Samacheer Kalvi Class 12 Maths Solutions Chapter 6 Applications of Vector Algebra Exercise 6.5

Chapter 6

Question 1. Find the parametric form of vector equation and Cartesian equations of straight line passing through (5, 2, 8) and is perpendicular to the straight lines \( \overline { r } = (\hat { i } + \hat { j } – \hat { k }) + s(2\hat { i } – 2\hat { j } + \hat { k }) \) and \( \overline { r } = (2\hat { i } – \hat { j } – 3\hat { k }) + t(\hat { i } + 2\hat { j } + 2\hat { k }) \).
Answer: Given point: \( A = (5, 2, 8) \), so the position vector is \( \vec{a} = 5\hat{i} + 2\hat{j} + 8\hat{k} \).
The line is perpendicular to two given lines, so its direction vector \( \vec{b} \) will be the cross product of the direction vectors of these two lines.
Let the direction vectors of the given lines be \( \vec{b_1} = 2\hat{i} - 2\hat{j} + \hat{k} \) and \( \vec{b_2} = \hat{i} + 2\hat{j} + 2\hat{k} \).
The direction vector of the required line is \( \vec{b} = \vec{b_1} \times \vec{b_2} \).
\[ \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 1 \\ 1 & 2 & 2 \end{vmatrix} \]
\( \vec{b} = \hat{i}((-2)(2) - (1)(2)) - \hat{j}((2)(2) - (1)(1)) + \hat{k}((2)(2) - (-2)(1)) \)
\( \vec{b} = \hat{i}(-4 - 2) - \hat{j}(4 - 1) + \hat{k}(4 + 2) \)
\( \vec{b} = -6\hat{i} - 3\hat{j} + 6\hat{k} \)
We can also take \( \vec{b} = -3(2\hat{i} + \hat{j} - 2\hat{k}) \), so a simpler direction vector for the line is \( \vec{b'} = 2\hat{i} + \hat{j} - 2\hat{k} \).
This vector is perpendicular to both given straight lines.

**Parametric form of vector equation:**
The equation of a line passing through a point \( \vec{a} \) and parallel to \( \vec{b'} \) is \( \overline{r} = \vec{a} + t\vec{b'} \).
\( \overline{r} = (5\hat{i} + 2\hat{j} + 8\hat{k}) + t(2\hat{i} + \hat{j} - 2\hat{k}) \).

**Cartesian equation:**
If \( \overline{r} = x\hat{i} + y\hat{j} + z\hat{k} \), then comparing coefficients gives:
\( x\hat{i} + y\hat{j} + z\hat{k} = (5 + 2t)\hat{i} + (2 + t)\hat{j} + (8 - 2t)\hat{k} \)
\( x = 5 + 2t \implies \frac{x-5}{2} = t \)
\( y = 2 + t \implies \frac{y-2}{1} = t \)
\( z = 8 - 2t \implies \frac{z-8}{-2} = t \)
So, the Cartesian form is: \( \frac{x-5}{2} = \frac{y-2}{1} = \frac{z-8}{-2} \). This form clearly shows the point and direction of the line.
In simple words: First, we find a direction vector for the new line by crossing the two given direction vectors. This new vector tells us which way the line goes. Then, we use the given point and this new direction to write both the vector and Cartesian equations for the line.

🎯 Exam Tip: Remember that a line perpendicular to two other lines has a direction vector found by taking the cross product of the direction vectors of those two lines. This is a key step in such problems.

Question 2. Show that the lines \( \overline { r } = (6\hat { i } + \hat { j } + 2\hat { k }) + s(\hat { i } + 2\hat { j } – 3\hat { k }) \) and \( \overline { r } = (3\hat { i } + 2\hat { j } – 2\hat { k }) + t(2\hat { i } + 4\hat { j } – 5\hat { k }) \) are skew lines and hence find the shortest distance between them.
Answer: Let the first line be \( L_1 \) and the second line be \( L_2 \).
For \( L_1 \): \( \vec{a_1} = 6\hat{i} + \hat{j} + 2\hat{k} \) and \( \vec{b_1} = \hat{i} + 2\hat{j} - 3\hat{k} \).
For \( L_2 \): \( \vec{a_2} = 3\hat{i} + 2\hat{j} - 2\hat{k} \) and \( \vec{b_2} = 2\hat{i} + 4\hat{j} - 5\hat{k} \).

**Check if lines are parallel:**
We check if \( \vec{b_1} \) is a scalar multiple of \( \vec{b_2} \).
\( \vec{b_1} = \hat{i} + 2\hat{j} - 3\hat{k} \)
\( \vec{b_2} = 2\hat{i} + 4\hat{j} - 5\hat{k} \)
If \( \vec{b_1} = k\vec{b_2} \), then \( 1 = 2k \), \( 2 = 4k \), \( -3 = -5k \).
From the first two, \( k = \frac{1}{2} \). But for the third, \( k = \frac{3}{5} \). Since \( k \) is not consistent, \( \vec{b_1} \) is not a scalar multiple of \( \vec{b_2} \).
Therefore, the lines are not parallel.

**Check if lines intersect (and are thus coplanar):**
If the lines are not parallel, they are either intersecting or skew. To be skew, they must not intersect and not be parallel. The condition for two lines to be skew is that \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) \neq 0 \).

First, calculate \( \vec{a_2} - \vec{a_1} \):
\( \vec{a_2} - \vec{a_1} = (3\hat{i} + 2\hat{j} - 2\hat{k}) - (6\hat{i} + \hat{j} + 2\hat{k}) \)
\( \vec{a_2} - \vec{a_1} = -3\hat{i} + \hat{j} - 4\hat{k} \).

Next, calculate \( \vec{b_1} \times \vec{b_2} \):
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix} \]
\( \vec{b_1} \times \vec{b_2} = \hat{i}((2)(-5) - (-3)(4)) - \hat{j}((1)(-5) - (-3)(2)) + \hat{k}((1)(4) - (2)(2)) \)
\( \vec{b_1} \times \vec{b_2} = \hat{i}(-10 + 12) - \hat{j}(-5 + 6) + \hat{k}(4 - 4) \)
\( \vec{b_1} \times \vec{b_2} = 2\hat{i} - \hat{j} + 0\hat{k} = 2\hat{i} - \hat{j} \).

Now, calculate the scalar triple product: \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) \):
\( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-3\hat{i} + \hat{j} - 4\hat{k}) \cdot (2\hat{i} - \hat{j} + 0\hat{k}) \)
\( = (-3)(2) + (1)(-1) + (-4)(0) \)
\( = -6 - 1 + 0 = -7 \).
Since \( -7 \neq 0 \), the lines are skew lines.

**Shortest distance between skew lines:**
The formula for the shortest distance \( \delta \) between skew lines is \( \delta = \frac{ |(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})| }{ |\vec{b_1} \times \vec{b_2}| } \).
We already found \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = -7 \).
Now, find the magnitude of \( \vec{b_1} \times \vec{b_2} \):
\( |\vec{b_1} \times \vec{b_2}| = |2\hat{i} - \hat{j}| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{4 + 1} = \sqrt{5} \).
So, the shortest distance \( \delta = \frac{ |-7| }{ \sqrt{5} } = \frac{ 7 }{ \sqrt{5} } \) units.
To rationalize the denominator, we can write \( \delta = \frac{7\sqrt{5}}{5} \) units. This is the minimum separation between the two lines.
In simple words: First, we checked if the lines were parallel by looking at their direction vectors. They were not. Then, we checked if they intersect using a special dot product calculation. Since that calculation was not zero, it means the lines don't meet and are not parallel, so they are skew. Finally, we used a formula involving these same values to find the shortest distance between them.

🎯 Exam Tip: Remember the two key conditions: for parallel lines, direction vectors are proportional; for skew lines, they are not parallel AND the scalar triple product \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) \) is not zero. Calculating this product is often the most complex part.

Question 3. If the two lines \( \frac { x-1 }{ 2 } = \frac { y+1 }{ 3 } = \frac { z-1 }{ 4 } \) and \( \frac { x-3 }{ 1 } = \frac { y-m }{ 2 } = z \) intersect at a point, find the value of m.
Answer: Let the first line be \( L_1 \) and the second line be \( L_2 \).
For \( L_1 \), the point \( (x_1, y_1, z_1) = (1, -1, 1) \) and direction vector \( \vec{d_1} = (2, 3, 4) \).
For \( L_2 \), the point \( (x_2, y_2, z_2) = (3, m, 0) \) (since \( z = \frac{z-0}{1} \)) and direction vector \( \vec{d_2} = (1, 2, 1) \).
The condition for two lines to intersect is that the scalar triple product of \( (\vec{a_2} - \vec{a_1}) \), \( \vec{d_1} \), and \( \vec{d_2} \) must be zero. This is equivalent to setting the determinant of the matrix formed by these vectors to zero.
\[ \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ d_{1x} & d_{1y} & d_{1z} \\ d_{2x} & d_{2y} & d_{2z} \end{vmatrix} = 0 \]
Substitute the given values:
\( x_2 - x_1 = 3 - 1 = 2 \)
\( y_2 - y_1 = m - (-1) = m + 1 \)
\( z_2 - z_1 = 0 - 1 = -1 \)
\[ \begin{vmatrix} 2 & m+1 & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{vmatrix} = 0 \]
Expand the determinant:
\( 2((3)(1) - (4)(2)) - (m+1)((2)(1) - (4)(1)) + (-1)((2)(2) - (3)(1)) = 0 \)
\( 2(3 - 8) - (m+1)(2 - 4) - 1(4 - 3) = 0 \)
\( 2(-5) - (m+1)(-2) - 1(1) = 0 \)
\( -10 + 2(m+1) - 1 = 0 \)
\( -10 + 2m + 2 - 1 = 0 \)
\( 2m - 9 = 0 \)
\( 2m = 9 \)
\( m = \frac{9}{2} \). The value of m ensures that the two lines meet at a single point.
In simple words: When two lines meet, there's a special mathematical rule that says a certain determinant must be zero. We set up this determinant using points and direction numbers from both lines. Then, we solved the equation to find the unknown value of 'm'.

🎯 Exam Tip: The condition for two lines to intersect is that their shortest distance is zero. This is calculated using the scalar triple product of the vector connecting points on the lines and their direction vectors. Setting this determinant to zero is crucial for solving such problems.

Question 4. Show that the lines \( \frac { x-3 }{ 3 } = \frac { y-3 }{ -1 }, z − 1 = 0 \) and \( \frac { x-6 }{ 2 } = \frac { z-1 }{ 3 }, y − 2 = 0 \) intersect. Also find the point of intersection.
Answer: We need to show that these two lines meet at a point.

**(i) For the first line:**
The equations are \( \frac{x-3}{3} = \frac{y-3}{-1} \) and \( z-1 = 0 \).
Let \( \frac{x-3}{3} = \frac{y-3}{-1} = t \).
Then \( x = 3t + 3 \)
\( y = -t + 3 \)
And from \( z-1 = 0 \), we get \( z = 1 \).
So, any point on the first line can be written as \( (3t+3, -t+3, 1) \). Let's call this Point 1.

**(ii) For the second line:**
The equations are \( \frac{x-6}{2} = \frac{z-1}{3} \) and \( y-2 = 0 \).
Let \( \frac{x-6}{2} = \frac{z-1}{3} = s \).
Then \( x = 2s + 6 \)
From \( y-2 = 0 \), we get \( y = 2 \).
And \( z = 3s + 1 \).
So, any point on the second line can be written as \( (2s+6, 2, 3s+1) \). Let's call this Point 2.

**To find the intersection point, we set the coordinates equal:**
Compare the x-coordinates: \( 3t+3 = 2s+6 \implies 3t - 2s = 3 \) (Equation A)
Compare the y-coordinates: \( -t+3 = 2 \implies -t = -1 \implies t = 1 \).
Compare the z-coordinates: \( 1 = 3s+1 \implies 3s = 0 \implies s = 0 \).

**Verify consistency:**
Substitute \( t=1 \) and \( s=0 \) into Equation A:
\( 3(1) - 2(0) = 3 - 0 = 3 \).
Since \( 3 = 3 \), the values of \( t \) and \( s \) are consistent. This confirms that the lines intersect.

**Point of intersection:**
Substitute \( t=1 \) into the coordinates of Point 1:
\( x = 3(1) + 3 = 6 \)
\( y = -(1) + 3 = 2 \)
\( z = 1 \)
So, the point of intersection is \( (6, 2, 1) \). We can also check by substituting \( s=0 \) into the coordinates of Point 2: \( x = 2(0)+6=6 \), \( y=2 \), \( z=3(0)+1=1 \). Both give the same point, confirming our answer.
In simple words: We wrote down general points for each line using 't' and 's' as variables. Then, we made the x, y, and z parts of these points equal to each other to find if they share a common spot. Solving these equations gave us values for 't' and 's', which means the lines indeed cross. Using these values, we found the exact spot where they meet.

🎯 Exam Tip: When proving intersection, always find the parameters (like 't' and 's') from two coordinate equations and then verify them with the third. If all three equations are satisfied, the lines intersect.

Question 5. Show that the straight lines \( x + 1 = 2y = -12z \) and \( x = y + 2 = 6z – 6 \) are skew and hence find the shortest distance between them.
Answer: We need to convert both lines into standard vector form or Cartesian form \( \frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n} \).

**For the first line:** \( x + 1 = 2y = -12z \).
Divide by the least common multiple of 1, 2, and -12, which is 12 (or -12 for convenience).
\( \frac{x+1}{12} = \frac{2y}{12} = \frac{-12z}{12} \)
\( \frac{x+1}{12} = \frac{y}{6} = \frac{z}{-1} \).
So, for the first line: \( \vec{a_1} = -\hat{i} + 0\hat{j} + 0\hat{k} = -\hat{i} \) and \( \vec{b_1} = 12\hat{i} + 6\hat{j} - \hat{k} \).

**For the second line:** \( x = y + 2 = 6z – 6 \).
Rewrite \( x = y+2 \) as \( x = \frac{y+2}{1} \).
Rewrite \( y+2 = 6z-6 \). Divide by 6:
\( \frac{y+2}{6} = \frac{6z-6}{6} = \frac{z-1}{1} \).
So, \( \frac{x}{6} = \frac{y+2}{6} = \frac{z-1}{1} \).
For the second line: \( \vec{a_2} = 0\hat{i} - 2\hat{j} + \hat{k} = -2\hat{j} + \hat{k} \) and \( \vec{b_2} = 6\hat{i} + 6\hat{j} + \hat{k} \).

**Check for parallelism:**
We check if \( \vec{b_1} \) is a scalar multiple of \( \vec{b_2} \).
\( \vec{b_1} = 12\hat{i} + 6\hat{j} - \hat{k} \)
\( \vec{b_2} = 6\hat{i} + 6\hat{j} + \hat{k} \)
For them to be parallel, \( \frac{12}{6} = \frac{6}{6} = \frac{-1}{1} \). This gives \( 2 = 1 = -1 \), which is false. So, the vectors are not parallel.
Therefore, the lines are not parallel.

**Check if lines are skew and find shortest distance:**
We need to calculate \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) \).
First, \( \vec{a_2} - \vec{a_1} = (-2\hat{j} + \hat{k}) - (-\hat{i}) = \hat{i} - 2\hat{j} + \hat{k} \).

Next, calculate \( \vec{b_1} \times \vec{b_2} \):
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 12 & 6 & -1 \\ 6 & 6 & 1 \end{vmatrix} \]
\( \vec{b_1} \times \vec{b_2} = \hat{i}((6)(1) - (-1)(6)) - \hat{j}((12)(1) - (-1)(6)) + \hat{k}((12)(6) - (6)(6)) \)
\( \vec{b_1} \times \vec{b_2} = \hat{i}(6 + 6) - \hat{j}(12 + 6) + \hat{k}(72 - 36) \)
\( \vec{b_1} \times \vec{b_2} = 12\hat{i} - 18\hat{j} + 36\hat{k} \).

Now, calculate the scalar triple product: \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) \):
\( (\hat{i} - 2\hat{j} + \hat{k}) \cdot (12\hat{i} - 18\hat{j} + 36\hat{k}) \)
\( = (1)(12) + (-2)(-18) + (1)(36) \)
\( = 12 + 36 + 36 = 84 \).
Since \( 84 \neq 0 \), the lines are skew.

**Shortest distance:**
\( \delta = \frac{ |(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})| }{ |\vec{b_1} \times \vec{b_2}| } \).
Numerator is \( |84| = 84 \).
Denominator: \( |\vec{b_1} \times \vec{b_2}| = |12\hat{i} - 18\hat{j} + 36\hat{k}| \)
\( = \sqrt{12^2 + (-18)^2 + 36^2} = \sqrt{144 + 324 + 1296} = \sqrt{1764} \).
Let's simplify \( 12\hat{i} - 18\hat{j} + 36\hat{k} = 6(2\hat{i} - 3\hat{j} + 6\hat{k}) \).
So, \( |\vec{b_1} \times \vec{b_2}| = 6 \sqrt{2^2 + (-3)^2 + 6^2} = 6 \sqrt{4 + 9 + 36} = 6 \sqrt{49} = 6 \times 7 = 42 \).
The shortest distance \( \delta = \frac{ 84 }{ 42 } = 2 \) units. This is the minimum separation between the two lines.
In simple words: First, we changed the lines into a standard format that shows a point on the line and its direction. Then, we checked if they are parallel; they were not. Next, we did a special vector calculation that showed they don't meet, proving they are skew lines. Finally, we used these calculated values in a formula to find the shortest distance between them.

🎯 Exam Tip: Converting complex line equations into the standard symmetric form or vector form is the first and most critical step. Errors here will propagate through the entire calculation, so be extra careful with algebra and signs.

Question 6. Find the parametric form of vector equation of the straight line passing through (-1, 2, 1) and parallel to the straight line \( \overline { r } = (2\hat { i } + 3\hat { j } – \hat { k }) + t(\hat { i } – 2\hat { j } + \hat { k }) \) and hence find the shortest distance between the lines.
Answer: Let the new line be \( L_1 \) and the given line be \( L_2 \).

**For line \( L_1 \) (the new line):**
It passes through the point \( (-1, 2, 1) \), so \( \vec{a_1} = -\hat{i} + 2\hat{j} + \hat{k} \).
It is parallel to the given line \( L_2 \), so its direction vector will be the same as \( L_2 \)'s direction vector.
From \( L_2 = (2\hat{i} + 3\hat{j} – \hat{k}) + t(\hat{i} – 2\hat{j} + \hat{k}) \), the direction vector is \( \vec{b_1} = \hat{i} - 2\hat{j} + \hat{k} \).

**Parametric form of vector equation for \( L_1 \):**
\( \overline{r} = \vec{a_1} + s\vec{b_1} \)
\( \overline{r} = (-\hat{i} + 2\hat{j} + \hat{k}) + s(\hat{i} - 2\hat{j} + \hat{k}) \). This is the equation of the new line.

**For line \( L_2 \) (the given line):**
It passes through the point \( (2, 3, -1) \), so \( \vec{a_2} = 2\hat{i} + 3\hat{j} - \hat{k} \).
Its direction vector is \( \vec{b_2} = \hat{i} - 2\hat{j} + \hat{k} \).

**Shortest distance between the two lines:**
Since \( \vec{b_1} = \vec{b_2} \), the two lines are parallel. The formula for the shortest distance \( \delta \) between two parallel lines is \( \delta = \frac{ |(\vec{a_2} - \vec{a_1}) \times \vec{b}| }{ |\vec{b}| } \), where \( \vec{b} \) is the common direction vector.

First, calculate \( \vec{a_2} - \vec{a_1} \):
\( \vec{a_2} - \vec{a_1} = (2\hat{i} + 3\hat{j} - \hat{k}) - (-\hat{i} + 2\hat{j} + \hat{k}) \)
\( = 2\hat{i} + 3\hat{j} - \hat{k} + \hat{i} - 2\hat{j} - \hat{k} \)
\( = 3\hat{i} + \hat{j} - 2\hat{k} \).

Next, calculate \( (\vec{a_2} - \vec{a_1}) \times \vec{b} \), using \( \vec{b} = \hat{i} - 2\hat{j} + \hat{k} \):
\[ (\vec{a_2} - \vec{a_1}) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & -2 & 1 \end{vmatrix} \]
\( = \hat{i}((1)(1) - (-2)(-2)) - \hat{j}((3)(1) - (-2)(1)) + \hat{k}((3)(-2) - (1)(1)) \)
\( = \hat{i}(1 - 4) - \hat{j}(3 + 2) + \hat{k}(-6 - 1) \)
\( = -3\hat{i} - 5\hat{j} - 7\hat{k} \).

Now, find the magnitude of this cross product:
\( |(\vec{a_2} - \vec{a_1}) \times \vec{b}| = |-3\hat{i} - 5\hat{j} - 7\hat{k}| = \sqrt{(-3)^2 + (-5)^2 + (-7)^2} \)
\( = \sqrt{9 + 25 + 49} = \sqrt{83} \).

Finally, find the magnitude of the direction vector \( \vec{b} \):
\( |\vec{b}| = |\hat{i} - 2\hat{j} + \hat{k}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \).

The shortest distance \( \delta = \frac{ \sqrt{83} }{ \sqrt{6} } = \sqrt{\frac{83}{6}} \) units. This is the constant separation between the two parallel lines.
In simple words: First, we wrote the equation for the new line, using the given point and the direction from the parallel line. Since the two lines are parallel, we used a special formula to find the distance between them. This involved calculating a vector from one point to another, crossing it with the common direction vector, and then dividing by the length of the direction vector.

🎯 Exam Tip: Carefully distinguish between formulas for skew lines and parallel lines. For parallel lines, the direction vectors are identical or proportional, leading to a simpler shortest distance formula that uses a cross product involving only one direction vector.

Question 7. Find the foot of the perpendicular drawn from the point (5, 4, 2) to the line \( \frac { x+1 }{ 2 } = \frac { y-3 }{ 3 } = \frac { z-1 }{ -1 } \). Also, find the equation of the perpendicular.
Answer: Let the given point be \( D = (5, 4, 2) \).
Let the given line be \( L \), with equation \( \frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{-1} \).
Any point F on the line L can be represented by setting the ratios equal to a parameter, say \( t \).
\( \frac{x+1}{2} = t \implies x = 2t - 1 \)
\( \frac{y-3}{3} = t \implies y = 3t + 3 \)
\( \frac{z-1}{-1} = t \implies z = -t + 1 \)
So, the coordinates of any point F on the line are \( (2t - 1, 3t + 3, -t + 1) \).

Now, we find the direction vector of the line segment DF.
\( \vec{DF} = F - D = (2t - 1 - 5)\hat{i} + (3t + 3 - 4)\hat{j} + (-t + 1 - 2)\hat{k} \)
\( \vec{DF} = (2t - 6)\hat{i} + (3t - 1)\hat{j} + (-t - 1)\hat{k} \).

The direction vector of the given line \( L \) is \( \vec{b} = 2\hat{i} + 3\hat{j} - \hat{k} \).
Since DF is perpendicular to line L, their direction vectors must be orthogonal. This means their dot product is zero.
\( \vec{DF} \cdot \vec{b} = 0 \)
\( ((2t - 6)\hat{i} + (3t - 1)\hat{j} + (-t - 1)\hat{k}) \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = 0 \)
\( (2t - 6)(2) + (3t - 1)(3) + (-t - 1)(-1) = 0 \)
\( 4t - 12 + 9t - 3 + t + 1 = 0 \)
\( (4t + 9t + t) + (-12 - 3 + 1) = 0 \)
\( 14t - 14 = 0 \)
\( 14t = 14 \)
\( t = 1 \).

**Foot of the perpendicular (F):**
Substitute \( t=1 \) back into the coordinates of F:
\( x = 2(1) - 1 = 1 \)
\( y = 3(1) + 3 = 6 \)
\( z = -(1) + 1 = 0 \)
So, the foot of the perpendicular is \( F = (1, 6, 0) \). This is the point on the line closest to D.

**Equation of the perpendicular (line DF):**
The line DF passes through \( D=(5, 4, 2) \) and \( F=(1, 6, 0) \).
The direction vector of DF is \( \vec{DF} = (1-5)\hat{i} + (6-4)\hat{j} + (0-2)\hat{k} \)
\( \vec{DF} = -4\hat{i} + 2\hat{j} - 2\hat{k} \).
Using point D and direction vector \( \vec{DF} \), the Cartesian equation of the perpendicular is:
\( \frac{x-5}{-4} = \frac{y-4}{2} = \frac{z-2}{-2} \). We can simplify the direction ratios by dividing by -2, giving (2, -1, 1).
So, the equation of the perpendicular is \( \frac{x-5}{2} = \frac{y-4}{-1} = \frac{z-2}{1} \). This line represents the shortest path from point D to the given line.
In simple words: First, we found a general point on the given line. Then, we made a vector from the given outside point to this general point. Since this vector must be straight down (perpendicular) to the line, we used the dot product rule (making it zero) to find the exact value of 't'. This 't' helped us find the precise point on the line where the perpendicular meets it. Finally, we wrote the equation for this perpendicular line using the two points we now know.

🎯 Exam Tip: The key idea here is that the vector connecting the given point to the foot of the perpendicular (which lies on the line) must be perpendicular to the direction vector of the given line. This orthogonality condition, expressed as a zero dot product, allows you to find the parameter 't'.