} + (7-7)\hat{j} + (5 - (-3))\hat{k} \)
\( = 3\hat{i} + 0\hat{j} + 8\hat{k} \)
Next, find \( \vec{b_1} \times \vec{b_2} \):
\[ \vec{b_1} \times \vec{b_2} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & -5 & -5 \\ 7 & 1 & 3 \end{array} \right| \]
\( = \hat{i}((-5)(3) - (-5)(1)) - \hat{j}((4)(3) - (-5)(7)) + \hat{k}((4)(1) - (-5)(7)) \)
\( = \hat{i}(-15 + 5) - \hat{j}(12 + 35) + \hat{k}(4 + 35) \)
\( = -10\hat{i} - 47\hat{j} + 39\hat{k} \)
Now, find \( |\vec{b_1} \times \vec{b_2}| \):
\( |\vec{b_1} \times \vec{b_2}| = \sqrt{(-10)^2 + (-47)^2 + (39)^2} \)
\( = \sqrt{100 + 2209 + 1521} = \sqrt{3830} \).
Finally, calculate the scalar triple product \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) \):
\( = (3\hat{i} + 0\hat{j} + 8\hat{k}) \cdot (-10\hat{i} - 47\hat{j} + 39\hat{k}) \)
\( = (3)(-10) + (0)(-47) + (8)(39) \)
\( = -30 + 0 + 312 = 282 \).
Since \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 282 \), which is not zero, the shortest distance between the lines is not zero.
Thus, the given lines do not intersect.

(iii) Given lines are:
Line 1: \( \vec{r} = (3 - t) \hat{i} + (4 + 2 t) \hat{j} + (t - 2) \hat{k} \)
This can be rewritten as: \( \vec{r} = (3\hat{i} + 4\hat{j} - 2\hat{k}) + t(-\hat{i} + 2\hat{j} + \hat{k}) \)
Line 2: \( \vec{r} = (1 + s) \hat{i} + (3s - 7) \hat{j} + (2s - 2) \hat{k} \)
This can be rewritten as: \( \vec{r} = (\hat{i} - 7\hat{j} - 2\hat{k}) + s(\hat{i} + 3\hat{j} + 2\hat{k}) \)
Comparing with the general form \( \vec{r} = \vec{a_1} + t\vec{b_1} \) and \( \vec{r} = \vec{a_2} + s\vec{b_2} \):
\( \vec{a_1} = 3\hat{i} + 4\hat{j} - 2\hat{k} \)
\( \vec{b_1} = -\hat{i} + 2\hat{j} + \hat{k} \)
\( \vec{a_2} = \hat{i} - 7\hat{j} - 2\hat{k} \)
\( \vec{b_2} = \hat{i} + 3\hat{j} + 2\hat{k} \)
First, find \( \vec{a_2} - \vec{a_1} \):
\( \vec{a_2} - \vec{a_1} = (\hat{i} - 7\hat{j} - 2\hat{k}) - (3\hat{i} + 4\hat{j} - 2\hat{k}) \)
\( = (1-3)\hat{i} + (-7-4)\hat{j} + (-2 - (-2))\hat{k} \)
\( = -2\hat{i} - 11\hat{j} + 0\hat{k} \)
Next, find \( \vec{b_1} \times \vec{b_2} \):
\[ \vec{b_1} \times \vec{b_2} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 1 \\ 1 & 3 & 2 \end{array} \right| \]
\( = \hat{i}((2)(2) - (1)(3)) - \hat{j}((-1)(2) - (1)(1)) + \hat{k}((-1)(3) - (2)(1)) \)
\( = \hat{i}(4 - 3) - \hat{j}(-2 - 1) + \hat{k}(-3 - 2) \)
\( = \hat{i} + 3\hat{j} - 5\hat{k} \)
Now, find \( |\vec{b_1} \times \vec{b_2}| \):
\( |\vec{b_1} \times \vec{b_2}| = \sqrt{1^2 + 3^2 + (-5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35} \).
Finally, calculate the scalar triple product \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) \):
\( = (-2\hat{i} - 11\hat{j} + 0\hat{k}) \cdot (\hat{i} + 3\hat{j} - 5\hat{k}) \)
\( = (-2)(1) + (-11)(3) + (0)(-5) \)
\( = -2 - 33 + 0 = -35 \).
Since \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = -35 \), which is not zero, the shortest distance between the lines is not zero.
Thus, the given lines do not intersect.
The shortest distance is \( S.D. = \frac{|-35|}{\sqrt{35}} = \frac{35}{\sqrt{35}} = \sqrt{35} \) units.
In simple words: For each pair of lines, we calculate their shortest distance. If the shortest distance is zero, it means the lines intersect. If the distance is not zero, the lines do not intersect. We use vector algebra to find the starting points and direction vectors, then apply a specific formula to find this distance.

🎯 Exam Tip: Always state the condition for intersection (shortest distance = 0) clearly. For lines in Cartesian form, converting them to vector form first can sometimes simplify calculations for shortest distance problems.

 

Question 10. Find the shortest distance between the following pairs of parallel lines:
(i) \( \vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k}) \) and \( \vec{r} = (2 \hat{i} - \hat{j} - \hat{k}) + \mu(-\hat{i} + \hat{j} - \hat{k}) \)
(ii) \( \vec{r} = (\hat{i} + \hat{j}) + \lambda(2 \hat{i} - \hat{j} + \hat{k}) \) and \( \vec{r} = (2 \hat{i} + \hat{j} - \hat{k}) + \mu(4 \hat{i} - 2 \hat{j} + 2 \hat{k}) \)
Answer:
(i) Given lines are:
Line 1: \( \vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k}) \)
Line 2: \( \vec{r} = (2 \hat{i} - \hat{j} - \hat{k}) + \mu(-\hat{i} + \hat{j} - \hat{k}) \)
Notice that the direction vector of Line 2, \( -\hat{i} + \hat{j} - \hat{k} \), is \( -1 \) times the direction vector of Line 1, \( \hat{i} - \hat{j} + \hat{k} \). This means the lines are parallel.
For parallel lines, the shortest distance formula is: \( S.D. = \frac{|(\vec{a_2} - \vec{a_1}) \times \vec{b}|}{|\vec{b}|} \), where \( \vec{b} \) is the common direction vector.
Here, we can take \( \vec{b} = \hat{i} - \hat{j} + \hat{k} \).
From the equations:
\( \vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k} \)
\( \vec{a_2} = 2\hat{i} - \hat{j} - \hat{k} \)
First, find \( \vec{a_2} - \vec{a_1} \):
\( \vec{a_2} - \vec{a_1} = (2\hat{i} - \hat{j} - \hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) \)
\( = (2-1)\hat{i} + (-1-2)\hat{j} + (-1-3)\hat{k} \)
\( = \hat{i} - 3\hat{j} - 4\hat{k} \)
Next, find \( (\vec{a_2} - \vec{a_1}) \times \vec{b} \):
\[ (\vec{a_2} - \vec{a_1}) \times \vec{b} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & -4 \\ 1 & -1 & 1 \end{array} \right| \]
\( = \hat{i}((-3)(1) - (-4)(-1)) - \hat{j}((1)(1) - (-4)(1)) + \hat{k}((1)(-1) - (-3)(1)) \)
\( = \hat{i}(-3 - 4) - \hat{j}(1 + 4) + \hat{k}(-1 + 3) \)
\( = -7\hat{i} - 5\hat{j} + 2\hat{k} \)
Now, find \( |(\vec{a_2} - \vec{a_1}) \times \vec{b}| \):
\( |(\vec{a_2} - \vec{a_1}) \times \vec{b}| = \sqrt{(-7)^2 + (-5)^2 + 2^2} \)
\( = \sqrt{49 + 25 + 4} = \sqrt{78} \).
Finally, find \( |\vec{b}| \):
\( |\vec{b}| = |\hat{i} - \hat{j} + \hat{k}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \).
The shortest distance (S.D.) is:
\( S.D. = \frac{\sqrt{78}}{\sqrt{3}} = \sqrt{\frac{78}{3}} = \sqrt{26} \) units.

(ii) Given lines are:
Line 1: \( \vec{r} = (\hat{i} + \hat{j}) + \lambda(2 \hat{i} - \hat{j} + \hat{k}) \)
Line 2: \( \vec{r} = (2 \hat{i} + \hat{j} - \hat{k}) + \mu(4 \hat{i} - 2 \hat{j} + 2 \hat{k}) \)
Notice that the direction vector of Line 2, \( 4 \hat{i} - 2 \hat{j} + 2 \hat{k} \), is \( 2 \) times the direction vector of Line 1, \( 2 \hat{i} - \hat{j} + \hat{k} \). This means the lines are parallel.
We use \( \vec{b} = 2\hat{i} - \hat{j} + \hat{k} \) as the common direction vector.
From the equations:
\( \vec{a_1} = \hat{i} + \hat{j} \)
\( \vec{a_2} = 2\hat{i} + \hat{j} - \hat{k} \)
First, find \( \vec{a_2} - \vec{a_1} \):
\( \vec{a_2} - \vec{a_1} = (2\hat{i} + \hat{j} - \hat{k}) - (\hat{i} + \hat{j}) \)
\( = (2-1)\hat{i} + (1-1)\hat{j} + (-1-0)\hat{k} \)
\( = \hat{i} + 0\hat{j} - \hat{k} \)
Next, find \( (\vec{a_2} - \vec{a_1}) \times \vec{b} \):
\[ (\vec{a_2} - \vec{a_1}) \times \vec{b} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & -1 \\ 2 & -1 & 1 \end{array} \right| \]
\( = \hat{i}((0)(1) - (-1)(-1)) - \hat{j}((1)(1) - (-1)(2)) + \hat{k}((1)(-1) - (0)(2)) \)
\( = \hat{i}(0 - 1) - \hat{j}(1 + 2) + \hat{k}(-1 - 0) \)
\( = -\hat{i} - 3\hat{j} - \hat{k} \)
Now, find \( |(\vec{a_2} - \vec{a_1}) \times \vec{b}| \):
\( |(\vec{a_2} - \vec{a_1}) \times \vec{b}| = \sqrt{(-1)^2 + (-3)^2 + (-1)^2} \)
\( = \sqrt{1 + 9 + 1} = \sqrt{11} \).
Finally, find \( |\vec{b}| \):
\( |\vec{b}| = |2\hat{i} - \hat{j} + \hat{k}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \).
The shortest distance (S.D.) is:
\( S.D. = \frac{\sqrt{11}}{\sqrt{6}} \) units.
In simple words: When lines are parallel, we use a simpler formula for the shortest distance. We pick one starting point from each line and find the vector connecting them. We also choose one of the shared direction vectors. The formula involves taking the cross product of these two vectors, then finding its length, and dividing by the length of the direction vector. This directly gives the distance between the parallel lines.

🎯 Exam Tip: Always first check if the lines are parallel by comparing their direction vectors. If they are scalar multiples of each other, use the formula for parallel lines; otherwise, use the formula for skew lines. This saves time and ensures the correct method is applied.

 

Question 11. Define the line of shortest distance between two skew lines. Find the shortest distance and the vector equation of the line of shortest distance between the lines given by:
(i) \( \vec{r} = (8 + 3\lambda) \hat{i} - (9 + 16\lambda) \hat{j} + (10 + 7\lambda) \hat{k} \) and \( \vec{r} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} + \mu(3 \hat{i} + 8 \hat{j} - 5 \hat{k}) \)
(ii) \( \vec{r} = (3 \hat{i} + 8 \hat{j} + 3 \hat{k}) + \lambda(3 \hat{i} - \hat{j} + \hat{k}) \) and \( \vec{r} = (-3\mu - 3) \hat{i} + (2\mu - 7) \hat{j} + (4\mu + 6) \hat{k} \)
Answer:
A line of shortest distance is a special line that connects two skew lines (lines that are not parallel and do not intersect). If we have two skew lines, say 'l' and 'm', there is only one unique line, 'p', which intersects both 'l' and 'm' and is perpendicular to both. This line 'p' represents the shortest possible distance between the two skew lines. This line connects the points on the skew lines that are closest to each other.
(i) To find the shortest distance and vector equation for the first pair of lines: First, identify the position vectors \( \vec{a_1}, \vec{a_2} \) and direction vectors \( \vec{b_1}, \vec{b_2} \) from the given equations. For the first line \( \vec{r} = (8 \hat{i} - 9 \hat{j} + 10 \hat{k}) + \lambda(3 \hat{i} - 16 \hat{j} + 7 \hat{k}) \), we have \( \vec{a_1} = 8 \hat{i} - 9 \hat{j} + 10 \hat{k} \) and \( \vec{b_1} = 3 \hat{i} - 16 \hat{j} + 7 \hat{k} \). For the second line \( \vec{r} = (15 \hat{i} + 29 \hat{j} + 5 \hat{k}) + \mu(3 \hat{i} + 8 \hat{j} - 5 \hat{k}) \), we have \( \vec{a_2} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} \) and \( \vec{b_2} = 3 \hat{i} + 8 \hat{j} - 5 \hat{k} \). Next, calculate the difference between the position vectors: \( \vec{a_2} - \vec{a_1} = (15 - 8)\hat{i} + (29 - (-9))\hat{j} + (5 - 10)\hat{k} = 7 \hat{i} + 38 \hat{j} - 5 \hat{k} \). Then, find the cross product of the direction vectors: \( \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix} \) \( \implies \vec{b_1} \times \vec{b_2} = \hat{i}((-16)(-5) - (7)(8)) - \hat{j}((3)(-5) - (7)(3)) + \hat{k}((3)(8) - (-16)(3)) \) \( \implies \vec{b_1} \times \vec{b_2} = \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 + 48) \) \( \implies \vec{b_1} \times \vec{b_2} = 24 \hat{i} + 36 \hat{j} + 72 \hat{k} = 12(2 \hat{i} + 3 \hat{j} + 6 \hat{k}) \). Now, find the magnitude of this cross product: \( |\vec{b_1} \times \vec{b_2}| = \sqrt{24^2 + 36^2 + 72^2} = \sqrt{576 + 1296 + 5184} = \sqrt{7056} = 84 \). The shortest distance (SD) between two skew lines is given by the formula: \( SD = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} \) Calculate the dot product in the numerator: \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (7 \hat{i} + 38 \hat{j} - 5 \hat{k}) \cdot (24 \hat{i} + 36 \hat{j} + 72 \hat{k}) \) \( \implies (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (7)(24) + (38)(36) + (-5)(72) \) \( \implies (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 168 + 1368 - 360 = 1176 \). Substitute these values into the SD formula: \( SD = \frac{|1176|}{84} = 14 \) units. This is the minimum separation between the lines. To find the vector equation of the line of shortest distance, we need the coordinates of points P and Q, where the shortest distance occurs. By converting the line equations to Cartesian form and solving, we find the parameters \( t = -1 \) for line 1 and \( r = -2 \) for line 2. Substitute \( t = -1 \) into the Cartesian form of line 1 \( (x, y, z) = (3t+8, -16t-9, 7t+10) \): \( P = (3(-1)+8, -16(-1)-9, 7(-1)+10) = (5, 7, 3) \). Substitute \( r = -2 \) into the Cartesian form of line 2 \( (x, y, z) = (3r+15, 8r+29, -5r+5) \): \( Q = (3(-2)+15, 8(-2)+29, -5(-2)+5) = (9, 13, 15) \). The vector equation of the line of shortest distance passing through P(5,7,3) and Q(9,13,15) is: \( \vec{R} = \vec{P} + \lambda' (\vec{Q} - \vec{P}) \) \( \implies \vec{R} = (5 \hat{i} + 7 \hat{j} + 3 \hat{k}) + \lambda' ((9-5)\hat{i} + (13-7)\hat{j} + (15-3)\hat{k}) \) \( \implies \vec{R} = (5 \hat{i} + 7 \hat{j} + 3 \hat{k}) + \lambda' (4\hat{i} + 6\hat{j} + 12\hat{k}) \). We can simplify the direction vector \( (4\hat{i} + 6\hat{j} + 12\hat{k}) \) to \( (2\hat{i} + 3\hat{j} + 6\hat{k}) \) since it's just a direction. So, the vector equation of the line of shortest distance is \( \vec{R} = (5 \hat{i} + 7 \hat{j} + 3 \hat{k}) + \lambda' (2\hat{i} + 3\hat{j} + 6\hat{k}) \).
(ii) To find the shortest distance and vector equation for the second pair of lines: First, rewrite the given line equations in the standard vector form \( \vec{r} = \vec{a} + \lambda \vec{b} \). For the first line \( \vec{r} = (3 \hat{i} + 8 \hat{j} + 3 \hat{k}) + \lambda(3 \hat{i} - \hat{j} + \hat{k}) \): \( \vec{a_1} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k} \) and \( \vec{b_1} = 3 \hat{i} - \hat{j} + \hat{k} \). For the second line \( \vec{r} = (-3\mu - 3) \hat{i} + (2\mu - 7) \hat{j} + (4\mu + 6) \hat{k} \): We can rewrite it as \( \vec{r} = (-3 \hat{i} - 7 \hat{j} + 6 \hat{k}) + \mu(-3 \hat{i} + 2 \hat{j} + 4 \hat{k}) \). So, \( \vec{a_2} = -3 \hat{i} - 7 \hat{j} + 6 \hat{k} \) and \( \vec{b_2} = -3 \hat{i} + 2 \hat{j} + 4 \hat{k} \). Next, find the difference between the position vectors: \( \vec{a_2} - \vec{a_1} = (-3 - 3)\hat{i} + (-7 - 8)\hat{j} + (6 - 3)\hat{k} = -6 \hat{i} - 15 \hat{j} + 3 \hat{k} \). Then, find the cross product of the direction vectors: \( \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix} \) \( \implies \vec{b_1} \times \vec{b_2} = \hat{i}((-1)(4) - (1)(2)) - \hat{j}((3)(4) - (1)(-3)) + \hat{k}((3)(2) - (-1)(-3)) \) \( \implies \vec{b_1} \times \vec{b_2} = \hat{i}(-4 - 2) - \hat{j}(12 + 3) + \hat{k}(6 - 3) \) \( \implies \vec{b_1} \times \vec{b_2} = -6 \hat{i} - 15 \hat{j} + 3 \hat{k} \). Now, find the magnitude of this cross product: \( |\vec{b_1} \times \vec{b_2}| = \sqrt{(-6)^2 + (-15)^2 + 3^2} = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30} \). The shortest distance (SD) between two skew lines is given by the formula: \( SD = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} \) Calculate the dot product in the numerator: \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-6 \hat{i} - 15 \hat{j} + 3 \hat{k}) \cdot (-6 \hat{i} - 15 \hat{j} + 3 \hat{k}) \) \( \implies (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-6)(-6) + (-15)(-15) + (3)(3) \) \( \implies (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 36 + 225 + 9 = 270 \). Substitute these values into the SD formula: \( SD = \frac{|270|}{\sqrt{270}} = \sqrt{270} = 3\sqrt{30} \) units. This distance represents the shortest gap between the lines. To find the vector equation of the line of shortest distance, we need the coordinates of points P and Q where the shortest distance occurs. By converting the line equations to Cartesian form and solving, we find the parameters \( t = 0 \) for line 1 and \( r = 0 \) for line 2. Substitute \( t = 0 \) into the Cartesian form of line 1 \( (x, y, z) = (3t+3, -t+8, t+3) \): \( P = (3(0)+3, -(0)+8, (0)+3) = (3, 8, 3) \). Substitute \( r = 0 \) into the Cartesian form of line 2 \( (x, y, z) = (-3r-3, 2r-7, 4r+6) \): \( Q = (-3(0)-3, 2(0)-7, 4(0)+6) = (-3, -7, 6) \). The vector equation of the line of shortest distance passing through P(3,8,3) and Q(-3,-7,6) is: \( \vec{R} = \vec{P} + \mu' (\vec{Q} - \vec{P}) \) \( \implies \vec{R} = (3 \hat{i} + 8 \hat{j} + 3 \hat{k}) + \mu' ((-3-3)\hat{i} + (-7-8)\hat{j} + (6-3)\hat{k}) \) \( \implies \vec{R} = (3 \hat{i} + 8 \hat{j} + 3 \hat{k}) + \mu' (-6\hat{i} - 15\hat{j} + 3\hat{k}) \). We can simplify the direction vector \( (-6\hat{i} - 15\hat{j} + 3\hat{k}) \) to \( (2\hat{i} + 5\hat{j} - \hat{k}) \) by dividing by -3. So, the vector equation of the line of shortest distance is \( \vec{R} = (3 \hat{i} + 8 \hat{j} + 3 \hat{k}) + \mu' (2\hat{i} + 5\hat{j} - \hat{k}) \).In simple words: When two lines in space are not parallel and don't meet, a special line connects them at their closest points. We find this shortest distance by using vector math, which involves calculating differences between starting points and cross products of direction vectors. To find the equation of this connecting line, we first figure out the exact points on each original line where this shortest connection happens.

🎯 Exam Tip: When dealing with skew lines, correctly identifying \( \vec{a_1}, \vec{b_1}, \vec{a_2}, \vec{b_2} \) is crucial. Remember that the shortest distance is always a positive value, and the vector equation of the shortest distance line requires finding the specific points of intersection with the given lines.