RBSE Solutions Class 12 Maths Chapter 14 Three Dimensional Geometry Exercise 14.2

Question 1. Find the equation of the line which passes through the points (5, 7,9) and is parallel to the following axis :
(i) A-axis
(ii) F-axis
(iii) Z-axis
Answer:
Let the position vector of the point (5, 7, 9) be \( \vec{a} = 5\hat{i} + 7\hat{j} + 9\hat{k} \).
(i) For the line parallel to the X-axis:
The direction vector for the X-axis is \( \vec{b} = 1\hat{i} + 0\hat{j} + 0\hat{k} \).
The vector equation of the line is \( \vec{r} = \vec{a} + \lambda \vec{b} \).
\( \implies \vec{r} = (5\hat{i} + 7\hat{j} + 9\hat{k}) + \lambda (1\hat{i} + 0\hat{j} + 0\hat{k}) \)
\( \implies \vec{r} = (5+\lambda)\hat{i} + 7\hat{j} + 9\hat{k} \)
For the Cartesian equation, let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
Comparing the coefficients:
\( x = 5 + \lambda \implies x - 5 = \lambda \)
\( y = 7 \implies y - 7 = 0 \)
\( z = 9 \implies z - 9 = 0 \)
So, the Cartesian equation of the line is \( \frac{x-5}{1} = \frac{y-7}{0} = \frac{z-9}{0} \). This form means \( y-7=0 \) and \( z-9=0 \), with x varying freely.

(ii) For the line parallel to the F-axis (assuming this means the Y-axis based on the solution steps):
The direction vector for the Y-axis is \( \vec{b} = 0\hat{i} + 1\hat{j} + 0\hat{k} \).
The vector equation of the line is \( \vec{r} = \vec{a} + \lambda \vec{b} \).
\( \implies \vec{r} = (5\hat{i} + 7\hat{j} + 9\hat{k}) + \lambda (0\hat{i} + 1\hat{j} + 0\hat{k}) \)
\( \implies \vec{r} = 5\hat{i} + (7+\lambda)\hat{j} + 9\hat{k} \)
For the Cartesian equation, let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
Comparing the coefficients:
\( x = 5 \implies x - 5 = 0 \)
\( y = 7 + \lambda \implies y - 7 = \lambda \)
\( z = 9 \implies z - 9 = 0 \)
So, the Cartesian equation of the line is \( \frac{x-5}{0} = \frac{y-7}{1} = \frac{z-9}{0} \). This form means \( x-5=0 \) and \( z-9=0 \), with y varying freely.

(iii) For the line parallel to the Z-axis:
The direction vector for the Z-axis is \( \vec{b} = 0\hat{i} + 0\hat{j} + 1\hat{k} \).
The vector equation of the line is \( \vec{r} = \vec{a} + \lambda \vec{b} \).
\( \implies \vec{r} = (5\hat{i} + 7\hat{j} + 9\hat{k}) + \lambda (0\hat{i} + 0\hat{j} + 1\hat{k}) \)
\( \implies \vec{r} = 5\hat{i} + 7\hat{j} + (9+\lambda)\hat{k} \)
For the Cartesian equation, let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
Comparing the coefficients:
\( x = 5 \implies x - 5 = 0 \)
\( y = 7 \implies y - 7 = 0 \)
\( z = 9 + \lambda \implies z - 9 = \lambda \)
So, the Cartesian equation of the line is \( \frac{x-5}{0} = \frac{y-7}{0} = \frac{z-9}{1} \). This form means \( x-5=0 \) and \( y-7=0 \), with z varying freely.
In simple words: To find the equation of a line, you need a point it passes through and its direction. When a line is parallel to an axis, its direction vector is simply the unit vector of that axis. Then, you use the general formulas for vector and Cartesian equations.

🎯 Exam Tip: Remember that a line parallel to an axis means its direction ratios are (1,0,0) for X-axis, (0,1,0) for Y-axis, and (0,0,1) for Z-axis. This simplifies finding the direction vector.

Question 2. Find the equation of the line in vector and in cartesian form that passes through the point with vector \( 2\hat{i} - 3\hat{j} + 4\hat{k} \) and is parallel to the vector \( 3\hat{i} + 4\hat{j} - 5\hat{k} \).
Answer:
Let the position vector of the given point be \( \vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k} \).
Let the line be parallel to the vector \( \vec{b} = 3\hat{i} + 4\hat{j} - 5\hat{k} \).
(i) Vector equation of the line:
The general formula for a line passing through a point \( \vec{a} \) and parallel to a vector \( \vec{b} \) is \( \vec{r} = \vec{a} + \lambda \vec{b} \).
Substitute the given values:
\( \implies \vec{r} = (2\hat{i} - 3\hat{j} + 4\hat{k}) + \lambda (3\hat{i} + 4\hat{j} - 5\hat{k}) \)
\( \implies \vec{r} = (2+3\lambda)\hat{i} + (-3+4\lambda)\hat{j} + (4-5\lambda)\hat{k} \)

(ii) Cartesian equation of the line:
Let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
Comparing the coefficients from the vector equation:
\( x = 2 + 3\lambda \implies \frac{x-2}{3} = \lambda \)
\( y = -3 + 4\lambda \implies \frac{y+3}{4} = \lambda \)
\( z = 4 - 5\lambda \implies \frac{z-4}{-5} = \lambda \)
Since all these expressions are equal to \( \lambda \), the Cartesian equation of the line is:
\( \frac{x-2}{3} = \frac{y+3}{4} = \frac{z-4}{-5} \)
In simple words: We use specific formulas for lines in 3D space. The vector form shows how to get to any point on the line by starting at a fixed point and moving in the direction of the line. The Cartesian form gives the same information but using x, y, z coordinates.

🎯 Exam Tip: Always clearly identify the position vector of the point (\(\vec{a}\)) and the direction vector (\(\vec{b}\)) from the question before applying the standard formulas.

Question 3. Find the equation of the line parallel to the vector \( 2\hat{i} - \hat{j} + 3\hat{k} \) and passes through the point (5,-2,4).
Answer:
The line passes through the point (5, -2, 4). So, its position vector is \( \vec{a} = 5\hat{i} - 2\hat{j} + 4\hat{k} \).
The line is parallel to the vector \( \vec{b} = 2\hat{i} - \hat{j} + 3\hat{k} \). This vector gives the direction of the line.
(i) Vector equation of the line:
Using the formula \( \vec{r} = \vec{a} + \lambda \vec{b} \):
\( \implies \vec{r} = (5\hat{i} - 2\hat{j} + 4\hat{k}) + \lambda (2\hat{i} - \hat{j} + 3\hat{k}) \)
\( \implies \vec{r} = (5+2\lambda)\hat{i} + (-2-\lambda)\hat{j} + (4+3\lambda)\hat{k} \)

(ii) Cartesian equation of the line:
Let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
Comparing the coefficients from the vector equation:
\( x = 5 + 2\lambda \implies \frac{x-5}{2} = \lambda \)
\( y = -2 - \lambda \implies \frac{y+2}{-1} = \lambda \)
\( z = 4 + 3\lambda \implies \frac{z-4}{3} = \lambda \)
The Cartesian equation of the line is:
\( \frac{x-5}{2} = \frac{y+2}{-1} = \frac{z-4}{3} \)
In simple words: When a line is parallel to a vector, that vector is its direction. We use the given point and this direction to write the line's equation in both vector and coordinate forms.

🎯 Exam Tip: Make sure to correctly handle the negative sign in the direction ratio for the y-component, i.e., \( -\hat{j} \) corresponds to a direction ratio of -1.

Question 4. Find the equation of the line passes through the point (2,-1,1) and is parallel to the line \( \frac{x-3}{2} = \frac{y+1}{7} = \frac{z-2}{-3} \).
Answer:
The line passes through the point (2, -1, 1). So, its position vector is \( \vec{a} = 2\hat{i} - \hat{j} + \hat{k} \).
Since the required line is parallel to the given line \( \frac{x-3}{2} = \frac{y+1}{7} = \frac{z-2}{-3} \), their direction ratios will be the same.
The direction vector of the given line (and thus for our required line) is \( \vec{m} = 2\hat{i} + 7\hat{j} - 3\hat{k} \). This is our \( \vec{b} \).
(i) Vector equation of the line:
Using the formula \( \vec{r} = \vec{a} + \lambda \vec{b} \):
\( \implies \vec{r} = (2\hat{i} - \hat{j} + \hat{k}) + \lambda (2\hat{i} + 7\hat{j} - 3\hat{k}) \)
\( \implies \vec{r} = (2+2\lambda)\hat{i} + (-1+7\lambda)\hat{j} + (1-3\lambda)\hat{k} \)

(ii) Cartesian equation of the line:
Let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
Comparing the coefficients from the vector equation:
\( x = 2 + 2\lambda \implies \frac{x-2}{2} = \lambda \)
\( y = -1 + 7\lambda \implies \frac{y+1}{7} = \lambda \)
\( z = 1 - 3\lambda \implies \frac{z-1}{-3} = \lambda \)
The Cartesian equation of the line is:
\( \frac{x-2}{2} = \frac{y+1}{7} = \frac{z-1}{-3} \)
In simple words: Parallel lines share the same direction. So, we can take the direction numbers from the given line's equation and use them with the new point to find the equation of the desired line.

🎯 Exam Tip: When lines are parallel, their direction ratios are proportional. In this case, they are exactly the same, which simplifies finding the direction vector.

Question 5. If cartesian equation of line is \( \frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2} \), then find vector equation of line.
Answer:
The given Cartesian equation of the line is \( \frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2} \).
From this equation, we can identify:
The line passes through the point \( (x_1, y_1, z_1) = (5, -4, 6) \). So, the position vector of this point is \( \vec{a} = 5\hat{i} - 4\hat{j} + 6\hat{k} \).
The direction ratios of the line are (a, b, c) = (3, 7, 2). So, the direction vector is \( \vec{b} = 3\hat{i} + 7\hat{j} + 2\hat{k} \).
The vector equation of a line is given by the formula \( \vec{r} = \vec{a} + \lambda \vec{b} \).
Substitute the values of \( \vec{a} \) and \( \vec{b} \) into the formula:
\( \implies \vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + \lambda (3\hat{i} + 7\hat{j} + 2\hat{k}) \)
This is the required vector equation of the line. This formula shows that any point on the line can be reached by starting at \( \vec{a} \) and moving some distance (\( \lambda \)) in the direction of \( \vec{b} \).
In simple words: To change a Cartesian line equation into a vector equation, just pick out the point the line goes through and its direction numbers. Then, use the standard vector form: point plus lambda times direction.

🎯 Exam Tip: Remember that in the Cartesian form \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \), the point is \( (x_1, y_1, z_1) \) and the direction ratios are \( (a, b, c) \). Be careful with signs, especially for \( y+4 \), which means \( y - (-4) \).

Question 6. Find the equation of a line in cartesian form which passes through (1, 2, 3) and is parallel to \( \frac{x-2}{-1} = \frac{y+3}{7} = \frac{2z-6}{3} \).
Answer:
The line passes through the point \( (x_1, y_1, z_1) = (1, 2, 3) \).
The given line is \( \frac{x-2}{-1} = \frac{y+3}{7} = \frac{2z-6}{3} \).
To find the direction ratios of the given line, we need to rewrite its Z-component in the standard form \( \frac{z-z_1}{c} \).
\( \frac{2z-6}{3} = \frac{2(z-3)}{3} = \frac{z-3}{3/2} \)
So, the given line is \( \frac{x-2}{-1} = \frac{y+3}{7} = \frac{z-3}{3/2} \).
The direction ratios of the given line are \( (-1, 7, 3/2) \).
Since the required line is parallel to this given line, its direction ratios will also be proportional to \( (-1, 7, 3/2) \). We can multiply these ratios by 2 to get integer direction ratios: \( (-2, 14, 3) \). Let's use \( (a,b,c) = (-2, 14, 3) \).
The Cartesian equation of a line passing through \( (x_1, y_1, z_1) \) with direction ratios \( (a, b, c) \) is:
\( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \)
Substitute the point (1, 2, 3) and direction ratios (-2, 14, 3):
\( \implies \frac{x-1}{-2} = \frac{y-2}{14} = \frac{z-3}{3} \)
This is the required Cartesian equation of the line. The line is defined by these proportions.
In simple words: First, fix the given line's equation so the 'z' part is standard. Then, since our new line is parallel, it uses the same direction numbers. We combine these direction numbers with the point our new line goes through to write its equation.

🎯 Exam Tip: Always normalize the given Cartesian equation to the standard form \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \) to correctly extract the direction ratios, especially if coefficients other than 1 appear with x, y, or z.

Question 7. Coordinates of three vertices of parallelogram ABCD are A(4, 5, 10), В(2, 3, 4) and C(1, 2, – 1). Find vector and cartesian equation of AB and AC. Also Find coordinates of D.
Answer:
Let the origin be O. The position vectors of points A, B, and C are respectively:
\( \vec{a} = 4\hat{i} + 5\hat{j} + 10\hat{k} \)
\( \vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k} \)
\( \vec{c} = 1\hat{i} + 2\hat{j} - 1\hat{k} \)

**1. Equation of Line AB:**
The line AB passes through point A \( (4, 5, 10) \) and point B \( (2, 3, 4) \).
Position vector of A: \( \vec{a} = 4\hat{i} + 5\hat{j} + 10\hat{k} \)
Direction vector of line AB: \( \vec{b} - \vec{a} = (2\hat{i} + 3\hat{j} + 4\hat{k}) - (4\hat{i} + 5\hat{j} + 10\hat{k}) \)
\( \implies \vec{b} - \vec{a} = (2-4)\hat{i} + (3-5)\hat{j} + (4-10)\hat{k} = -2\hat{i} - 2\hat{j} - 6\hat{k} \)
(i) Vector equation of line AB:
\( \vec{r} = \vec{a} + \lambda (\vec{b} - \vec{a}) \)
\( \implies \vec{r} = (4\hat{i} + 5\hat{j} + 10\hat{k}) + \lambda (-2\hat{i} - 2\hat{j} - 6\hat{k}) \)
\( \implies \vec{r} = (4-2\lambda)\hat{i} + (5-2\lambda)\hat{j} + (10-6\lambda)\hat{k} \)
(ii) Cartesian equation of line AB:
Let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
Comparing coefficients:
\( x = 4 - 2\lambda \implies \frac{x-4}{-2} = \lambda \)
\( y = 5 - 2\lambda \implies \frac{y-5}{-2} = \lambda \)
\( z = 10 - 6\lambda \implies \frac{z-10}{-6} = \lambda \)
The Cartesian equation of line AB is \( \frac{x-4}{-2} = \frac{y-5}{-2} = \frac{z-10}{-6} \). We can simplify the direction ratios by dividing by -2:
\( \implies \frac{x-4}{1} = \frac{y-5}{1} = \frac{z-10}{3} \)

**2. Equation of Line AC:**
The line AC passes through point A \( (4, 5, 10) \) and point C \( (1, 2, -1) \).
Position vector of A: \( \vec{a} = 4\hat{i} + 5\hat{j} + 10\hat{k} \)
Direction vector of line AC: \( \vec{c} - \vec{a} = (1\hat{i} + 2\hat{j} - 1\hat{k}) - (4\hat{i} + 5\hat{j} + 10\hat{k}) \)
\( \implies \vec{c} - \vec{a} = (1-4)\hat{i} + (2-5)\hat{j} + (-1-10)\hat{k} = -3\hat{i} - 3\hat{j} - 11\hat{k} \)
(i) Vector equation of line AC:
\( \vec{r} = \vec{a} + \mu (\vec{c} - \vec{a}) \)
\( \implies \vec{r} = (4\hat{i} + 5\hat{j} + 10\hat{k}) + \mu (-3\hat{i} - 3\hat{j} - 11\hat{k}) \)
\( \implies \vec{r} = (4-3\mu)\hat{i} + (5-3\mu)\hat{j} + (10-11\mu)\hat{k} \)
(ii) Cartesian equation of line AC:
Let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
Comparing coefficients:
\( x = 4 - 3\mu \implies \frac{x-4}{-3} = \mu \)
\( y = 5 - 3\mu \implies \frac{y-5}{-3} = \mu \)
\( z = 10 - 11\mu \implies \frac{z-10}{-11} = \mu \)
The Cartesian equation of line AC is \( \frac{x-4}{-3} = \frac{y-5}{-3} = \frac{z-10}{-11} \). We can simplify by dividing by -1:
\( \implies \frac{x-4}{3} = \frac{y-5}{3} = \frac{z-10}{11} \)

**3. Coordinates of D:**
In a parallelogram ABCD, the diagonals bisect each other. This means the midpoint of AC is the same as the midpoint of BD.
Midpoint of AC: \( P = \left( \frac{x_A+x_C}{2}, \frac{y_A+y_C}{2}, \frac{z_A+z_C}{2} \right) \)
\( P = \left( \frac{4+1}{2}, \frac{5+2}{2}, \frac{10+(-1)}{2} \right) = \left( \frac{5}{2}, \frac{7}{2}, \frac{9}{2} \right) \)
Let the coordinates of D be \( (x_D, y_D, z_D) \).
Midpoint of BD: \( Q = \left( \frac{x_B+x_D}{2}, \frac{y_B+y_D}{2}, \frac{z_B+z_D}{2} \right) \)
\( Q = \left( \frac{2+x_D}{2}, \frac{3+y_D}{2}, \frac{4+z_D}{2} \right) \)
Since P and Q are the same point:
\( \frac{2+x_D}{2} = \frac{5}{2} \implies 2+x_D = 5 \implies x_D = 3 \)
\( \frac{3+y_D}{2} = \frac{7}{2} \implies 3+y_D = 7 \implies y_D = 4 \)
\( \frac{4+z_D}{2} = \frac{9}{2} \implies 4+z_D = 9 \implies z_D = 5 \)
Therefore, the coordinates of point D are (3, 4, 5). Finding the fourth vertex involves understanding the properties of parallelograms.
In simple words: First, we find the equations for lines AB and AC by using the points they pass through to get a direction and a starting point. Then, for the fourth corner D, we use the rule that the middle point of the lines connecting opposite corners (diagonals) is the same for a parallelogram.

🎯 Exam Tip: For problems involving parallelograms, always remember the property that diagonals bisect each other. This means their midpoints coincide, which is key to finding unknown vertex coordinates.

Question 8. Cartesian equation of a line is \( 3x + 1 = 6y - 2 = 1 - z \). Find the point from where it passes and also find its direction ratios and vector equation.
Answer:
The given Cartesian equation of the line is \( 3x + 1 = 6y - 2 = 1 - z \).
To find the point and direction ratios, we need to rewrite this equation in the standard form \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \).
Let each part be equal to \( \lambda \).
\( 3x + 1 = \lambda \implies 3(x + \frac{1}{3}) = \lambda \implies x + \frac{1}{3} = \frac{\lambda}{3} \implies \frac{x - (-\frac{1}{3})}{1/3} = \lambda \)
\( 6y - 2 = \lambda \implies 6(y - \frac{2}{6}) = \lambda \implies y - \frac{1}{3} = \frac{\lambda}{6} \implies \frac{y - \frac{1}{3}}{1/6} = \lambda \)
\( 1 - z = \lambda \implies -(z - 1) = \lambda \implies z - 1 = -\lambda \implies \frac{z - 1}{-1} = \lambda \)
So, the standard Cartesian equation is: \( \frac{x - (-\frac{1}{3})}{1/3} = \frac{y - \frac{1}{3}}{1/6} = \frac{z - 1}{-1} \)
From this, we can identify:
The line passes through the point \( (x_1, y_1, z_1) = (-\frac{1}{3}, \frac{1}{3}, 1) \). This is the base point.
The direction ratios are \( (a, b, c) = (\frac{1}{3}, \frac{1}{6}, -1) \).
To get simpler integer direction ratios, we can multiply them by the least common multiple of the denominators (6):
\( (6 \times \frac{1}{3}, 6 \times \frac{1}{6}, 6 \times -1) = (2, 1, -6) \).
So, the direction vector is \( \vec{b} = 2\hat{i} + \hat{j} - 6\hat{k} \).

(i) Point from where it passes: \( (-\frac{1}{3}, \frac{1}{3}, 1) \)

(ii) Direction ratios: \( (2, 1, -6) \)

(iii) Vector equation of the line:
The position vector of the point is \( \vec{a} = -\frac{1}{3}\hat{i} + \frac{1}{3}\hat{j} + 1\hat{k} \).
Using the formula \( \vec{r} = \vec{a} + \lambda \vec{b} \):
\( \implies \vec{r} = (-\frac{1}{3}\hat{i} + \frac{1}{3}\hat{j} + \hat{k}) + \lambda (2\hat{i} + \hat{j} - 6\hat{k}) \)
This equation describes the path of the line in vector form.
In simple words: First, we change the given equation into a standard form. From this, we can easily see the point where the line starts and the numbers that show its direction. Then, we use these to write the line's equation using vectors.

🎯 Exam Tip: Always ensure the coefficients of \( x, y, \) and \( z \) are 1 and that the terms are in the form \( (x-x_1) \), etc. before identifying \( x_1, y_1, z_1 \) and direction ratios from a Cartesian equation. Be especially careful with negative signs, like in \( 1-z \).

Question 9. Find an equation passes through (1,2,3) and is parallel to the vector \( 3\hat{i} + 2\hat{j} - 2\hat{k} \).
Answer:
The line passes through the point (1, 2, 3). So, its position vector is \( \vec{a} = 1\hat{i} + 2\hat{j} + 3\hat{k} \).
The line is parallel to the vector \( \vec{b} = 3\hat{i} + 2\hat{j} - 2\hat{k} \). This vector represents the direction of the line.
(i) Vector equation of the line:
Using the general formula \( \vec{r} = \vec{a} + \lambda \vec{b} \):
\( \implies \vec{r} = (1\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda (3\hat{i} + 2\hat{j} - 2\hat{k}) \)
\( \implies \vec{r} = (1+3\lambda)\hat{i} + (2+2\lambda)\hat{j} + (3-2\lambda)\hat{k} \)

(ii) Cartesian equation of the line:
Let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
Comparing the coefficients from the vector equation:
\( x = 1 + 3\lambda \implies \frac{x-1}{3} = \lambda \)
\( y = 2 + 2\lambda \implies \frac{y-2}{2} = \lambda \)
\( z = 3 - 2\lambda \implies \frac{z-3}{-2} = \lambda \)
The Cartesian equation of the line is:
\( \frac{x-1}{3} = \frac{y-2}{2} = \frac{z-3}{-2} \)
In simple words: To find the line's equation, we take the given point as its starting location and the given parallel vector as its direction. We then write this information in both vector and coordinate forms.

🎯 Exam Tip: Always make sure to use the coordinates of the *given point* for \( (x_1, y_1, z_1) \) and the components of the *parallel vector* for the direction ratios \( (a, b, c) \) in the respective formulas.

Question 10. Find the equation of line in vector and 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:
The line passes through a point whose position vector is \( \vec{a} = 2\hat{i} - \hat{j} + 4\hat{k} \).
The line is in the direction of the vector \( \vec{b} = \hat{i} + 2\hat{j} - \hat{k} \). This is our direction vector.
(i) Vector equation of the line:
Using the general formula \( \vec{r} = \vec{a} + \lambda \vec{b} \):
\( \implies \vec{r} = (2\hat{i} - \hat{j} + 4\hat{k}) + \lambda (\hat{i} + 2\hat{j} - \hat{k}) \)
\( \implies \vec{r} = (2+\lambda)\hat{i} + (-1+2\lambda)\hat{j} + (4-\lambda)\hat{k} \)

(ii) Cartesian equation of the line:
Let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
Comparing the coefficients from the vector equation:
\( x = 2 + \lambda \implies \frac{x-2}{1} = \lambda \)
\( y = -1 + 2\lambda \implies \frac{y+1}{2} = \lambda \)
\( z = 4 - \lambda \implies \frac{z-4}{-1} = \lambda \)
The Cartesian equation of the line is:
\( \frac{x-2}{1} = \frac{y+1}{2} = \frac{z-4}{-1} \)
In simple words: We are given a starting point and a direction for the line. We plug these directly into the vector equation formula, and then convert that into the Cartesian form by matching the x, y, and z components.

🎯 Exam Tip: Pay attention to the terms "position vector" and "direction vector" in the question to correctly identify \( \vec{a} \) and \( \vec{b} \) before applying the formulas.

Question 11. Find the cartesian equation of a line passes through the point (- 2, 4, -5) and is parallel to \( \frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6} \).
Answer:
The line passes through the point \( (x_1, y_1, z_1) = (-2, 4, -5) \).
The required line is parallel to the given line \( \frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6} \).
Since parallel lines have the same direction ratios, we can take the direction ratios directly from the given line.
The direction ratios of the given line are \( (a, b, c) = (3, 5, 6) \).
The Cartesian equation of a line passing through a point \( (x_1, y_1, z_1) \) with direction ratios \( (a, b, c) \) is given by:
\( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \)
Substitute the point \( (-2, 4, -5) \) and the direction ratios \( (3, 5, 6) \):
\( \implies \frac{x - (-2)}{3} = \frac{y - 4}{5} = \frac{z - (-5)}{6} \)
\( \implies \frac{x+2}{3} = \frac{y-4}{5} = \frac{z+5}{6} \)
This is the required Cartesian equation of the line. It describes all points on the line.
In simple words: Because the two lines are parallel, they point in the same direction. We use the direction numbers from the given line and combine them with the point our new line passes through to write its equation.

🎯 Exam Tip: Remember to correctly handle the signs for the coordinates of the passing point, especially when they are negative, like \( x - (-2) \) becomes \( x+2 \).

Question 12. The cartesian equation of a line is \( \frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2} \). Write its vector form.
Answer:
The given Cartesian equation of the line is \( \frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2} \).
From this equation, we can identify:
The line passes through the point \( (x_1, y_1, z_1) = (5, -4, 6) \). The position vector of this point is \( \vec{a} = 5\hat{i} - 4\hat{j} + 6\hat{k} \).
The direction ratios of the line are \( (a, b, c) = (3, 7, 2) \). The direction vector is \( \vec{b} = 3\hat{i} + 7\hat{j} + 2\hat{k} \).
The vector equation of a line is given by the formula \( \vec{r} = \vec{a} + \lambda \vec{b} \).
Substitute the values of \( \vec{a} \) and \( \vec{b} \) into the formula:
\( \implies \vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + \lambda (3\hat{i} + 7\hat{j} + 2\hat{k}) \)
This is the required vector equation of the line. It shows how to reach any point on the line using a starting point and a direction.
In simple words: To convert a Cartesian equation to a vector equation, just pull out the point the line passes through and its direction numbers. Then, use the simple formula: position vector of the point, plus a scalar (lambda) times the direction vector.

🎯 Exam Tip: Always pay close attention to the signs in the Cartesian equation. For example, \( y+4 \) means the y-coordinate of the passing point is -4, not 4.

Question 13. Find the equation of line in vector and cartesian form that passes through origin and (5, -2, 3).
Answer:
The line passes through two points: the origin O(0, 0, 0) and point P(5, -2, 3).
Let \( \vec{a} \) be the position vector of the first point (origin): \( \vec{a} = 0\hat{i} + 0\hat{j} + 0\hat{k} = \vec{0} \).
Let \( \vec{b} \) be the position vector of the second point (5, -2, 3): \( \vec{b} = 5\hat{i} - 2\hat{j} + 3\hat{k} \).
The direction vector for the line passing through two points \( \vec{a} \) and \( \vec{b} \) is \( \vec{b} - \vec{a} \).
\( \vec{b} - \vec{a} = (5\hat{i} - 2\hat{j} + 3\hat{k}) - (0\hat{i} + 0\hat{j} + 0\hat{k}) = 5\hat{i} - 2\hat{j} + 3\hat{k} \).

(i) Vector equation of the line:
Using the formula \( \vec{r} = \vec{a} + \lambda (\vec{b} - \vec{a}) \) with \( \vec{a} = \vec{0} \):
\( \implies \vec{r} = \vec{0} + \lambda (5\hat{i} - 2\hat{j} + 3\hat{k}) \)
\( \implies \vec{r} = \lambda (5\hat{i} - 2\hat{j} + 3\hat{k}) \)

(ii) Cartesian equation of the line:
The line passes through \( (x_1, y_1, z_1) = (0, 0, 0) \).
The direction ratios are \( (a, b, c) = (5, -2, 3) \).
Using the formula \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \):
\( \implies \frac{x-0}{5} = \frac{y-0}{-2} = \frac{z-0}{3} \)
\( \implies \frac{x}{5} = \frac{y}{-2} = \frac{z}{3} \)
In simple words: When a line passes through the origin and another point, the vector from the origin to that point gives the line's direction. We use this direction and the origin as the starting point to write both the vector and Cartesian equations.

🎯 Exam Tip: When a line passes through the origin, the position vector \( \vec{a} \) becomes \( \vec{0} \), simplifying the vector equation to \( \vec{r} = \lambda \vec{b} \) and the Cartesian equation to \( \frac{x}{a} = \frac{y}{b} = \frac{z}{c} \).

Question 14. Find the equation of line in vector and cartesian form that passes through the points (3, -2,-5) and (3,-2,6).
Answer:
Let the two points be A(3, -2, -5) and B(3, -2, 6).
The position vector of point A is \( \vec{a} = 3\hat{i} - 2\hat{j} - 5\hat{k} \).
The position vector of point B is \( \vec{b} = 3\hat{i} - 2\hat{j} + 6\hat{k} \).
The direction vector of the line passing through points A and B is \( \vec{b} - \vec{a} \).
\( \vec{b} - \vec{a} = (3\hat{i} - 2\hat{j} + 6\hat{k}) - (3\hat{i} - 2\hat{j} - 5\hat{k}) \)
\( \implies \vec{b} - \vec{a} = (3-3)\hat{i} + (-2 - (-2))\hat{j} + (6 - (-5))\hat{k} \)
\( \implies \vec{b} - \vec{a} = 0\hat{i} + 0\hat{j} + 11\hat{k} \).
So, the direction vector \( \vec{d} = 11\hat{k} \). This means the line is parallel to the Z-axis.

(i) Vector equation of the line:
Using the formula for a line passing through a point \( \vec{a} \) with direction vector \( \vec{d} \): \( \vec{r} = \vec{a} + \lambda \vec{d} \)
\( \implies \vec{r} = (3\hat{i} - 2\hat{j} - 5\hat{k}) + \lambda (11\hat{k}) \)
\( \implies \vec{r} = 3\hat{i} - 2\hat{j} + (-5+11\lambda)\hat{k} \)

(ii) Cartesian equation of the line:
The line passes through \( (x_1, y_1, z_1) = (3, -2, -5) \).
The direction ratios are \( (a, b, c) = (0, 0, 11) \). We can simplify these to \( (0, 0, 1) \).
Using the formula \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \):
\( \implies \frac{x-3}{0} = \frac{y-(-2)}{0} = \frac{z-(-5)}{1} \)
\( \implies \frac{x-3}{0} = \frac{y+2}{0} = \frac{z+5}{1} \)
This equation shows that \( x-3=0 \) and \( y+2=0 \), while \( z \) can vary. This describes a line parallel to the Z-axis.
In simple words: We find the direction of the line by subtracting the two given points' position vectors. Then, we use one of the points and this direction to write the line's equation in both vector and Cartesian forms. The line happens to be parallel to the Z-axis since its x and y coordinates don't change.

🎯 Exam Tip: When two coordinates of the direction vector are zero, it means the line is parallel to the axis corresponding to the non-zero component. For example, \( (0,0,c) \) means parallel to the Z-axis. In the Cartesian equation, a zero in the denominator means the numerator must be zero, not that division by zero is allowed.

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