Samacheer Kalvi Class 12 Maths Solutions Chapter 6 Applications of Vector Algebra Exercise 6.4

Question 1. Find the non-parametric form of vector equation and Cartesian equations of the straight line passing through the point with position vector \( 4\hat { i } + 3\hat {j} - 7\hat {k} \) and parallel to the vector \( 2\hat { i } - 6\hat { j } + 7\hat { k } \).
Answer:
Given a point with position vector \( \vec{a} = 4\hat{i} + 3\hat{j} - 7\hat{k} \). This means the coordinates are \( (x_1, y_1, z_1) = (4, 3, -7) \).
The line is parallel to a vector \( \vec{b} = 2\hat{i} - 6\hat{j} + 7\hat{k} \). So, the direction ratios are \( (l, m, n) = (2, -6, 7) \).
The non-parametric vector equation of the straight line is given by \( (\vec{r} - \vec{a}) \times \vec{b} = 0 \).
Substituting the values, we get: \( [\vec{r} - (4\hat{i} + 3\hat{j} - 7\hat{k})] \times (2\hat{i} - 6\hat{j} + 7\hat{k}) = 0 \).
The Cartesian equations of the straight line are \( \frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n} \).
Using the given point and direction ratios:
\( \implies \frac{x - 4}{2} = \frac{y - 3}{-6} = \frac{z - (-7)}{7} \)
\( \implies \frac{x - 4}{2} = \frac{y - 3}{-6} = \frac{z + 7}{7} \).
In simple words: To find the equation of a line, we need a point it passes through and a vector it runs parallel to. We use these two pieces of information to write the vector form and then the coordinate form (Cartesian equations).

🎯 Exam Tip: Remember that the non-parametric vector equation uses the cross product, while the parametric vector equation is \( \vec{r} = \vec{a} + t\vec{b} \). Make sure to match the correct form to the question.

Question 2. Find the parametric form of vector equation and Cartesian equations of the straight line passing through the point (-2, 3, 4) and parallel to the straight line \( \frac {x-1 }{ -4 } = \frac { y+3 }{ 5 } = \frac { 8-z }{ 6 } \).
Answer:
The straight line passes through the point \( (-2, 3, 4) \). So, the position vector of this point is \( \vec{a} = -2\hat{i} + 3\hat{j} + 4\hat{k} \). Here, \( (x_1, y_1, z_1) = (-2, 3, 4) \).
The line is parallel to another straight line given by \( \frac {x-1 }{ -4 } = \frac { y+3 }{ 5 } = \frac { 8-z }{ 6 } \).
We need to rewrite the third term to match the standard form \( \frac{z - z_1}{n} \):
\( \frac{8-z}{6} = \frac{-(z-8)}{6} = \frac{z-8}{-6} \).
So, the given parallel line is \( \frac {x-1 }{ -4 } = \frac { y+3 }{ 5 } = \frac { z-8 }{ -6 } \).
The direction ratios of this parallel line are \( (l, m, n) = (-4, 5, -6) \). So, the direction vector is \( \vec{b} = -4\hat{i} + 5\hat{j} - 6\hat{k} \).
The parametric vector equation of the required straight line is \( \vec{r} = \vec{a} + t\vec{b} \), where \( t \in \mathbb{R} \).
Substituting the values of \( \vec{a} \) and \( \vec{b} \):
\( \vec{r} = (-2\hat{i} + 3\hat{j} + 4\hat{k}) + t(-4\hat{i} + 5\hat{j} - 6\hat{k}) \), where \( t \in \mathbb{R} \).
The Cartesian equation of the straight line is \( \frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n} \).
Substituting the point \( (-2, 3, 4) \) and direction ratios \( (-4, 5, -6) \):
\( \frac{x - (-2)}{-4} = \frac{y - 3}{5} = \frac{z - 4}{-6} \)
\( \implies \frac{x + 2}{-4} = \frac{y - 3}{5} = \frac{z - 4}{-6} \).
In simple words: When a line is parallel to another line, it means they share the same direction. We use the direction numbers from the second line along with the given point to create the new line's equations.

🎯 Exam Tip: Always convert the given equation of the parallel line to the standard form \( \frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n} \) before extracting direction ratios, especially when terms like \( (8-z) \) are present.

Question 3. Find the points where the straight line passes through (6, 7, 4) and (8, 4, 9) cuts the xz and yz planes.
Answer:
The straight line passes through points \( P_1(6, 7, 4) \) and \( P_2(8, 4, 9) \).
First, find the direction ratios of the line. These are \( (x_2 - x_1, y_2 - y_1, z_2 - z_1) \).
Direction ratios \( = (8 - 6, 4 - 7, 9 - 4) = (2, -3, 5) \).
So the straight line is parallel to the vector \( \vec{b} = 2\hat{i} - 3\hat{j} + 5\hat{k} \).
The parametric vector equation can be written using either point. Using \( P_1(6, 7, 4) \):
\( \vec{r} = (6\hat{i} + 7\hat{j} + 4\hat{k}) + t(2\hat{i} - 3\hat{j} + 5\hat{k}) \).
(Or, using \( P_2(8, 4, 9) \): \( \vec{r} = (8\hat{i} + 4\hat{j} + 9\hat{k}) + t(2\hat{i} - 3\hat{j} + 5\hat{k}) \)).
The Cartesian equation of the straight line passing through two points is \( \frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1} \).
Using point \( (6, 7, 4) \) and direction ratios \( (2, -3, 5) \):
\( \frac{x - 6}{2} = \frac{y - 7}{-3} = \frac{z - 4}{5} \). Let this ratio be equal to \( t \).
So, any arbitrary point on the line can be represented as \( (x, y, z) = (2t + 6, -3t + 7, 5t + 4) \).

(i) **Intersection with the xz-plane:**
In the xz-plane, the y-coordinate is always 0. So, we set \( y = 0 \).
\( -3t + 7 = 0 \)
\( 3t = 7 \)
\( t = \frac{7}{3} \).
Now substitute \( t = \frac{7}{3} \) back into the general point coordinates:
\( x = 2(\frac{7}{3}) + 6 = \frac{14}{3} + \frac{18}{3} = \frac{32}{3} \)
\( z = 5(\frac{7}{3}) + 4 = \frac{35}{3} + \frac{12}{3} = \frac{47}{3} \).
The point where the line cuts the xz-plane is \( (\frac{32}{3}, 0, \frac{47}{3}) \).

(ii) **Intersection with the yz-plane:**
In the yz-plane, the x-coordinate is always 0. So, we set \( x = 0 \).
\( 2t + 6 = 0 \)
\( 2t = -6 \)
\( t = -3 \).
Now substitute \( t = -3 \) back into the general point coordinates:
\( y = -3(-3) + 7 = 9 + 7 = 16 \)
\( z = 5(-3) + 4 = -15 + 4 = -11 \).
The point where the line cuts the yz-plane is \( (0, 16, -11) \).
In simple words: To find where a line crosses a plane (like the xz-plane), we use the special property of that plane (like y=0 for xz-plane). We find a 'time' value (t) for that intersection and then put it back into the line's equation to get the exact point.

🎯 Exam Tip: Remember that for the xz-plane, y=0; for the yz-plane, x=0; and for the xy-plane, z=0. This is a common point of confusion.

Question 4. Find the direction cosines of the straight line passing through the points (5, 6, 7) and (7, 9, 13). Also, find the parametric form of vector equation and Cartesian equations of the straight line passing through two given points.
Answer:
Given points are \( P_1(5, 6, 7) \) and \( P_2(7, 9, 13) \).
First, find the direction ratios \( (l, m, n) \) of the line. These are \( (x_2 - x_1, y_2 - y_1, z_2 - z_1) \).
Direction ratios \( = (7 - 5, 9 - 6, 13 - 7) = (2, 3, 6) \).
So, the straight line is parallel to the vector \( \vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k} \).
Next, calculate the magnitude of the direction vector:
\( |\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \).
The direction cosines are \( (\frac{l}{|\vec{b}|}, \frac{m}{|\vec{b}|}, \frac{n}{|\vec{b}|}) \).
Direction cosines \( = (\frac{2}{7}, \frac{3}{7}, \frac{6}{7}) \).

Now, find the equations of the line passing through these two points.
Using point \( P_1(5, 6, 7) \), the position vector is \( \vec{a} = 5\hat{i} + 6\hat{j} + 7\hat{k} \).
The parametric form of the vector equation of the straight line is \( \vec{r} = \vec{a} + t\vec{b} \), where \( t \in \mathbb{R} \).
Substituting the values:
\( \vec{r} = (5\hat{i} + 6\hat{j} + 7\hat{k}) + t(2\hat{i} + 3\hat{j} + 6\hat{k}) \).
(Alternatively, using point \( P_2(7, 9, 13) \): \( \vec{r} = (7\hat{i} + 9\hat{j} + 13\hat{k}) + t(2\hat{i} + 3\hat{j} + 6\hat{k}) \), where \( t \in \mathbb{R} \)).
The Cartesian equation of the straight line is \( \frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n} \).
Using point \( (5, 6, 7) \) and direction ratios \( (2, 3, 6) \):
\( \frac{x - 5}{2} = \frac{y - 6}{3} = \frac{z - 7}{6} \).
(Alternatively, using point \( (7, 9, 13) \) and direction ratios \( (2, 3, 6) \): \( \frac{x - 7}{2} = \frac{y - 9}{3} = \frac{z - 13}{6} \)).
In simple words: First, find the "steps" to go from one point to the next, which gives the direction ratios. Then, divide these steps by their total length to get the direction cosines. Finally, use one point and these direction steps to write the line's equations.

🎯 Exam Tip: The direction ratios are specific to the direction vector, but the direction cosines are unique and represent the cosines of the angles the line makes with the coordinate axes. Make sure to calculate the magnitude correctly.

Question 5. Find the acute angle between the following lines.
(i) \( \overline { r } = (4\hat { i } - \hat {j }) + t(\hat { i } + 2\hat { j } - 2\hat { k }), \overline { r } = (\hat { i } - 2\hat { j } + 4\hat { k }) + s(-\hat { i } - 2\hat { j } + 2\hat { k }) \)
(ii) \( \frac {x+4 }{ 3 } = \frac { y-7 }{ 4 } = \frac { z+5 }{ 5 }, \overline { r } = 4\hat { k } + t(2\hat { i } + \hat { j } + \hat { k }) \)
(iii) \( 2x = 3y = -z \) and \( 6x = -y = -4z \).
Answer:
(i) For the first line, the direction vector is \( \vec{b_1} = \hat{i} + 2\hat{j} - 2\hat{k} \).
For the second line, the direction vector is \( \vec{b_2} = -\hat{i} - 2\hat{j} + 2\hat{k} \).
We can see that \( \vec{b_2} = -1(\hat{i} + 2\hat{j} - 2\hat{k}) = -\vec{b_1} \). This means the lines are parallel or the same, running in opposite directions.
To find the angle \( \theta \) between the lines, we use the formula \( \cos \theta = \frac { |\vec{b_1} \cdot \vec{b_2}| }{ |\vec{b_1}| |\vec{b_2}| } \).
First, calculate the dot product:
\( \vec{b_1} \cdot \vec{b_2} = (1)(-1) + (2)(-2) + (-2)(2) = -1 - 4 - 4 = -9 \).
Next, find the magnitudes of the vectors:
\( |\vec{b_1}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \).
\( |\vec{b_2}| = \sqrt{(-1)^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \).
Now, calculate \( \cos \theta \):
\( \cos \theta = \frac{ |-9| }{ (3)(3) } = \frac{9}{9} = 1 \).
Since \( \cos \theta = 1 \), the angle \( \theta = 0^\circ \). This shows the lines are parallel.

(ii) For the first line, given in Cartesian form \( \frac {x+4 }{ 3 } = \frac { y-7 }{ 4 } = \frac { z+5 }{ 5 } \), the direction vector is \( \vec{b_1} = 3\hat{i} + 4\hat{j} + 5\hat{k} \).
For the second line, given in vector form \( \overline { r } = 4\hat { k } + t(2\hat { i } + \hat { j } + \hat { k }) \), the direction vector is \( \vec{b_2} = 2\hat{i} + \hat{j} + \hat{k} \).
Calculate the dot product:
\( \vec{b_1} \cdot \vec{b_2} = (3)(2) + (4)(1) + (5)(1) = 6 + 4 + 5 = 15 \).
Find the magnitudes of the vectors:
\( |\vec{b_1}| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2} \).
\( |\vec{b_2}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \).
Now, calculate \( \cos \theta \):
\( \cos \theta = \frac{ |\vec{b_1} \cdot \vec{b_2}| }{ |\vec{b_1}| |\vec{b_2}| } = \frac{15}{ (5\sqrt{2})(\sqrt{6}) } = \frac{15}{ 5\sqrt{12} } = \frac{3}{ \sqrt{12} } = \frac{3}{ 2\sqrt{3} } = \frac{3\sqrt{3}}{ 2 \times 3 } = \frac{\sqrt{3}}{2} \).
Since \( \cos \theta = \frac{\sqrt{3}}{2} \), the acute angle \( \theta = \frac{\pi}{6} \) or \( 30^\circ \).

(iii) For the first line, \( 2x = 3y = -z \). We can write this as:
\( \frac{x}{1/2} = \frac{y}{1/3} = \frac{z}{-1} \).
To make it simpler with integers, we can divide by the common multiple of denominators or just use the given numbers directly.
Let \( 2x = 3y = -z = k \). Then \( x = k/2, y = k/3, z = -k \).
So, the direction ratios for the first line are proportional to \( (\frac{1}{2}, \frac{1}{3}, -1) \). We can multiply by 6 to get integer ratios: \( (3, 2, -6) \).
So, \( \vec{b_1} = 3\hat{i} + 2\hat{j} - 6\hat{k} \).
For the second line, \( 6x = -y = -4z \). We can write this as:
\( \frac{x}{1/6} = \frac{y}{-1} = \frac{z}{-1/4} \).
To simplify to integers, multiply by 12: \( (2, -12, -3) \).
So, \( \vec{b_2} = 2\hat{i} - 12\hat{j} - 3\hat{k} \).
Now, calculate the dot product:
\( \vec{b_1} \cdot \vec{b_2} = (3)(2) + (2)(-12) + (-6)(-3) = 6 - 24 + 18 = 0 \).
Since the dot product \( \vec{b_1} \cdot \vec{b_2} = 0 \), the lines are perpendicular.
Therefore, \( \cos \theta = 0 \), which means the acute angle \( \theta = \frac{\pi}{2} \) or \( 90^\circ \).
In simple words: To find the angle between lines, first pull out their direction vectors from the equations. Then, use the dot product formula, which involves multiplying the corresponding parts of the vectors and dividing by their lengths. A dot product of zero means the lines are at a right angle.

🎯 Exam Tip: Remember that \( \cos \theta = \frac{ |\vec{b_1} \cdot \vec{b_2}| }{ |\vec{b_1}| |\vec{b_2}| } \). For lines in symmetric form like \( 2x = 3y = -z \), convert them to \( \frac{x}{a} = \frac{y}{b} = \frac{z}{c} \) to easily find the direction ratios \( (a,b,c) \). Pay attention to negative signs and fractions when converting. If the dot product is zero, the angle is 90 degrees.

Question 6. The vertices of \( \triangle ABC \) are A(7, 2, 1), B(6, 0, 3), and C(4, 2, 4). Find \( \angle ABC \).
Answer:
To find \( \angle ABC \), we need to find the angle between vectors \( \vec{BA} \) and \( \vec{BC} \). The angle is formed at vertex B.
First, find the vector \( \vec{BA} \). It is found by \( \vec{OA} - \vec{OB} \).
\( \vec{BA} = (7\hat{i} + 2\hat{j} + \hat{k}) - (6\hat{i} + 0\hat{j} + 3\hat{k}) \)
\( \vec{BA} = (7-6)\hat{i} + (2-0)\hat{j} + (1-3)\hat{k} \)
\( \vec{BA} = \hat{i} + 2\hat{j} - 2\hat{k} \).
Next, find the vector \( \vec{BC} \). It is found by \( \vec{OC} - \vec{OB} \).
\( \vec{BC} = (4\hat{i} + 2\hat{j} + 4\hat{k}) - (6\hat{i} + 0\hat{j} + 3\hat{k}) \)
\( \vec{BC} = (4-6)\hat{i} + (2-0)\hat{j} + (4-3)\hat{k} \)
\( \vec{BC} = -2\hat{i} + 2\hat{j} + \hat{k} \).
To find the angle \( \theta \) between \( \vec{BA} \) and \( \vec{BC} \), we use the dot product formula:
\( \cos \theta = \frac{ \vec{BA} \cdot \vec{BC} }{ |\vec{BA}| |\vec{BC}| } \).
Calculate the dot product:
\( \vec{BA} \cdot \vec{BC} = (1)(-2) + (2)(2) + (-2)(1) = -2 + 4 - 2 = 0 \).
Since the dot product \( \vec{BA} \cdot \vec{BC} = 0 \), the vectors are perpendicular.
Therefore, \( \cos \theta = 0 \), which means \( \angle ABC = \frac{\pi}{2} \) or \( 90^\circ \). The angle at B is a right angle.
In simple words: To find an angle inside a triangle, we first find the two vectors that meet at that angle's corner. Then, we multiply these vectors together using the dot product and divide by their lengths. If the dot product is zero, the angle is a right angle.

🎯 Exam Tip: When finding the angle at a vertex, say B, always use vectors starting FROM B, such as \( \vec{BA} \) and \( \vec{BC} \). Do not use \( \vec{AB} \) and \( \vec{BC} \), as this would not represent the internal angle at B.

Question 7. If the straight line joining the points (2, 1, 4) and (a - 1, 4, -1) is parallel to the line joining the points (0, 2, b - 1) and (5, 3, -2) find the values of a and b.
Answer:
For the first line, let the points be \( P_1(2, 1, 4) \) and \( P_2(a - 1, 4, -1) \).
The direction ratios \( (l_1, m_1, n_1) \) are \( (x_2 - x_1, y_2 - y_1, z_2 - z_1) \).
\( l_1 = (a - 1) - 2 = a - 3 \)
\( m_1 = 4 - 1 = 3 \)
\( n_1 = -1 - 4 = -5 \).
So, the direction ratios for the first line are \( (a - 3, 3, -5) \).

For the second line, let the points be \( P_3(0, 2, b - 1) \) and \( P_4(5, 3, -2) \).
The direction ratios \( (l_2, m_2, n_2) \) are \( (x_4 - x_3, y_4 - y_3, z_4 - z_3) \).
\( l_2 = 5 - 0 = 5 \)
\( m_2 = 3 - 2 = 1 \)
\( n_2 = -2 - (b - 1) = -2 - b + 1 = -b - 1 \).
So, the direction ratios for the second line are \( (5, 1, -b - 1) \).

Since the two straight lines are parallel, their direction ratios must be proportional. This means:
\( \frac{l_1}{l_2} = \frac{m_1}{m_2} = \frac{n_1}{n_2} \).
\( \implies \frac{a - 3}{5} = \frac{3}{1} = \frac{-5}{-b - 1} \).
Now we solve for \( a \) and \( b \) by taking two pairs of equalities:
From \( \frac{a - 3}{5} = \frac{3}{1} \):
\( a - 3 = 3 \times 5 \)
\( a - 3 = 15 \)
\( a = 15 + 3 \)
\( a = 18 \).
From \( \frac{3}{1} = \frac{-5}{-b - 1} \):
\( 3(-b - 1) = -5 \)
\( -3b - 3 = -5 \)
\( -3b = -5 + 3 \)
\( -3b = -2 \)
\( b = \frac{-2}{-3} \)
\( b = \frac{2}{3} \).
Thus, the values are \( a = 18 \) and \( b = \frac{2}{3} \).
In simple words: When two lines are parallel, their directions are the same. This means the ratios of their direction components must be equal. By setting up these equal ratios, we can find the unknown values.

🎯 Exam Tip: Remember that "parallel" means direction ratios are proportional, while "perpendicular" means their dot product is zero. Be careful with signs when calculating direction ratios and solving equations.

Question 8. If the straight lines \( \frac {x-5 }{ 5m+2 } = \frac { 2-y }{ 5 } = \frac { 1-z }{ -1 } \) and \( x = \frac { 2y+1 }{ 4 } = \frac { 1-z }{ -3 } \) are perpendicular to each other, find the value of m.
Answer:
First, we need to find the direction ratios for both lines by writing them in the standard Cartesian form \( \frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n} \).

**For the first line:** \( \frac {x-5 }{ 5m+2 } = \frac { 2-y }{ 5 } = \frac { 1-z }{ -1 } \)
We need to adjust the terms with \( y \) and \( z \):
\( \frac{2-y}{5} = \frac{-(y-2)}{5} = \frac{y-2}{-5} \)
\( \frac{1-z}{-1} = \frac{-(z-1)}{-1} = \frac{z-1}{1} \)
So, the first line's equation in standard form is: \( \frac {x-5 }{ 5m+2 } = \frac { y-2 }{ -5 } = \frac { z-1 }{ 1 } \).
The direction ratios for the first line are \( (l_1, m_1, n_1) = (5m+2, -5, 1) \).

**For the second line:** \( x = \frac { 2y+1 }{ 4 } = \frac { 1-z }{ -3 } \)
We need to adjust the terms with \( x, y \) and \( z \):
\( x = \frac{x-0}{1} \)
\( \frac{2y+1}{4} = \frac{2(y+1/2)}{4} = \frac{y+1/2}{2} \)
\( \frac{1-z}{-3} = \frac{-(z-1)}{-3} = \frac{z-1}{3} \)
So, the second line's equation in standard form is: \( \frac {x-0 }{ 1 } = \frac { y+1/2 }{ 2 } = \frac { z-1 }{ 3 } \).
The direction ratios for the second line are \( (l_2, m_2, n_2) = (1, 2, 3) \).

Since the two lines are perpendicular, the dot product of their direction vectors must be zero:
\( l_1 l_2 + m_1 m_2 + n_1 n_2 = 0 \).
\( (5m+2)(1) + (-5)(2) + (1)(3) = 0 \)
\( 5m + 2 - 10 + 3 = 0 \)
\( 5m - 5 = 0 \)
\( 5m = 5 \)
\( m = 1 \).
The value of \( m \) is 1.
In simple words: First, rewrite both line equations in a standard format to clearly see their direction numbers. If lines are perpendicular, multiplying their direction numbers together and adding the results will always give zero. We use this rule to solve for 'm'.

🎯 Exam Tip: Always be cautious when extracting direction ratios from non-standard forms like \( (2-y) \) or \( (2y+1) \). Ensure the variables \( x, y, z \) have a coefficient of 1 and are subtracted from constants (e.g., \( y-y_1 \)).

Question 9. Show that the points (2, 3, 4), (-1, 4, 5) and (8, 1, 2) are collinear.
Answer:
Let the given points be A(2, 3, 4), B(-1, 4, 5), and C(8, 1, 2).
To show that the points are collinear, we need to show that two vectors formed by these points are parallel (i.e., one is a scalar multiple of the other) and share a common point.

First, find the vector \( \vec{AB} \). It is found by \( \vec{OB} - \vec{OA} \).
\( \vec{AB} = (-1\hat{i} + 4\hat{j} + 5\hat{k}) - (2\hat{i} + 3\hat{j} + 4\hat{k}) \)
\( \vec{AB} = (-1-2)\hat{i} + (4-3)\hat{j} + (5-4)\hat{k} \)
\( \vec{AB} = -3\hat{i} + \hat{j} + \hat{k} \).

Next, find the vector \( \vec{BC} \). It is found by \( \vec{OC} - \vec{OB} \).
\( \vec{BC} = (8\hat{i} + \hat{j} + 2\hat{k}) - (-1\hat{i} + 4\hat{j} + 5\hat{k}) \)
\( \vec{BC} = (8-(-1))\hat{i} + (1-4)\hat{j} + (2-5)\hat{k} \)
\( \vec{BC} = (8+1)\hat{i} + (-3)\hat{j} + (-3)\hat{k} \)
\( \vec{BC} = 9\hat{i} - 3\hat{j} - 3\hat{k} \).

Now, we check if \( \vec{BC} \) is a scalar multiple of \( \vec{AB} \).
\( \vec{BC} = 9\hat{i} - 3\hat{j} - 3\hat{k} = -3(-3\hat{i} + \hat{j} + \hat{k}) \).
We can see that \( \vec{BC} = -3\vec{AB} \).
Since \( \vec{BC} \) is a scalar multiple of \( \vec{AB} \), the vectors \( \vec{AB} \) and \( \vec{BC} \) are parallel. Also, they share a common point B.
Therefore, the points A, B, and C are collinear.
In simple words: To prove points lie on the same line, we pick a point and make two vectors from it to the other two points. If these two vectors are parallel (meaning one is just a scaled version of the other), then all three points must be on the same straight line.

🎯 Exam Tip: To show collinearity, always demonstrate that one vector is a scalar multiple of another (e.g., \( \vec{AB} = k\vec{BC} \)) AND that they share a common point. Showing just proportionality of direction ratios is not enough; the common point is crucial.