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

Question 1. Find the vector equation of a line which passes through the vector \( 2 \hat{i} + 4 \hat{j} + 5 \hat{k} \) and is parallel to the vector \( 2 \hat{i} + 2 \hat{j} - 3 \hat{k} \). Find the Cartesian form also.
Answer: We know that the vector equation of a line that passes through a point with position vector \( \vec{a} \) and is parallel to a vector \( \vec{b} \) is given by the formula \( \vec{r} = \vec{a} + \lambda\vec{b} \), where \( \lambda \) is a scalar constant.
Here, the position vector of the point is \( \vec{a} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \) (from the working, assuming a typo in the question or the starting point in the solution is \( 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \) instead of \( 2 \hat{i} + 4 \hat{j} + 5 \hat{k} \)) and the parallel vector is \( \vec{b} = 2 \hat{i} + 2 \hat{j} - 3 \hat{k} \).
So, the required vector equation of the line is:
\( \vec{r} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} + \lambda(2 \hat{i} + 2 \hat{j} - 3 \hat{k}) \)
To find the Cartesian form, let \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \).
So, \( x \hat{i} + y \hat{j} + z \hat{k} = (3 + 2\lambda) \hat{i} + (4 + 2\lambda) \hat{j} + (5 - 3\lambda) \hat{k} \)
Comparing the coefficients of \( \hat{i}, \hat{j}, \hat{k} \) on both sides, we get:
\( x = 3 + 2\lambda \)
\( \implies \frac{x-3}{2} = \lambda \)
\( y = 4 + 2\lambda \)
\( \implies \frac{y-4}{2} = \lambda \)
\( z = 5 - 3\lambda \)
\( \implies \frac{z-5}{-3} = \lambda \)
Since all these expressions are equal to \( \lambda \), we can set them equal to each other:
\( \frac{x-3}{2} = \frac{y-4}{2} = \frac{z-5}{-3} \)
This is the required equation of the line in Cartesian form.
In simple words: First, we write the line's equation using vectors, which needs a starting point and a direction. Then, we change this vector form into a Cartesian form by comparing the parts of the \(x, y, z\) coordinates to get a set of ratios that define the line.

🎯 Exam Tip: Remember that a vector equation requires a point on the line and a direction vector. The Cartesian form is derived by equating the ratios of \(x, y, z\) components based on the parameter \( \lambda \). Be careful with signs when moving numbers to find \( \lambda \).

Question 2. Find the vector equation of a line which is parallel to the vector \( 2 \hat{i} – \hat{j} + 3 \hat{k} \) and which passes through the point (5, -2, 4). Reduce it to the Cartesian form.
Answer: We know that the vector equation of a line passing through a point with position vector \( \vec{a} \) and parallel to a vector \( \vec{b} \) is given by the formula \( \vec{r} = \vec{a} + \lambda \vec{b} \).
Here, the position vector of the point is \( \vec{a} = 5 \hat{i} - 2 \hat{j} + 4 \hat{k} \).
The line is parallel to the vector \( \vec{b} = 2 \hat{i} - \hat{j} + 3 \hat{k} \).
So, the required vector equation of the line is:
\( \vec{r} = 5 \hat{i} - 2 \hat{j} + 4 \hat{k} + \lambda(2 \hat{i} - \hat{j} + 3 \hat{k}) \)
To find the Cartesian form, let \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \).
So, \( x \hat{i} + y \hat{j} + z \hat{k} = (5 + 2\lambda) \hat{i} + (-2 - \lambda) \hat{j} + (4 + 3\lambda) \hat{k} \)
Comparing the coefficients of \( \hat{i}, \hat{j}, \hat{k} \) on both sides, we get:
\( x = 5 + 2\lambda \)
\( \implies \frac{x-5}{2} = \lambda \)
\( y = -2 - \lambda \)
\( \implies \frac{y+2}{-1} = \lambda \)
\( z = 4 + 3\lambda \)
\( \implies \frac{z-4}{3} = \lambda \)
Equating these expressions for \( \lambda \), we get the Cartesian equation:
\( \frac{x-5}{2} = \frac{y+2}{-1} = \frac{z-4}{3} \)
This is the required equation of the line in Cartesian form.
In simple words: We find the vector equation by combining the given point and the parallel direction. Then, we convert it to the Cartesian form by writing the \(x, y, z\) coordinates in terms of a variable \( \lambda \) and making them all equal.

🎯 Exam Tip: When converting from vector to Cartesian form, ensure that the coefficients of \( \lambda \) in the vector equation become the denominators in the Cartesian form, and the constant terms from the position vector are subtracted from \(x, y, z\) in the numerators.

Question 3. Find the vector and Cartesian equations of the line that passes through the points (3, -2, -5), (3, -2, 6).
Answer: Let the position vectors of the given points A and B be \( \vec{a} \) and \( \vec{b} \) respectively.
So, \( \vec{a} = 3 \hat{i} - 2 \hat{j} - 5 \hat{k} \)
And \( \vec{b} = 3 \hat{i} - 2 \hat{j} + 6 \hat{k} \)
The vector equation of a line passing through two points A and B with position vectors \( \vec{a} \) and \( \vec{b} \) is given by \( \vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a}) \).
First, let's find the direction vector \( \vec{b} - \vec{a} \):
\( \vec{b} - \vec{a} = (3 \hat{i} - 2 \hat{j} + 6 \hat{k}) - (3 \hat{i} - 2 \hat{j} - 5 \hat{k}) \)
\( = (3-3) \hat{i} + (-2-(-2)) \hat{j} + (6-(-5)) \hat{k} \)
\( = 0 \hat{i} + 0 \hat{j} + 11 \hat{k} \)
Now substitute this into the vector equation:
\( \vec{r} = 3 \hat{i} - 2 \hat{j} - 5 \hat{k} + \lambda(0 \hat{i} + 0 \hat{j} + 11 \hat{k}) \)
\( \implies \vec{r} = 3 \hat{i} - 2 \hat{j} + (11\lambda - 5) \hat{k} \)
This is the required vector equation.
To find the Cartesian equation, let \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \).
Comparing coefficients:
\( x = 3 + 0\lambda \implies x = 3 \implies \frac{x-3}{0} = \lambda \)
\( y = -2 + 0\lambda \implies y = -2 \implies \frac{y+2}{0} = \lambda \)
\( z = -5 + 11\lambda \implies \frac{z-(-5)}{11} = \lambda \)
Equating these expressions:
\( \frac{x-3}{0} = \frac{y+2}{0} = \frac{z+5}{11} \)
This is the required Cartesian equation. Note that division by zero here means the direction ratios are 0, 0, 11, indicating a line parallel to the z-axis.
In simple words: To find the equation of a line that goes through two points, we first find the vector from one point to the other to get the direction. Then we use one of the points and this direction to write the vector equation. After that, we turn it into the Cartesian form by splitting it into \(x, y, z\) parts.

🎯 Exam Tip: When calculating the direction vector \( \vec{b} - \vec{a} \), be very careful with negative signs. If a component of the direction vector is zero, it means the line is parallel to the plane defined by the other two axes, and its Cartesian form will have a zero in the denominator for that component.

Question 4. Find the equation of the line in vector and in Cartesian form that passes through the point with position vector \( 2 \hat{i} – \hat{j} + 4 \hat{k} \) and is in the direction \( \hat{i} + 2 \hat{j} – \hat{k} \).
Answer: We know that the vector equation of a line passing through a point with position vector \( \vec{a} \) and parallel to a vector \( \vec{b} \) is given by \( \vec{r} = \vec{a} + \lambda \vec{b} \).
Here, the position vector of the point is \( \vec{a} = 2 \hat{i} - \hat{j} + 4 \hat{k} \).
The direction vector is \( \vec{b} = \hat{i} + 2 \hat{j} - \hat{k} \).
So, the required vector equation is:
\( \vec{r} = 2 \hat{i} - \hat{j} + 4 \hat{k} + \lambda(\hat{i} + 2 \hat{j} - \hat{k}) \)
To find the Cartesian form, let \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \).
So, \( x \hat{i} + y \hat{j} + z \hat{k} = (2 + \lambda) \hat{i} + (-1 + 2\lambda) \hat{j} + (4 - \lambda) \hat{k} \)
Comparing the coefficients of \( \hat{i}, \hat{j}, \hat{k} \) on both sides:
\( x = 2 + \lambda \)
\( \implies \frac{x-2}{1} = \lambda \)
\( y = -1 + 2\lambda \)
\( \implies \frac{y+1}{2} = \lambda \)
\( z = 4 - \lambda \)
\( \implies \frac{z-4}{-1} = \lambda \)
Equating these expressions for \( \lambda \), we get the Cartesian equation:
\( \frac{x-2}{1} = \frac{y+1}{2} = \frac{z-4}{-1} \)
This is the required equation of the line in Cartesian form.
In simple words: We are given a point and the direction a line travels. We use these two pieces of information to write the line's vector equation. Then, we separate the \(x, y, z\) components to get the standard Cartesian form of the line.

🎯 Exam Tip: The direction vector's components directly become the denominators in the Cartesian equation, and the point's coordinates are subtracted from \(x, y, z\) in the numerators. Always double-check the signs when setting up these equations.

Question 5. The Cartesian equations of a line are \( 6x - 2 = 3y + 1 = 2z - 2 \). Find the vector equation of this line.
Answer: The given Cartesian equation of the line is \( 6x - 2 = 3y + 1 = 2z - 2 \).
To convert this to the standard symmetric form \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \), where \( (x_1, y_1, z_1) \) is a point on the line and \( (a, b, c) \) are its direction ratios, we need to make the coefficients of \(x, y, z\) equal to 1.
Divide each part by the coefficient of \(x, y, z\) respectively:
\( 6(x-\frac{2}{6}) = 3(y + \frac{1}{3}) = 2(z - \frac{2}{2}) \)
\( \implies 6(x-\frac{1}{3}) = 3(y + \frac{1}{3}) = 2(z - 1) \)
Now, divide the entire equation by the Least Common Multiple (LCM) of the coefficients (6, 3, 2), which is 6, to get 1 in the numerator for \(x, y, z\) terms:
\( \frac{6(x-\frac{1}{3})}{6} = \frac{3(y + \frac{1}{3})}{6} = \frac{2(z - 1)}{6} \)
\( \implies \frac{x-\frac{1}{3}}{1} = \frac{y + \frac{1}{3}}{2} = \frac{z - 1}{3} \)
From this standard form, we can identify:
A point on the line is \( (x_1, y_1, z_1) = (\frac{1}{3}, -\frac{1}{3}, 1) \). So, its position vector is \( \vec{a} = \frac{1}{3} \hat{i} - \frac{1}{3} \hat{j} + \hat{k} \).
The direction ratios of the line are proportional to \( (a, b, c) = (1, 2, 3) \). So, the direction vector is \( \vec{b} = \hat{i} + 2 \hat{j} + 3 \hat{k} \).
The vector equation of the line is \( \vec{r} = \vec{a} + \lambda \vec{b} \).
Substituting the values:
\( \vec{r} = \frac{1}{3} \hat{i} - \frac{1}{3} \hat{j} + \hat{k} + \lambda(\hat{i} + 2 \hat{j} + 3 \hat{k}) \)
This is the required vector equation of the line. This conversion from Cartesian to vector form shows how neatly coordinates can be represented.
In simple words: We are given the Cartesian equation of a line, but it's not in the easiest form. We change it so that \(x, y, z\) each have a coefficient of 1. From this new form, we can easily pick out a point on the line and its direction, which then lets us write the line's vector equation.

🎯 Exam Tip: To convert a Cartesian equation like \( Ax + B = Cy + D = Ez + F \) to standard symmetric form, first factor out A, C, E from each term, then divide the entire equation by the LCM of A, C, E to get unit coefficients for \(x, y, z\).

Question 6. Find the vector equation of a line passing through the point with position vector \( \hat{i} - 2 \hat{j} - 3 \hat{k} \) and parallel to the line joining the points with position vectors \( \hat{i} - \hat{j} + 4 \hat{k} \) and \( 2 \hat{i} + \hat{j} + 2 \hat{k} \). Also find the Cartesian form of the equation.
Answer: Let A be the point through which the line passes, with position vector \( \vec{a} = \hat{i} - 2 \hat{j} - 3 \hat{k} \).
Let the line be parallel to the line joining points B and C, with position vectors \( \vec{b} = \hat{i} - \hat{j} + 4 \hat{k} \) and \( \vec{c} = 2 \hat{i} + \hat{j} + 2 \hat{k} \).
The direction vector of the line joining B and C is \( \overrightarrow{BC} = \vec{c} - \vec{b} \).
\( \overrightarrow{BC} = (2 \hat{i} + \hat{j} + 2 \hat{k}) - (\hat{i} - \hat{j} + 4 \hat{k}) \)
\( = (2-1) \hat{i} + (1-(-1)) \hat{j} + (2-4) \hat{k} \)
\( = \hat{i} + 2 \hat{j} - 2 \hat{k} \)
Since the required line is parallel to \( \overrightarrow{BC} \), its direction vector \( \vec{d} \) will be \( \vec{d} = \hat{i} + 2 \hat{j} - 2 \hat{k} \).
The vector equation of a line passing through point A (position vector \( \vec{a} \)) and parallel to vector \( \vec{d} \) is \( \vec{r} = \vec{a} + \lambda \vec{d} \).
Substituting the values:
\( \vec{r} = \hat{i} - 2 \hat{j} - 3 \hat{k} + \lambda(\hat{i} + 2 \hat{j} - 2 \hat{k}) \)
This is the required vector equation.
To find the Cartesian form, let \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \).
\( x \hat{i} + y \hat{j} + z \hat{k} = (1 + \lambda) \hat{i} + (-2 + 2\lambda) \hat{j} + (-3 - 2\lambda) \hat{k} \)
Comparing the coefficients of \( \hat{i}, \hat{j}, \hat{k} \):
\( x = 1 + \lambda \implies \frac{x-1}{1} = \lambda \)
\( y = -2 + 2\lambda \implies \frac{y+2}{2} = \lambda \)
\( z = -3 - 2\lambda \implies \frac{z+3}{-2} = \lambda \)
Equating these expressions for \( \lambda \):
\( \frac{x-1}{1} = \frac{y+2}{2} = \frac{z+3}{-2} \)
This is the required Cartesian equation of the line. Lines in 3D space are always determined by a point and a direction.
In simple words: First, we find the direction of the line by subtracting the position vectors of the two given points. Then, we use this direction and the first given point to write the line's vector equation. Finally, we convert this vector equation into its Cartesian form by matching the \(x, y, z\) parts.

🎯 Exam Tip: When a line is parallel to another line, it means they share the same direction vector. Always calculate the direction vector correctly first (final point's position vector minus initial point's position vector) before forming the equation.

Question 7. Show that the points, whose position vectors are given by \( -4 \hat{i} + 2 \hat{j} – 3 \hat{k} \), \( \hat{i} + 3 \hat{j} – 2 \hat{k} \) and \( (-9 \hat{i} + \hat{j} – 4 \hat{k}) \) are collinear.
Answer: Let the given points be A, B, and C, with position vectors:
\( \vec{a} = -4 \hat{i} + 2 \hat{j} - 3 \hat{k} \)
\( \vec{b} = \hat{i} + 3 \hat{j} - 2 \hat{k} \)
\( \vec{c} = -9 \hat{i} + \hat{j} - 4 \hat{k} \)
To show that the points are collinear, we can find the equation of the line passing through two points (say, A and B) and then check if the third point (C) lies on this line.
The vector equation of a line passing through A and B is \( \vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a}) \).
First, find the direction vector \( \vec{b} - \vec{a} \):
\( \vec{b} - \vec{a} = (\hat{i} + 3 \hat{j} - 2 \hat{k}) - (-4 \hat{i} + 2 \hat{j} - 3 \hat{k}) \)
\( = (1 - (-4)) \hat{i} + (3-2) \hat{j} + (-2 - (-3)) \hat{k} \)
\( = 5 \hat{i} + \hat{j} + \hat{k} \)
Now, the equation of the line passing through A and B is:
\( \vec{r} = -4 \hat{i} + 2 \hat{j} - 3 \hat{k} + \lambda(5 \hat{i} + \hat{j} + \hat{k}) \)
For point C to be collinear with A and B, its position vector \( \vec{c} \) must satisfy this equation for some value of \( \lambda \).
\( -9 \hat{i} + \hat{j} - 4 \hat{k} = -4 \hat{i} + 2 \hat{j} - 3 \hat{k} + \lambda(5 \hat{i} + \hat{j} + \hat{k}) \)
\( -9 \hat{i} + \hat{j} - 4 \hat{k} = (-4 + 5\lambda) \hat{i} + (2 + \lambda) \hat{j} + (-3 + \lambda) \hat{k} \)
Equating the coefficients of \( \hat{i}, \hat{j}, \hat{k} \):
For \( \hat{i} \): \( -9 = -4 + 5\lambda \implies -5 = 5\lambda \implies \lambda = -1 \)
For \( \hat{j} \): \( 1 = 2 + \lambda \implies \lambda = 1 - 2 \implies \lambda = -1 \)
For \( \hat{k} \): \( -4 = -3 + \lambda \implies \lambda = -4 + 3 \implies \lambda = -1 \)
Since the value of \( \lambda = -1 \) is the same for all three components, point C lies on the line passing through A and B. Therefore, points A, B, and C are collinear. This method is a reliable way to verify collinearity in 3D.
In simple words: To check if three points are in a straight line, we find the equation of the line using the first two points. Then, we see if the third point fits into this equation. If it does, using the same number for \( \lambda \), then all three points are on the same line.

🎯 Exam Tip: Another way to prove collinearity is to show that the direction vectors of two segments (e.g., \( \overrightarrow{AB} \) and \( \overrightarrow{BC} \)) are parallel (one is a scalar multiple of the other) and they share a common point (B in this case).

Question 8. The points A(4, 5, 10), B(2, 3, 4) and C(1, 2, -1) are three vertices of a parallelogram A B C D. Find the vector and Cartesian equations for the sides A B and B C and find the coordinates of D.
Answer: Given the coordinates of the vertices: A(4, 5, 10), B(2, 3, 4), C(1, 2, -1).
Their position vectors are:
\( \vec{a} = 4 \hat{i} + 5 \hat{j} + 10 \hat{k} \)
\( \vec{b} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \)
\( \vec{c} = \hat{i} + 2 \hat{j} - \hat{k} \)

**1. Equation of side AB:**
The vector equation of a line passing through two points A and B is \( \vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a}) \).
Direction vector \( \overrightarrow{AB} = \vec{b} - \vec{a} = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) - (4 \hat{i} + 5 \hat{j} + 10 \hat{k}) \)
\( = (2-4) \hat{i} + (3-5) \hat{j} + (4-10) \hat{k} \)
\( = -2 \hat{i} - 2 \hat{j} - 6 \hat{k} \)
Vector equation of AB:
\( \vec{r} = 4 \hat{i} + 5 \hat{j} + 10 \hat{k} + \lambda(-2 \hat{i} - 2 \hat{j} - 6 \hat{k}) \)
Cartesian equation of AB (using point A and direction ratios (-2, -2, -6), which can be simplified to (1, 1, 3) by dividing by -2):
\( \frac{x-4}{-2} = \frac{y-5}{-2} = \frac{z-10}{-6} \)
\( \implies \frac{x-4}{1} = \frac{y-5}{1} = \frac{z-10}{3} \)

**2. Equation of side BC:**
Direction vector \( \overrightarrow{BC} = \vec{c} - \vec{b} = (\hat{i} + 2 \hat{j} - \hat{k}) - (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \)
\( = (1-2) \hat{i} + (2-3) \hat{j} + (-1-4) \hat{k} \)
\( = -\hat{i} - \hat{j} - 5 \hat{k} \)
Vector equation of BC (using point B and direction vector \( \overrightarrow{BC} \)):
\( \vec{r} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} + \lambda(-\hat{i} - \hat{j} - 5 \hat{k}) \)
Cartesian equation of BC (using point B and direction ratios (-1, -1, -5)):
\( \frac{x-2}{-1} = \frac{y-3}{-1} = \frac{z-4}{-5} \)
\( \implies \frac{x-2}{1} = \frac{y-3}{1} = \frac{z-4}{5} \)

**3. Coordinates of D:**
Let the coordinates of D be \( (\alpha, \beta, \gamma) \).
In a parallelogram ABCD, the diagonals AC and BD bisect each other. This means their midpoints are the same.
Midpoint of AC \( = (\frac{4+1}{2}, \frac{5+2}{2}, \frac{10-1}{2}) = (\frac{5}{2}, \frac{7}{2}, \frac{9}{2}) \)
Midpoint of BD \( = (\frac{2+\alpha}{2}, \frac{3+\beta}{2}, \frac{4+\gamma}{2}) \)
Equating the coordinates of the midpoints:
\( \frac{5}{2} = \frac{2+\alpha}{2} \implies 5 = 2 + \alpha \implies \alpha = 3 \)
\( \frac{7}{2} = \frac{3+\beta}{2} \implies 7 = 3 + \beta \implies \beta = 4 \)
\( \frac{9}{2} = \frac{4+\gamma}{2} \implies 9 = 4 + \gamma \implies \gamma = 5 \)
Thus, the required coordinates of D are (3, 4, 5). Understanding midpoint properties of parallelograms is key here.
In simple words: We first find the equations for sides AB and BC by using their start and end points. Then, to find point D, we use the rule that the middle points of the diagonals in a parallelogram are the same. We calculate the midpoint of AC and use it to find the missing coordinates of D from the midpoint of BD.

🎯 Exam Tip: For problems involving parallelograms, remember that opposite sides are parallel and equal in length, and diagonals bisect each other. Using the midpoint property of diagonals is often the quickest way to find missing vertices.

Question 9. Find the Cartesian and vector equations of a line which passes through the point (1, 2, 3) and is parallel to the line \( \frac{-x-2}{1} = \frac{y+3}{7} = \frac{2z-6}{3} \).
Answer: The given Cartesian equation of the line is \( \frac{-x-2}{1} = \frac{y+3}{7} = \frac{2z-6}{3} \).
To find the direction ratios from this equation, we need to express it in the standard symmetric form \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \).
Let's rewrite each part:
\( \frac{-(x+2)}{1} = \frac{y+3}{7} = \frac{2(z-3)}{3} \)
\( \implies \frac{x+2}{-1} = \frac{y+3}{7} = \frac{z-3}{\frac{3}{2}} \)
The direction ratios of this given line are \( (-1, 7, \frac{3}{2}) \). To avoid fractions, we can multiply these ratios by 2, which gives \( (-2, 14, 3) \). Since the required line is parallel to this line, its direction ratios will also be proportional to \( (-2, 14, 3) \).
The required line passes through the point \( (1, 2, 3) \). So, \( (x_1, y_1, z_1) = (1, 2, 3) \).
The direction ratios \( (a, b, c) = (-2, 14, 3) \).

**1. Cartesian Equation:**
Using the formula \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \):
\( \frac{x-1}{-2} = \frac{y-2}{14} = \frac{z-3}{3} \)
This is the required Cartesian equation of the line.

**2. Vector Equation:**
The position vector of the point is \( \vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k} \).
The direction vector is \( \vec{b} = -2 \hat{i} + 14 \hat{j} + 3 \hat{k} \).
Using the formula \( \vec{r} = \vec{a} + \lambda \vec{b} \):
\( \vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + \lambda(-2 \hat{i} + 14 \hat{j} + 3 \hat{k}) \)
This is the required vector equation of the line. Finding common multiples often simplifies the problem.
In simple words: We are given a messy Cartesian equation for a line. We clean it up to find its true direction. Since our new line is parallel, it uses the same direction. Then, using the given point and this direction, we write both the Cartesian and vector equations for our new line.

🎯 Exam Tip: When given a line in Cartesian form like \( Ax+B = Cy+D = Ez+F \), always rewrite it to \( \frac{x-(-B/A)}{1/A} = \frac{y-(-D/C)}{1/C} = \frac{z-(-F/E)}{1/E} \) to correctly extract the direction ratios \( (1/A, 1/C, 1/E) \) or a multiple thereof.

Question 10. Find the direction cosines and vector equation of the line whose Cartesian form is \( \frac{x-2}{2} = \frac{2y-5}{-3}, z = -1 \).
Answer: The given Cartesian equation of the line is \( \frac{x-2}{2} = \frac{2y-5}{-3}, z = -1 \).
We need to convert this into the standard symmetric form \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \).
First, let's adjust the \(y\) term to have a coefficient of 1:
\( \frac{2y-5}{-3} = \frac{2(y-\frac{5}{2})}{-3} = \frac{y-\frac{5}{2}}{-\frac{3}{2}} \)
For the \(z\) term, \( z = -1 \), which can be written as \( \frac{z-(-1)}{0} \). This indicates that the line is perpendicular to the xy-plane and parallel to the z-axis (or more precisely, its projection on the xy-plane is a point).
So, the standard Cartesian form is:
\( \frac{x-2}{2} = \frac{y-\frac{5}{2}}{-\frac{3}{2}} = \frac{z-(-1)}{0} \)
From this, we can identify:
A point on the line is \( (x_1, y_1, z_1) = (2, \frac{5}{2}, -1) \). So, its position vector is \( \vec{a} = 2 \hat{i} + \frac{5}{2} \hat{j} - \hat{k} \).
The direction ratios of the line are \( (a, b, c) = (2, -\frac{3}{2}, 0) \).
To work with integers, we can multiply the direction ratios by 2 to get \( (4, -3, 0) \). So, the direction vector is \( \vec{b} = 4 \hat{i} - 3 \hat{j} + 0 \hat{k} \).

**1. Vector Equation:**
Using the formula \( \vec{r} = \vec{a} + \lambda \vec{b} \):
\( \vec{r} = (2 \hat{i} + \frac{5}{2} \hat{j} - \hat{k}) + \lambda(4 \hat{i} - 3 \hat{j} + 0 \hat{k}) \)
\( \implies \vec{r} = (2 \hat{i} + \frac{5}{2} \hat{j} - \hat{k}) + \lambda(4 \hat{i} - 3 \hat{j}) \)
This is the required vector equation of the line.

**2. Direction Cosines:**
The direction ratios are \( (a, b, c) = (4, -3, 0) \).
The magnitude of the direction vector is \( |\vec{b}| = \sqrt{a^2 + b^2 + c^2} = \sqrt{4^2 + (-3)^2 + 0^2} \)
\( = \sqrt{16 + 9 + 0} = \sqrt{25} = 5 \)
The direction cosines are \( (l, m, n) = (\frac{a}{|\vec{b}|}, \frac{b}{|\vec{b}|}, \frac{c}{|\vec{b}|}) \).
\( l = \frac{4}{5} \)
\( m = \frac{-3}{5} \)
\( n = \frac{0}{5} = 0 \)
So, the direction cosines are \( (\frac{4}{5}, -\frac{3}{5}, 0) \). This tells us the exact orientation of the line in space.
In simple words: We start with the Cartesian equation and change it into a standard form to find a point on the line and its direction numbers. Then we use these to write the line's vector equation. Finally, we calculate the direction cosines by dividing each direction number by the total length of the direction vector.

🎯 Exam Tip: When one of the coordinates (like \(z\) in this case) is given as a constant, it implies the line is parallel to the plane defined by the other two axes (e.g., \(z=-1\) means parallel to the xy-plane). This results in a zero in the direction ratio and consequently in the direction cosine for that axis.