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

Question 1. Find the equations of a line passing through the point (-1, 2, 3) and having direction ratios proportional to -4, 5, 6.
Answer: We know the equation of a line passing through a point \( (x_1, y_1, z_1) \) and having direction ratios \( \) is given by \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \).
Here, the given point is \( (-1, 2, 3) \) and the direction ratios are \( <-4, 5, 6> \).
Substituting these values into the formula:
\( \frac{x-(-1)}{-4} = \frac{y-2}{5} = \frac{z-3}{6} \)
\( \implies \frac{x+1}{-4} = \frac{y-2}{5} = \frac{z-3}{6} \)
This is the required equation of the line. Direction ratios define the line's specific orientation in three-dimensional space.
In simple words: To write the equation of a line, we use a special formula that needs one point the line passes through and its direction numbers. We just put the given point and direction numbers into the formula to get the line's equation.

🎯 Exam Tip: Remember to correctly handle the signs when substituting negative coordinates into the formula, especially for terms like \( x - (-1) \) which becomes \( x+1 \).

Question 2. Find the equations of a line passing through the point (2, -3, 0) and having direction cosines \( \frac{-1}{7}, \frac{4}{7}, \frac{-6}{7} \).
Answer: The equation of a line passing through a point \( (x_1, y_1, z_1) \) and having direction cosines \( \) is given by \( \frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n} \).
Here, the given point is \( (2, -3, 0) \) and the direction cosines are \( <\frac{-1}{7}, \frac{4}{7}, \frac{-6}{7}> \).
Substituting these values:
\( \frac{x-2}{\frac{-1}{7}} = \frac{y-(-3)}{\frac{4}{7}} = \frac{z-0}{\frac{-6}{7}} \)
\( \implies \frac{x-2}{\frac{-1}{7}} = \frac{y+3}{\frac{4}{7}} = \frac{z}{\frac{-6}{7}} \)
We can simplify this by multiplying the denominators by 7:
\( \implies \frac{x-2}{-1} = \frac{y+3}{4} = \frac{z}{-6} \)
This is the required equation of the line. Direction cosines are a special type of direction ratio that are normalized, meaning the sum of their squares is one.
In simple words: When we know a point a line goes through and its direction cosines (which are like special direction numbers), we can write the line's equation using a simple formula. We put the point's numbers and the direction cosines into the formula. We can then make the equation simpler by clearing the fractions.

🎯 Exam Tip: When direction cosines are given, the denominators in the symmetric form can be multiplied by a common factor to get integer direction ratios, simplifying the appearance of the equation.

Question 3. Find the equations of a line passing through the points (2, 3, 4) and (4, 6, 5).
Answer: We know that the equation of a line passing through two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) is given by \( \frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1} \).
Here, the given points are \( (x_1, y_1, z_1) = (2, 3, 4) \) and \( (x_2, y_2, z_2) = (4, 6, 5) \).
First, calculate the denominators:
\( x_2-x_1 = 4-2 = 2 \)
\( y_2-y_1 = 6-3 = 3 \)
\( z_2-z_1 = 5-4 = 1 \)
Now, substitute these values and the first point into the formula:
\( \frac{x-2}{2} = \frac{y-3}{3} = \frac{z-4}{1} \)
This is the required equation of the line. This form of the equation is very useful for visualizing a line segment between two points.
In simple words: To find the equation of a line that goes through two given points, we use a formula that subtracts the coordinates of the points. We subtract the x's, y's, and z's separately for the bottom parts of the fractions, and then use the coordinates of one point for the top parts.

🎯 Exam Tip: When using the two-point form, ensure you consistently use the coordinates of either the first or second point for \( x_1, y_1, z_1 \) in the numerators, while the denominators are always the differences \( (x_2-x_1) \), etc.

Question 4. Find the equations of a line passing through the points (3, -2, -5) and (3, -2, 6).
Answer: The equation of a line passing through two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) is given by \( \frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1} \).
Here, the given points are \( (x_1, y_1, z_1) = (3, -2, -5) \) and \( (x_2, y_2, z_2) = (3, -2, 6) \).
First, calculate the denominators:
\( x_2-x_1 = 3-3 = 0 \)
\( y_2-y_1 = -2-(-2) = -2+2 = 0 \)
\( z_2-z_1 = 6-(-5) = 6+5 = 11 \)
Now, substitute these values and the first point into the formula:
\( \frac{x-3}{0} = \frac{y-(-2)}{0} = \frac{z-(-5)}{11} \)
\( \implies \frac{x-3}{0} = \frac{y+2}{0} = \frac{z+5}{11} \)
This form indicates that \( x-3=0 \) (so \( x=3 \)) and \( y+2=0 \) (so \( y=-2 \)). The line is parallel to the z-axis, showing no change in x or y coordinates.
In simple words: When a line goes through two points where the x-coordinates are the same and the y-coordinates are the same, it means the line is straight up and down (parallel to the z-axis). We write its equation to show that x and y do not change, while z changes.

🎯 Exam Tip: When the denominator in the symmetric form is zero, it implies that the line is parallel to the corresponding axis, and the numerator must also be zero for the equation to hold, meaning the coordinate is constant.

Question 5. Find the coordinates of the point, where the line through (5, 1, 6) and (3, 4, 1) crosses the (i) yz-plane (ii) the xy-plane (iii) the zx-plane.
Answer: First, we find the equation of the line passing through the points \( (5, 1, 6) \) and \( (3, 4, 1) \).
Using the two-point formula \( \frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1} \):
\( \frac{x-5}{3-5} = \frac{y-1}{4-1} = \frac{z-6}{1-6} \)
\( \implies \frac{x-5}{-2} = \frac{y-1}{3} = \frac{z-6}{-5} \) (Let this be equation (1))
(i) **Crossing the yz-plane:** A point on the yz-plane has its x-coordinate equal to 0. So, we set \( x=0 \) in equation (1):
\( \frac{0-5}{-2} = \frac{y-1}{3} = \frac{z-6}{-5} \)
\( \implies \frac{5}{2} = \frac{y-1}{3} = \frac{z-6}{-5} \)
From \( \frac{y-1}{3} = \frac{5}{2} \):
\( 2(y-1) = 15 \)
\( 2y-2 = 15 \)
\( 2y = 17 \)
\( y = \frac{17}{2} \)
From \( \frac{z-6}{-5} = \frac{5}{2} \):
\( 2(z-6) = -25 \)
\( 2z-12 = -25 \)
\( 2z = -13 \)
\( z = -\frac{13}{2} \)
Thus, the coordinates of the point where the line crosses the yz-plane are \( (0, \frac{17}{2}, -\frac{13}{2}) \).
(ii) **Crossing the xy-plane:** A point on the xy-plane has its z-coordinate equal to 0. So, we set \( z=0 \) in equation (1):
\( \frac{x-5}{-2} = \frac{y-1}{3} = \frac{0-6}{-5} \)
\( \implies \frac{x-5}{-2} = \frac{y-1}{3} = \frac{6}{5} \)
From \( \frac{x-5}{-2} = \frac{6}{5} \):
\( 5(x-5) = -12 \)
\( 5x-25 = -12 \)
\( 5x = 13 \)
\( x = \frac{13}{5} \)
From \( \frac{y-1}{3} = \frac{6}{5} \):
\( 5(y-1) = 18 \)
\( 5y-5 = 18 \)
\( 5y = 23 \)
\( y = \frac{23}{5} \)
Thus, the coordinates of the point where the line crosses the xy-plane are \( (\frac{13}{5}, \frac{23}{5}, 0) \).
(iii) **Crossing the zx-plane:** A point on the zx-plane has its y-coordinate equal to 0. So, we set \( y=0 \) in equation (1):
\( \frac{x-5}{-2} = \frac{0-1}{3} = \frac{z-6}{-5} \)
\( \implies \frac{x-5}{-2} = -\frac{1}{3} = \frac{z-6}{-5} \)
From \( \frac{x-5}{-2} = -\frac{1}{3} \):
\( 3(x-5) = 2 \)
\( 3x-15 = 2 \)
\( 3x = 17 \)
\( x = \frac{17}{3} \)
From \( \frac{z-6}{-5} = -\frac{1}{3} \):
\( 3(z-6) = 5 \)
\( 3z-18 = 5 \)
\( 3z = 23 \)
\( z = \frac{23}{3} \)
Thus, the coordinates of the point where the line crosses the zx-plane are \( (\frac{17}{3}, 0, \frac{23}{3}) \).
These intersection points are important for understanding where the line crosses the boundaries of the coordinate system, which is useful in many real-world applications like navigation.
In simple words: First, we write the equation for the line using the two points it passes through. Then, to find where it crosses a plane (like the yz-plane), we set the coordinate related to that plane to zero (like setting x to 0 for the yz-plane). We solve the line's equation for the other two coordinates to find the exact crossing point. We repeat this for each plane.

🎯 Exam Tip: Remember that a line crosses the yz-plane when \( x=0 \), the xy-plane when \( z=0 \), and the zx-plane when \( y=0 \). This is a crucial concept for finding intersection points.

Question 6. The cartesian equations of a line are \( 6x-2=3y+1=2z-2 \). Find the direction ratios.
Answer: The given equation of the line is \( 6x-2=3y+1=2z-2 \).
To find the direction ratios, we need to rewrite this equation into the standard symmetric form \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \), where \( a, b, c \) are the direction ratios.
We factor out the coefficients of \( x, y, z \):
\( 6(x-\frac{2}{6}) = 3(y+\frac{1}{3}) = 2(z-\frac{2}{2}) \)
\( \implies 6(x-\frac{1}{3}) = 3(y+\frac{1}{3}) = 2(z-1) \)
Now, we divide each part by a common factor (the LCM of the coefficients 6, 3, 2, which is 6) to bring the coefficients of \( (x-x_1) \), etc., to 1:
\( \frac{x-\frac{1}{3}}{1/6} = \frac{y+\frac{1}{3}}{1/3} = \frac{z-1}{1/2} \)
The denominators \( <1/6, 1/3, 1/2> \) are proportional to the direction ratios. To get integer direction ratios, we can multiply these by their LCM, which is 6:
\( a = 6 \times \frac{1}{6} = 1 \)
\( b = 6 \times \frac{1}{3} = 2 \)
\( c = 6 \times \frac{1}{2} = 3 \)
Thus, the direction ratios of the line are \( <1, 2, 3> \). The direction ratios tell us the "slope" of the line in 3D space, showing how much x, y, and z change for each step along the line.
In simple words: To find the direction numbers of a line given in a mixed up form, we need to change it so that x, y, and z are alone at the top of fractions. The numbers at the bottom of these fractions, once simplified, are the direction numbers of the line.

🎯 Exam Tip: Always ensure the coefficients of \( x, y, z \) in the numerator are +1 when converting to the symmetric form. If they are not, divide both the numerator and denominator by that coefficient.

Question 7. Find the cartesian equations of a line which
(i) passes through the point (1, 2, 3) and parallel to the line \( \frac{-x-2}{1} = \frac{y+3}{7} = \frac{2z-6}{3} \)
(ii) passes through the point (1,3,-2) and is parallel to the line given by \( \frac{x+1}{3} = \frac{y+4}{5} = \frac{z+3}{-6} \)
(iii) through the point (2, -1, 1) and parallel to the line joining the points (-1, 4, 1) and (1, 2, 2).
Answer: For each part, we first find the direction ratios of the given parallel line. Since parallel lines have proportional direction ratios, we use these for our new line.
The general equation of a line passing through \( (x_1, y_1, z_1) \) with direction ratios \( \) is \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \).
(i) **Line through (1, 2, 3) parallel to \( \frac{-x-2}{1} = \frac{y+3}{7} = \frac{2z-6}{3} \):**
First, simplify the equation of the given line to find its direction ratios:
\( \frac{-x-2}{1} = \frac{x+2}{-1} \)
\( \frac{2z-6}{3} = \frac{2(z-3)}{3} = \frac{z-3}{3/2} \)
So, the given line is \( \frac{x+2}{-1} = \frac{y+3}{7} = \frac{z-3}{3/2} \).
The direction ratios of this line are proportional to \( <-1, 7, 3/2> \). Multiplying by 2 to get integer ratios, we get \( <-2, 14, 3> \).
Since our required line is parallel, its direction ratios are also \( <-2, 14, 3> \).
It passes through \( (1, 2, 3) \). So, the equation is:
\( \frac{x-1}{-2} = \frac{y-2}{14} = \frac{z-3}{3} \)
(ii) **Line through (1, 3, -2) parallel to \( \frac{x+1}{3} = \frac{y+4}{5} = \frac{z+3}{-6} \):**
The given line is already in standard form. Its direction ratios are \( <3, 5, -6> \).
Since our required line is parallel, its direction ratios are also \( <3, 5, -6> \).
It passes through \( (1, 3, -2) \). So, the equation is:
\( \frac{x-1}{3} = \frac{y-3}{5} = \frac{z-(-2)}{-6} \)
\( \implies \frac{x-1}{3} = \frac{y-3}{5} = \frac{z+2}{-6} \)
(iii) **Line through (2, -1, 1) parallel to the line joining (-1, 4, 1) and (1, 2, 2):**
First, find the direction ratios of the line joining \( (-1, 4, 1) \) and \( (1, 2, 2) \):
\( a = x_2-x_1 = 1-(-1) = 2 \)
\( b = y_2-y_1 = 2-4 = -2 \)
\( c = z_2-z_1 = 2-1 = 1 \)
The direction ratios are \( <2, -2, 1> \).
Since our required line is parallel, its direction ratios are also \( <2, -2, 1> \).
It passes through \( (2, -1, 1) \). So, the equation is:
\( \frac{x-2}{2} = \frac{y-(-1)}{-2} = \frac{z-1}{1} \)
\( \implies \frac{x-2}{2} = \frac{y+1}{-2} = \frac{z-1}{1} \)
Parallel lines share the same direction, meaning their direction ratios will be proportional to each other, even if not exactly identical.
In simple words: To find the equation of a new line that is parallel to another line and passes through a certain point, we first find the direction numbers of the existing line. Then, we use these same direction numbers along with the given point to write the equation of our new line.

🎯 Exam Tip: Always make sure the equation of the parallel line is in its standard symmetric form before extracting direction ratios. If it's given in a non-standard form, perform the necessary algebraic steps to normalize it first.

Question 8. Prove that the points A(-1, 3, 2), B(-4, 2, -2) and C(5, 5, 10) are collinear.
Answer: To prove that three points are collinear, we can find the equation of the line passing through two of the points and then check if the third point satisfies this equation.
Let's find the equation of the line passing through points A(-1, 3, 2) and B(-4, 2, -2).
Using the two-point formula \( \frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1} \):
\( \frac{x-(-1)}{-4-(-1)} = \frac{y-3}{2-3} = \frac{z-2}{-2-2} \)
\( \implies \frac{x+1}{-3} = \frac{y-3}{-1} = \frac{z-2}{-4} \) (Let this be equation (1))
Now, we check if point C(5, 5, 10) lies on this line by substituting its coordinates into equation (1):
For the x-part: \( \frac{5+1}{-3} = \frac{6}{-3} = -2 \)
For the y-part: \( \frac{5-3}{-1} = \frac{2}{-1} = -2 \)
For the z-part: \( \frac{10-2}{-4} = \frac{8}{-4} = -2 \)
Since all three parts yield the same value (-2), point C(5, 5, 10) satisfies the equation of the line passing through A and B. Therefore, points A, B, and C are collinear. Collinear points are important in geometry for defining shapes and relationships, such as the vertices of a degenerate triangle.
In simple words: To show that three points are on the same straight line, we first find the equation of the line using the first two points. Then, we put the numbers of the third point into that line's equation. If the equation works out, meaning all parts are equal, then all three points are indeed on the same line.

🎯 Exam Tip: Another way to check for collinearity is to compute the direction ratios of AB and BC. If they are proportional, and points A, B, C share a common point (B), then they are collinear.

Question 9. Find the value of \( \lambda \), for which the points A(2, 1, 3), B(5, 0, 5) and C(-4, \( \lambda \),-1) are collinear.
Answer: Since points A, B, and C are collinear, point C must lie on the line passing through A and B.
First, let's find the equation of the line passing through A(2, 1, 3) and B(5, 0, 5).
Using the two-point formula \( \frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1} \):
\( \frac{x-2}{5-2} = \frac{y-1}{0-1} = \frac{z-3}{5-3} \)
\( \implies \frac{x-2}{3} = \frac{y-1}{-1} = \frac{z-3}{2} \) (Let this be equation (1))
Now, substitute the coordinates of point C(-4, \( \lambda \), -1) into equation (1):
\( \frac{-4-2}{3} = \frac{\lambda-1}{-1} = \frac{-1-3}{2} \)
\( \implies \frac{-6}{3} = \frac{\lambda-1}{-1} = \frac{-4}{2} \)
\( \implies -2 = \frac{\lambda-1}{-1} = -2 \)
From this, we can set up the equation for \( \lambda \):
\( \frac{\lambda-1}{-1} = -2 \)
\( \implies \lambda-1 = (-2) \times (-1) \)
\( \implies \lambda-1 = 2 \)
\( \implies \lambda = 2+1 \)
\( \implies \lambda = 3 \)
Thus, the value of \( \lambda \) for which the points are collinear is 3. This type of problem highlights how a single condition (collinearity) can be used to determine multiple unknown parameters in a geometric system.
In simple words: To find the missing number (lambda) that makes three points lie on one straight line, we first write the line's equation using the first two points. Since the third point is also on this line, we put its numbers into the equation. This gives us a simple equation to solve for lambda.

🎯 Exam Tip: When points are collinear, any two segments formed by these points (e.g., AB and BC) must have proportional direction ratios. This property is key to solving for unknown coordinates.

Question 10. Find the equations of a line passing through the point P(1, 2, 3) and having direction cosines \( \frac{2}{3}, -\frac{2}{3}, \frac{1}{3} \). Also find the coordinates of a point on the line at a distance of 6 units from P.
Answer: **Part 1: Equation of the line**
The line passes through point P(1, 2, 3) and has direction cosines \( <\frac{2}{3}, -\frac{2}{3}, \frac{1}{3}> \).
The general equation of a line passing through \( (x_1, y_1, z_1) \) with direction cosines \( \) is \( \frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n} \).
\( \frac{x-1}{2/3} = \frac{y-2}{-2/3} = \frac{z-3}{1/3} \)
Multiplying the denominators by 3 to simplify:
\( \implies \frac{x-1}{2} = \frac{y-2}{-2} = \frac{z-3}{1} \) (Let this be equation (1))
**Part 2: Coordinates of a point at a distance of 6 units from P**
Let the parameter for any point Q on the line be \( t \). So, \( \frac{x-1}{2} = \frac{y-2}{-2} = \frac{z-3}{1} = t \).
This gives the coordinates of Q as: \( x = 2t+1 \), \( y = -2t+2 \), \( z = t+3 \).
So, point Q is \( (2t+1, -2t+2, t+3) \).
We are given that the distance PQ is 6 units. The distance formula is \( \sqrt{(x_Q-x_P)^2 + (y_Q-y_P)^2 + (z_Q-z_P)^2} \).
\( \sqrt{((2t+1)-1)^2 + ((-2t+2)-2)^2 + ((t+3)-3)^2} = 6 \)
\( \sqrt{(2t)^2 + (-2t)^2 + (t)^2} = 6 \)
\( \sqrt{4t^2 + 4t^2 + t^2} = 6 \)
\( \sqrt{9t^2} = 6 \)
\( 3|t| = 6 \)
\( |t| = 2 \)
\( \implies t = \pm 2 \)
Now, we find the coordinates of Q for both values of \( t \):
When \( t=2 \):
\( Q_x = 2(2)+1 = 5 \)
\( Q_y = -2(2)+2 = -4+2 = -2 \)
\( Q_z = 2+3 = 5 \)
So, one point is \( (5, -2, 5) \).
When \( t=-2 \):
\( Q_x = 2(-2)+1 = -4+1 = -3 \)
\( Q_y = -2(-2)+2 = 4+2 = 6 \)
\( Q_z = -2+3 = 1 \)
So, the other point is \( (-3, 6, 1) \).
Thus, there are two points on the line at a distance of 6 units from P. The parameter \( t \) in the equation \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = t \) represents the signed distance from \( (x_1, y_1, z_1) \) along the line if \( (a, b, c) \) are direction cosines.
In simple words: First, we write the equation of the line using the given point and its direction numbers. Then, we describe any point on this line using a special letter 't'. We use the distance formula to say that the distance between the given point and this new point is 6. Solving this equation for 't' gives us two possible values, which means there are two points on the line that are 6 units away from the starting point.

🎯 Exam Tip: Remember that \( t \) can be positive or negative, indicating points on either side of the reference point P along the line. Always consider both possibilities when solving for distance-related problems.

Question 11. Find the values of p and q so that the points (p, q, 1), (-1, 4, -2) and (0, 2, -1) are collinear.
Answer: Let the given points be A(p, q, 1), B(-1, 4, -2), and C(0, 2, -1). If these three points are collinear, then point C must lie on the line passing through points A and B.
First, find the equation of the line passing through A(p, q, 1) and B(-1, 4, -2).
Using the two-point formula \( \frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1} \):
\( \frac{x-p}{-1-p} = \frac{y-q}{4-q} = \frac{z-1}{-2-1} \)
\( \implies \frac{x-p}{-1-p} = \frac{y-q}{4-q} = \frac{z-1}{-3} \) (Let this be equation (1))
Now, substitute the coordinates of point C(0, 2, -1) into equation (1):
\( \frac{0-p}{-1-p} = \frac{2-q}{4-q} = \frac{-1-1}{-3} \)
\( \implies \frac{-p}{-(1+p)} = \frac{2-q}{4-q} = \frac{-2}{-3} \)
\( \implies \frac{p}{1+p} = \frac{2-q}{4-q} = \frac{2}{3} \)
We now have two separate equations to solve for \( p \) and \( q \):
1. \( \frac{p}{1+p} = \frac{2}{3} \)
\( 3p = 2(1+p) \)
\( 3p = 2+2p \)
\( p = 2 \)
2. \( \frac{2-q}{4-q} = \frac{2}{3} \)
\( 3(2-q) = 2(4-q) \)
\( 6-3q = 8-2q \)
\( 6-8 = -2q+3q \)
\( -2 = q \)
Thus, the values are \( p=2 \) and \( q=-2 \). This method of checking collinearity is fundamental in analytic geometry and can be extended to determine if points are coplanar (lie on the same plane).
In simple words: To find the unknown numbers 'p' and 'q' that make three points lie on the same line, we first write the line's equation using the first two points, which will include 'p' and 'q'. Since the third point is also on this line, we put its numbers into the equation. This gives us two separate mini-equations that we can solve to find 'p' and 'q'.

🎯 Exam Tip: When setting up equations from collinearity, ensure you use the properties of equality (all ratios are equal) to form distinct equations that allow you to solve for each unknown variable independently.

Question 12. Find the equations of a line in standard form and hence find the coordinates of a point on it and its direction cosines : \( \frac{3-2x}{4} = \frac{2y-1}{2} = \frac{3+z}{2} \).
Answer: The given equation of the line is \( \frac{3-2x}{4} = \frac{2y-1}{2} = \frac{3+z}{2} \).
To convert this into the standard symmetric form \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \), we need to ensure the coefficients of \( x, y, z \) in the numerators are +1.
Let's rewrite each part:
\( \frac{3-2x}{4} = \frac{-(2x-3)}{4} = \frac{x-3/2}{-4/2} = \frac{x-3/2}{-2} \)
\( \frac{2y-1}{2} = \frac{2(y-1/2)}{2} = \frac{y-1/2}{1} \)
\( \frac{3+z}{2} = \frac{z-(-3)}{2} \)
So, the equation of the line in standard symmetric form is:
\( \frac{x-3/2}{-2} = \frac{y-1/2}{1} = \frac{z-(-3)}{2} \)
From this standard form, we can identify:
**Coordinates of a point on the line:** \( (x_1, y_1, z_1) = (\frac{3}{2}, \frac{1}{2}, -3) \).
**Direction ratios:** \( = <-2, 1, 2> \).
Now, to find the **direction cosines**, we first calculate the magnitude of the direction ratios:
Magnitude \( = \sqrt{a^2+b^2+c^2} = \sqrt{(-2)^2 + 1^2 + 2^2} \)
\( = \sqrt{4+1+4} = \sqrt{9} = 3 \).
The direction cosines \( \) are obtained by dividing each direction ratio by this magnitude:
\( l = \frac{-2}{3} \)
\( m = \frac{1}{3} \)
\( n = \frac{2}{3} \)
Thus, the direction cosines are \( <\frac{-2}{3}, \frac{1}{3}, \frac{2}{3}> \). Direction cosines are normalized direction ratios, ensuring that the vector representing the line's direction has a unit length, making them unique for a given line direction.
In simple words: To get the line's equation into a neat standard form, we change it so that x, y, and z are simple terms at the top of fractions. The numbers that remain in the numerator (like 3/2, 1/2, -3) tell us a point the line goes through. The numbers at the bottom of the fractions are the line's direction numbers. To find the direction cosines, we divide each direction number by the total 'size' of these direction numbers.

🎯 Exam Tip: Be careful with the signs when extracting the point \( (x_1, y_1, z_1) \) from the standard form; for example, \( z-(-3) \) implies \( z_1 = -3 \).

Question 13. Find the direction cosines of the line whose equations are \( \frac{x-2}{2} = \frac{2y-5}{-3}, z = -1 \).
Answer: The given equations of the line are \( \frac{x-2}{2} = \frac{2y-5}{-3} \) and \( z = -1 \).
To find the direction cosines, we first convert these into the standard symmetric form of a line.
For the x-part: \( \frac{x-2}{2} \) (already in standard form)
For the y-part: \( \frac{2y-5}{-3} = \frac{2(y-5/2)}{-3} = \frac{y-5/2}{-3/2} \)
For the z-part: The equation \( z = -1 \) means that \( z \) is constant. In the symmetric form, this implies the denominator for \( z \) is 0, and the numerator is \( z-(-1) \). So, \( z+1=0 \), or \( \frac{z-(-1)}{0} \).
Combining these, the line in standard symmetric form is:
\( \frac{x-2}{2} = \frac{y-5/2}{-3/2} = \frac{z-(-1)}{0} \)
From this, the direction ratios are \( <2, -3/2, 0> \).
To get integer direction ratios, we can multiply by 2:
Direction ratios \( = <4, -3, 0> \).
Now, we find the magnitude of these direction ratios:
Magnitude \( = \sqrt{a^2+b^2+c^2} = \sqrt{4^2 + (-3)^2 + 0^2} \)
\( = \sqrt{16+9+0} = \sqrt{25} = 5 \).
Finally, we find the direction cosines by dividing each direction ratio by the magnitude:
\( l = \frac{4}{5} \)
\( m = \frac{-3}{5} \)
\( n = \frac{0}{5} = 0 \)
Thus, the direction cosines of the line are \( <\frac{4}{5}, \frac{-3}{5}, 0> \). A direction cosine of zero for a specific axis means the line is perpendicular to that axis, running parallel to the plane formed by the other two axes.
In simple words: To find the direction cosines, we first change the given equations into a standard form for a line. The part where 'z' is a fixed number means its direction number is zero. We make sure 'x' and 'y' parts are also in the correct form. Then, we take the direction numbers (from the bottom of the fractions) and divide each by their total 'size' to get the direction cosines.

🎯 Exam Tip: An equation like \( z=k \) for a line means the line is parallel to the xy-plane and perpendicular to the z-axis, so its direction ratio along the z-axis is 0. Similarly for \( x=k \) or \( y=k \).

Question 14. Find the equations of a line through A(1,-1, 5) and parallel to the line \( \frac{x-2}{3} = \frac{y-5}{-2}, z = -1 \).
Answer: We need to find the equation of a line passing through A(1, -1, 5) and parallel to the line given by \( \frac{x-2}{3} = \frac{y-5}{-2} \) and \( z = -1 \).
First, let's find the direction ratios of the given parallel line. We convert its equation into standard symmetric form:
For the x-part: \( \frac{x-2}{3} \) (already in standard form)
For the y-part: \( \frac{y-5}{-2} \) (already in standard form)
For the z-part: The equation \( z = -1 \) implies that the line is parallel to the xy-plane. In the symmetric form, this means the denominator for the z-term is 0. We write it as \( \frac{z-(-1)}{0} \).
So, the standard symmetric equation of the parallel line is:
\( \frac{x-2}{3} = \frac{y-5}{-2} = \frac{z-(-1)}{0} \)
From this, the direction ratios of the parallel line are \( <3, -2, 0> \).
Since the required line is parallel to this line, it will have the same direction ratios: \( <3, -2, 0> \).
The required line passes through point A(1, -1, 5).
Using the formula \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \):
\( \frac{x-1}{3} = \frac{y-(-1)}{-2} = \frac{z-5}{0} \)
\( \implies \frac{x-1}{3} = \frac{y+1}{-2} = \frac{z-5}{0} \)
This is the required equation of the line. Lines that are parallel always share the same orientation in space, meaning their direction ratios are either identical or proportional.
In simple words: To write the equation of a new line that goes through point A and runs parallel to another given line, we first find the direction numbers of that second line. If 'z' is a fixed number, its direction number is zero. Then, we use these same direction numbers and the coordinates of point A to write the equation for our new line.

🎯 Exam Tip: Pay close attention to equations like \( z=k \) which indicate a zero in the corresponding direction ratio, meaning the line is parallel to the plane formed by the other two axes.

Question 15. The equation of a line is \( \frac{2x-5}{4} = \frac{y+4}{3} = \frac{6-z}{6} \). Find the direction cosines of a line parallel to the line.
Answer: The given equation of the line is \( \frac{2x-5}{4} = \frac{y+4}{3} = \frac{6-z}{6} \).
To find the direction cosines, we first need to convert this equation into the standard symmetric form \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \).
Let's rewrite each part to ensure \( x, y, z \) have a coefficient of +1:
For the x-part: \( \frac{2x-5}{4} = \frac{2(x-5/2)}{4} = \frac{x-5/2}{4/2} = \frac{x-5/2}{2} \)
For the y-part: \( \frac{y+4}{3} \) (already in standard form, as \( y-(-4) \))
For the z-part: \( \frac{6-z}{6} = \frac{-(z-6)}{6} = \frac{z-6}{-6} \)
So, the equation of the line in standard symmetric form is:
\( \frac{x-5/2}{2} = \frac{y-(-4)}{3} = \frac{z-6}{-6} \)
From this, the direction ratios of the line are \( = <2, 3, -6> \).
To find the direction cosines, we calculate the magnitude of these direction ratios:
Magnitude \( = \sqrt{a^2+b^2+c^2} = \sqrt{2^2 + 3^2 + (-6)^2} \)
\( = \sqrt{4+9+36} = \sqrt{49} = 7 \).
The direction cosines \( \) are obtained by dividing each direction ratio by this magnitude:
\( l = \frac{2}{7} \)
\( m = \frac{3}{7} \)
\( n = \frac{-6}{7} \)
Thus, the direction cosines of the line (or any line parallel to it) are \( <\frac{2}{7}, \frac{3}{7}, \frac{-6}{7}> \). Direction cosines are normalized direction ratios, providing a unique set of values that represent the exact orientation of a line in 3D space.
In simple words: To find the direction cosines of this line, we first need to change its equation into a standard form where 'x', 'y', and 'z' are simple terms at the top of fractions. The numbers at the bottom are the direction numbers. Then, we find the total 'size' of these direction numbers and divide each one by this size to get the direction cosines.

🎯 Exam Tip: Always prioritize making the coefficients of \( x, y, z \) positive 1 in the numerator. A common mistake is to overlook the negative sign in \( (6-z) \) and incorrectly take \( +6 \) as the denominator for \( z \).