OP Malhotra Class 12 Maths Solutions Chapter 24 The Plane Exercise 24 (A)

Question 1. Find the direction cosines of the normal to the plane
(i) 2x – 3y + 6z = 7
(ii) x + 2y + 2z - 1 = 0
Answer:
(i) Given equation of the plane is \( 2x - 3y + 6z = 7 \).
The direction ratios of the normal to this plane are \( <2, -3, 6> \).
To find the direction cosines, we divide each direction ratio by the magnitude of the normal vector, which is \( \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \).
So, the direction cosines of the normal to the given plane are:
\( \left< \frac{2}{\sqrt{2^2 + (-3)^2 + 6^2}}, \frac{-3}{\sqrt{2^2 + (-3)^2 + 6^2}}, \frac{6}{\sqrt{2^2 + (-3)^2 + 6^2}} \right> \)
\( \implies \left< \frac{2}{7}, \frac{-3}{7}, \frac{6}{7} \right> \)
(ii) Given equation of the plane is \( x + 2y + 2z - 1 = 0 \).
The direction ratios of the normal to this plane are \( <1, 2, 2> \).
The magnitude of the normal vector is \( \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \).
So, the direction cosines of the normal to the plane are:
\( \left< \frac{1}{\sqrt{1^2 + 2^2 + 2^2}}, \frac{2}{\sqrt{1^2 + 2^2 + 2^2}}, \frac{2}{\sqrt{1^2 + 2^2 + 2^2}} \right> \)
\( \implies \left< \frac{1}{3}, \frac{2}{3}, \frac{2}{3} \right> \)
In simple words: To find direction cosines of a plane's normal, take the coefficients of x, y, z (which are the direction ratios). Then, divide each of these numbers by the length of the vector formed by these coefficients. This length is found using the square root of the sum of their squares.

🎯 Exam Tip: Remember that direction cosines are always normalized by the magnitude of the direction ratios. Ensure the sign of each component is correctly maintained from the plane equation.

Question 2. Show that the normals to the planes \( x - y + z = 1 \), \( 3x + 2y - z + 2 = 0 \) are inclined to each other at an angle of 90°.
Answer:
The given equations of the planes are:
1. \( x - y + z = 1 \)
2. \( 3x + 2y - z + 2 = 0 \)
The direction ratios of the normal to plane (1) are \( \vec{n_1} = <1, -1, 1> \).
The direction ratios of the normal to plane (2) are \( \vec{n_2} = <3, 2, -1> \).
To find the angle between the normals, we can use the dot product formula. If the normals are perpendicular, their dot product should be zero.
Let \( a_1=1, b_1=-1, c_1=1 \) and \( a_2=3, b_2=2, c_2=-1 \).
Now, calculate the dot product \( \vec{n_1} \cdot \vec{n_2} = a_1a_2 + b_1b_2 + c_1c_2 \):
\( \vec{n_1} \cdot \vec{n_2} = (1)(3) + (-1)(2) + (1)(-1) \)
\( \implies 3 - 2 - 1 \)
\( \implies 0 \)
Since the dot product of their direction ratios is 0, it means the normals are perpendicular to each other. When two normal vectors are perpendicular, the angle between them is 90°. This proves the statement.
In simple words: First, find the "direction numbers" for the line that sticks straight out from each plane. Then, multiply the matching direction numbers together and add them up. If the final sum is zero, it means the lines sticking out from the planes are at a 90-degree angle to each other.

🎯 Exam Tip: To show two planes' normals are perpendicular, simply calculate the dot product of their direction ratios. If the result is zero, they are perpendicular.

Question 3. Find the intercepts made by the plane \( 2x - 3y + 4z = 12 \) on the coordinate axes.
Answer:
The given equation of the plane is \( 2x - 3y + 4z = 12 \).
To find the intercepts, we need to convert the equation into the intercept form, which is \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \), where a, b, and c are the x, y, and z-intercepts respectively. A helpful way to remember this is to divide the entire equation by the constant term on the right side.
Divide the entire equation by 12:
\( \frac{2x}{12} - \frac{3y}{12} + \frac{4z}{12} = \frac{12}{12} \)
\( \implies \frac{x}{6} - \frac{y}{4} + \frac{z}{3} = 1 \)
We can rewrite this as \( \frac{x}{6} + \frac{y}{-4} + \frac{z}{3} = 1 \).
Comparing this with the intercept form \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \), we get:
The x-intercept \( a = 6 \).
The y-intercept \( b = -4 \).
The z-intercept \( c = 3 \).
Thus, the intercepts made by the plane on the coordinate axes are 6, -4, and 3.
In simple words: To find where a plane cuts the x, y, and z lines, make the equation look like "x divided by a number, plus y divided by a number, plus z divided by a number, equals 1". The numbers under x, y, and z are your answers. To do this, simply divide every part of the plane's equation by the constant number on the right side.

🎯 Exam Tip: Always ensure the right-hand side of the plane equation is 1 before identifying the intercepts. Pay close attention to the signs of the intercepts as they indicate where the plane crosses the axes.

Question 4. Write the equation of each of the following planes in the intercept form:
(i) \( 2x - 3y + 4z = 12 \)
(ii) \( 3x + 2y - 5z = 15 \)
(iii) \( x + 3y + 4z = 12 \)
Answer:
(i) Given equation of plane is \( 2x - 3y + 4z = 12 \).
To convert it to the intercept form \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \), divide the entire equation by the constant term on the right side, which is 12.
\( \frac{2x}{12} - \frac{3y}{12} + \frac{4z}{12} = \frac{12}{12} \)
\( \implies \frac{x}{6} - \frac{y}{4} + \frac{z}{3} = 1 \)
This can also be written as \( \frac{x}{6} + \frac{y}{-4} + \frac{z}{3} = 1 \), which is the required intercept form of the plane.

(ii) Given equation of plane is \( 3x + 2y - 5z = 15 \).
Divide the entire equation by the constant term 15.
\( \frac{3x}{15} + \frac{2y}{15} - \frac{5z}{15} = \frac{15}{15} \)
\( \implies \frac{x}{5} + \frac{y}{15/2} - \frac{z}{3} = 1 \)
This can also be written as \( \frac{x}{5} + \frac{y}{15/2} + \frac{z}{-3} = 1 \), which is the required intercept form.

(iii) Given equation of plane is \( x + 3y + 4z = 12 \).
Divide the entire equation by the constant term 12.
\( \frac{x}{12} + \frac{3y}{12} + \frac{4z}{12} = \frac{12}{12} \)
\( \implies \frac{x}{12} + \frac{y}{4} + \frac{z}{3} = 1 \), which is the required intercept form of the plane.
In simple words: To write a plane's equation in "intercept form," you need to rearrange it so that the right side equals 1, and on the left side, you have 'x' divided by a number, 'y' divided by a number, and 'z' divided by a number, all added together. You do this by dividing every part of the original equation by the constant on the right side.

🎯 Exam Tip: When converting to intercept form, always divide by the constant on the right-hand side. Ensure the terms for x, y, and z are clearly in the \( \frac{x}{a} \) format, meaning their coefficients become 1 after division.

Question 5. Find the equation of the plane:
(i) which cuts the axes of x, y and z at (-2, 0, 0), (0, 3, 0), (0, 0, 5) respectively.
(ii) which cuts the axes of x and y at (3, 0, 0) and (0, -2, 0) and does not cut the z-axis.
(iii) which cuts the x-axis at (4, 0, 0) and does not cut the y-axis and the z-axis.
(iv) passing through the point (2, 3, 4) and making equal intercepts on the coordinates axes.
Answer:
(i) Let the equation of the plane be in intercept form: \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \).
The plane cuts the x-axis at \( (-2, 0, 0) \), so \( a = -2 \).
The plane cuts the y-axis at \( (0, 3, 0) \), so \( b = 3 \).
The plane cuts the z-axis at \( (0, 0, 5) \), so \( c = 5 \).
Substitute these values into the intercept form:
\( \frac{x}{-2} + \frac{y}{3} + \frac{z}{5} = 1 \)
This is the required equation of the plane.

(ii) Let the equation of the plane be \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \).
The plane cuts the x-axis at \( (3, 0, 0) \), so \( a = 3 \).
The plane cuts the y-axis at \( (0, -2, 0) \), so \( b = -2 \).
The plane does not cut the z-axis. This means the plane is parallel to the z-axis, so the z-intercept \( c \) is effectively infinite. In this case, the equation of the plane is \( \frac{x}{a} + \frac{y}{b} = 1 \).
Substitute the values of \( a \) and \( b \):
\( \frac{x}{3} + \frac{y}{-2} = 1 \)
This is the required equation of the plane.

(iii) Let the equation of the plane be \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \).
The plane cuts the x-axis at \( (4, 0, 0) \), so \( a = 4 \).
The plane does not cut the y-axis and z-axis. This means the plane is parallel to both the y-axis and z-axis, which implies it is parallel to the y-z plane. The equation of such a plane is of the form \( x = k \).
Since it cuts the x-axis at \( (4, 0, 0) \), the value of \( k \) must be 4.
So, the equation of the plane is \( x = 4 \).
This is the required equation of the plane.

(iv) Let the equation of the plane be \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \).
The plane makes equal intercepts on the coordinate axes, so \( a = b = c \).
Thus, the equation becomes \( \frac{x}{a} + \frac{y}{a} + \frac{z}{a} = 1 \), which simplifies to \( x + y + z = a \).
The plane passes through the point \( (2, 3, 4) \). Substitute these coordinates into the equation:
\( 2 + 3 + 4 = a \)
\( \implies a = 9 \)
Substitute the value of \( a \) back into the equation:
\( x + y + z = 9 \)
This is the required equation of the plane.
In simple words: To find the plane's equation, first look at how it crosses the x, y, and z lines. If it cuts all three, use the "intercept form" \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \). If it misses one axis, that means it's parallel to that axis, and its intercept for that axis is considered infinite, simplifying the equation. If it makes equal intercepts, put the same letter for all intercepts and then use the given point to find that letter's value.

🎯 Exam Tip: When a plane doesn't cut an axis, it's parallel to that axis, and the corresponding intercept is considered infinite, simplifying the intercept form (e.g., \( \frac{x}{a} + \frac{y}{b} = 1 \) for a plane parallel to the z-axis). For equal intercepts, remember to substitute the same variable for all intercepts.

Question 6. Find the equations of the two planes passing through the points (0, 4, -3) and (6, -4, 3), if the sum of their intercepts on the three axes is zero.
Answer:
Let the equation of the plane be \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \) (Equation 1).
It is given that the sum of its intercepts on the three axes is zero:
\( a + b + c = 0 \)
\( \implies c = -a - b \) (Equation 2)
The plane passes through the point \( (0, 4, -3) \). Substitute these coordinates into Equation 1:
\( \frac{0}{a} + \frac{4}{b} + \frac{-3}{c} = 1 \)
\( \implies \frac{4}{b} - \frac{3}{c} = 1 \)
\( \implies 4c - 3b = bc \) (Equation 3)
The plane also passes through the point \( (6, -4, 3) \). Substitute these coordinates into Equation 1:
\( \frac{6}{a} + \frac{-4}{b} + \frac{3}{c} = 1 \)
\( \implies \frac{6}{a} - \frac{4}{b} + \frac{3}{c} = 1 \) (Equation 4)
Now, substitute \( c = -a - b \) from Equation 2 into Equation 3:
\( 4(-a - b) - 3b = b(-a - b) \)
\( \implies -4a - 4b - 3b = -ab - b^2 \)
\( \implies -4a - 7b = -ab - b^2 \)
\( \implies b^2 + ab - 7b - 4a = 0 \) (Equation 5)
Next, substitute \( c = -a - b \) from Equation 2 into Equation 4:
\( \frac{6}{a} - \frac{4}{b} + \frac{3}{-a-b} = 1 \)
To solve this system, let's use a slightly different approach with Equation 4. From Equation 3, we have \( 1 = \frac{4}{b} - \frac{3}{c} \). Substitute this into Equation 4:
\( \frac{6}{a} - \frac{4}{b} + \frac{3}{c} = \frac{4}{b} - \frac{3}{c} \)
\( \implies \frac{6}{a} = \frac{8}{b} - \frac{6}{c} \) (Equation 4a)
This also needs substitution from Equation 2. Let's return to Equation 5 and use the values we found for \( a \), \( b \), \( c \) from the calculation steps in the source, as it appears they have been derived using a specific method.
From Equation 5, if we rearrange and factorize, we can find relationships between \( a \) and \( b \). The source's calculation implies \( b^2 - 4b - 12 = 0 \). Let's see if we can derive this or if it comes from some specific intermediate step.
Let's follow the source's flow more directly:
Substitute \( \frac{4}{b} - \frac{3}{c} = 1 \) into \( \frac{6}{a} - (\frac{4}{b} - \frac{3}{c}) = 1 \). This isn't exactly what the source does. The source appears to say "using eqn. (3) in eqn. (4); we have \( \frac{6}{a} - 1 = 1 \)". This implies that \( (\frac{4}{b} - \frac{3}{c}) = 1 \). This is correct from Eq. 3. So, the source is using Eq. 3 to simplify Eq. 4:
From Equation 4: \( \frac{6}{a} - (\frac{4}{b} - \frac{3}{c}) = 1 \)
Using Equation 3: \( \frac{6}{a} - (1) = 1 \)
\( \implies \frac{6}{a} = 2 \)
\( \implies a = 3 \)
Now we have \( a = 3 \). Substitute \( a = 3 \) into Equation 2:
\( c = -3 - b \)
Substitute \( a = 3 \) and \( c = -3 - b \) into Equation 3:
\( 4(-3 - b) - 3b = b(-3 - b) \)
\( \implies -12 - 4b - 3b = -3b - b^2 \)
\( \implies -12 - 7b = -3b - b^2 \)
\( \implies b^2 - 4b - 12 = 0 \)
Now, factorize this quadratic equation for \( b \):
\( (b - 6)(b + 2) = 0 \)
This gives two possible values for \( b \): \( b = 6 \) or \( b = -2 \).

Case 1: When \( b = 6 \)
Using \( a = 3 \) and \( c = -3 - b \):
\( c = -3 - 6 \)
\( \implies c = -9 \)
So, one set of intercepts is \( (a, b, c) = (3, 6, -9) \).
The equation of the plane for this case is:
\( \frac{x}{3} + \frac{y}{6} + \frac{z}{-9} = 1 \)

Case 2: When \( b = -2 \)
Using \( a = 3 \) and \( c = -3 - b \):
\( c = -3 - (-2) \)
\( \implies c = -3 + 2 \)
\( \implies c = -1 \)
So, the second set of intercepts is \( (a, b, c) = (3, -2, -1) \).
The equation of the plane for this case is:
\( \frac{x}{3} + \frac{y}{-2} + \frac{z}{-1} = 1 \)
Thus, the two required equations of the planes are \( \frac{x}{3} + \frac{y}{6} + \frac{z}{-9} = 1 \) and \( \frac{x}{3} + \frac{y}{-2} + \frac{z}{-1} = 1 \).
In simple words: First, assume the plane's equation is in "intercept form" (x/a + y/b + z/c = 1). We are told that a + b + c = 0. Use the two given points that the plane passes through to make two new equations. Solve these equations together with a + b + c = 0 to find the values for 'a', 'b', and 'c'. There will be two possible sets of values for 'a', 'b', 'c', which means two possible plane equations. Remember that a quadratic equation often gives two solutions.

🎯 Exam Tip: This problem involves multiple conditions. Start by writing the general intercept form of the plane and the given sum of intercepts. Substitute the given points one by one, simplify the resulting equations, and solve them simultaneously to find the values of a, b, and c. Expect multiple solutions as quadratic equations often arise.

Question 7. A plane meets the coordinate axes at A, B, C respectively such that the centroid of the triangle A B C is (1, -2, 3). Find the equation of the plane.
Answer:
Let the required equation of the plane be in intercept form: \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \).
When the plane meets the coordinate axes, the points A, B, C are:
A \( (a, 0, 0) \) (on the x-axis)
B \( (0, b, 0) \) (on the y-axis)
C \( (0, 0, c) \) (on the z-axis)
The centroid of a triangle with vertices \( (x_1, y_1, z_1) \), \( (x_2, y_2, z_2) \), and \( (x_3, y_3, z_3) \) is given by the formula:
\( \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right) \).
For triangle ABC, the centroid is:
\( \left( \frac{a + 0 + 0}{3}, \frac{0 + b + 0}{3}, \frac{0 + 0 + c}{3} \right) \)
\( \implies \left( \frac{a}{3}, \frac{b}{3}, \frac{c}{3} \right) \)
We are given that the centroid of triangle ABC is \( (1, -2, 3) \).
So, we can equate the coordinates:
\( \frac{a}{3} = 1 \implies a = 3 \)
\( \frac{b}{3} = -2 \implies b = -6 \)
\( \frac{c}{3} = 3 \implies c = 9 \)
Now, substitute the values of \( a, b, c \) back into the intercept form of the plane equation:
\( \frac{x}{3} + \frac{y}{-6} + \frac{z}{9} = 1 \)
To make it a standard form equation, we can find a common denominator, which is 18 in this case:
\( \frac{6x}{18} - \frac{3y}{18} + \frac{2z}{18} = 1 \)
\( \implies 6x - 3y + 2z = 18 \)
This is the required equation of the plane.
In simple words: First, imagine the plane cutting the x, y, and z lines at points A, B, and C. These points will have coordinates like (a,0,0), (0,b,0), and (0,0,c). Next, use the formula for a triangle's centroid, which is like finding the average of all the x-coordinates, y-coordinates, and z-coordinates. You are given the centroid's coordinates, so you can easily find the values of a, b, and c. Finally, put these values back into the plane's intercept form equation to get your answer.

🎯 Exam Tip: Remember the general formula for the centroid of a triangle. For a plane intersecting axes, the intercept values directly give the coordinates of the intersection points (e.g., (a,0,0)). Equating these with the given centroid coordinates is key to solving this type of problem.

Question 8.
(i) Find the equation of the plane the normal to which from the origin is of length of 5 and which makes angles, with the axes of x, y and z, equal to 120°, 45° and 120° respectively.
(ii) Find the equation of the plane the normal to which from the origin is of length 4 and which makes angles with the axes of x, y, z equal to 90°, 135°, 45° respectively.
(iii) Find the equation of the plane such that the foot of the normal to which from the origin is the point (2, 3, 1).
(iv) Find the equation of the plane such that the length of the perpendicular to which from the origin is equal to 2 units, and the angles \( \alpha, \beta, \gamma \) that this perpendicular makes with the axes of x, y, z are connected by the relation \( \frac{\cos \alpha}{-1} = \frac{\cos \beta}{4} = \frac{\cos \gamma}{4} \).
Answer:
(i) We are given that the length of the normal from the origin \( p = 5 \).
The normal makes angles \( 120^\circ, 45^\circ, 120^\circ \) with the x, y, and z axes respectively.
The direction cosines (l, m, n) of the normal are:
\( l = \cos 120^\circ = -\frac{1}{2} \)
\( m = \cos 45^\circ = \frac{1}{\sqrt{2}} \)
\( n = \cos 120^\circ = -\frac{1}{2} \)
The equation of a plane in normal form is given by \( lx + my + nz = p \).
Substitute the values of \( l, m, n, \) and \( p \):
\( \left(-\frac{1}{2}\right)x + \left(\frac{1}{\sqrt{2}}\right)y + \left(-\frac{1}{2}\right)z = 5 \)
Multiply the entire equation by \( -2\sqrt{2} \) to clear denominators and negative signs, making the leading coefficient positive:
\( \sqrt{2}x - 2y + \sqrt{2}z = -10\sqrt{2} \)
\( \implies \sqrt{2}x - 2y + \sqrt{2}z + 10\sqrt{2} = 0 \)
If we multiply by \( -2 \) only, it would be \( x - \sqrt{2}y + z = -10 \implies x - \sqrt{2}y + z + 10 = 0 \). This simpler form matches the source after a sign correction.

(ii) We are given that the length of the normal from the origin \( p = 4 \).
The normal makes angles \( 90^\circ, 135^\circ, 45^\circ \) with the x, y, and z axes respectively.
The direction cosines (l, m, n) of the normal are:
\( l = \cos 90^\circ = 0 \)
\( m = \cos 135^\circ = -\frac{1}{\sqrt{2}} \)
\( n = \cos 45^\circ = \frac{1}{\sqrt{2}} \)
The equation of a plane in normal form is \( lx + my + nz = p \).
Substitute the values of \( l, m, n, \) and \( p \):
\( (0)x + \left(-\frac{1}{\sqrt{2}}\right)y + \left(\frac{1}{\sqrt{2}}\right)z = 4 \)
\( \implies -\frac{1}{\sqrt{2}}y + \frac{1}{\sqrt{2}}z = 4 \)
Multiply the entire equation by \( \sqrt{2} \):
\( \implies -y + z = 4\sqrt{2} \)
\( \implies y - z + 4\sqrt{2} = 0 \)
This is the required equation of the plane.

(iii) The foot of the normal from the origin to the plane is the point \( N(2, 3, 1) \).
The line segment ON (from origin \( O(0,0,0) \) to \( N(2,3,1) \)) is the normal to the plane. Therefore, the direction ratios of the normal to the plane are \( <2-0, 3-0, 1-0> = <2, 3, 1> \).
The plane passes through the point N(2, 3, 1).
The equation of a plane passing through a point \( (x_1, y_1, z_1) \) with normal direction ratios \( \) is given by \( a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \).
Here, \( (x_1, y_1, z_1) = (2, 3, 1) \) and \( = <2, 3, 1> \).
Substitute these values:
\( 2(x - 2) + 3(y - 3) + 1(z - 1) = 0 \)
\( \implies 2x - 4 + 3y - 9 + z - 1 = 0 \)
\( \implies 2x + 3y + z - 14 = 0 \)
This is the required equation of the plane. A visual representation helps understand this geometry:

(iv) We are given that the length of the perpendicular from the origin \( p = 2 \) units.
The angles \( \alpha, \beta, \gamma \) that this perpendicular makes with the axes are connected by the relation \( \frac{\cos \alpha}{-1} = \frac{\cos \beta}{4} = \frac{\cos \gamma}{8} \).
Let \( \frac{\cos \alpha}{-1} = \frac{\cos \beta}{4} = \frac{\cos \gamma}{8} = k \).
Then \( \cos \alpha = -k, \cos \beta = 4k, \cos \gamma = 8k \).
We know that for direction cosines, \( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \).
Substitute the values:
\( (-k)^2 + (4k)^2 + (8k)^2 = 1 \)
\( \implies k^2 + 16k^2 + 64k^2 = 1 \)
\( \implies 81k^2 = 1 \)
\( \implies k^2 = \frac{1}{81} \)
\( \implies k = \pm \frac{1}{9} \)
So, the direction cosines \( (l, m, n) \) can be:
Case 1: Using \( k = \frac{1}{9} \)
\( l = \cos \alpha = -1 \cdot \frac{1}{9} = -\frac{1}{9} \)
\( m = \cos \beta = 4 \cdot \frac{1}{9} = \frac{4}{9} \)
\( n = \cos \gamma = 8 \cdot \frac{1}{9} = \frac{8}{9} \)
The equation of the plane is \( lx + my + nz = p \):
\( -\frac{1}{9}x + \frac{4}{9}y + \frac{8}{9}z = 2 \)
Multiply by 9:
\( -x + 4y + 8z = 18 \)
\( \implies x - 4y - 8z + 18 = 0 \)
Case 2: Using \( k = -\frac{1}{9} \)
\( l = \cos \alpha = -1 \cdot (-\frac{1}{9}) = \frac{1}{9} \)
\( m = \cos \beta = 4 \cdot (-\frac{1}{9}) = -\frac{4}{9} \)
\( n = \cos \gamma = 8 \cdot (-\frac{1}{9}) = -\frac{8}{9} \)
The equation of the plane is \( lx + my + nz = p \):
\( \frac{1}{9}x - \frac{4}{9}y - \frac{8}{9}z = 2 \)
Multiply by 9:
\( x - 4y - 8z = 18 \)
\( \implies x - 4y - 8z - 18 = 0 \)
Thus, there are two possible equations for the plane: \( x - 4y - 8z + 18 = 0 \) and \( x - 4y - 8z - 18 = 0 \).
In simple words: (i) If you know how far the plane is from the center (origin) and the angles its straight-out line (normal) makes with the axes, use the formula "lx + my + nz = p" to find the plane's equation. 'l', 'm', 'n' are the cosines of those angles, and 'p' is the distance. (ii) Same as (i), just with different angles and distance. (iii) If you know the exact point where the normal line from the origin touches the plane, this point gives you the direction numbers for the normal. Then, use the formula for a plane through a point with a given normal direction. (iv) When the angles' cosines are related by a ratio, first find the value of that ratio using the fact that the sum of the squares of direction cosines is 1. This will give you two sets of direction cosines (because of \( \pm \) square root), leading to two possible plane equations.

🎯 Exam Tip: For problems involving the normal form of a plane \( (lx + my + nz = p) \), always correctly calculate the direction cosines \( (l, m, n) \) from the given angles. If the foot of the normal is given, treat it as a point on the plane and use the vector from the origin to that point as the normal vector's direction ratios.

Question 9. Convert each of the following equations to the normal form and then determine the direction cosines and the length of the normal from the origin:
(i) \( 2x - 2y + z - 12 = 0 \)
(ii) \( 9x + 6y - 2z + 7 = 0 \)
Answer:
(i) Given equation of the plane is \( 2x - 2y + z - 12 = 0 \).
Rewrite it as \( 2x - 2y + z = 12 \).
To convert to normal form \( lx + my + nz = p \), we need to divide the entire equation by the magnitude of the normal vector \( \vec{n} = <2, -2, 1> \).
The magnitude is \( \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \).
Divide the equation by 3:
\( \frac{2x}{3} - \frac{2y}{3} + \frac{z}{3} = \frac{12}{3} \)
\( \implies \frac{2}{3}x - \frac{2}{3}y + \frac{1}{3}z = 4 \)
This is the required normal form of the plane.
Comparing with \( lx + my + nz = p \):
The direction cosines of the normal are \( l = \frac{2}{3}, m = -\frac{2}{3}, n = \frac{1}{3} \).
The length of the normal from the origin is \( p = 4 \).

(ii) Given equation of the plane is \( 9x + 6y - 2z + 7 = 0 \).
Rewrite it as \( 9x + 6y - 2z = -7 \).
To ensure the length \( p \) is positive, we make the right-hand side positive by multiplying by -1:
\( -9x - 6y + 2z = 7 \).
Now, find the magnitude of the normal vector \( \vec{n} = <-9, -6, 2> \).
The magnitude is \( \sqrt{(-9)^2 + (-6)^2 + 2^2} = \sqrt{81 + 36 + 4} = \sqrt{121} = 11 \).
Divide the equation by 11:
\( \frac{-9x}{11} - \frac{6y}{11} + \frac{2z}{11} = \frac{7}{11} \)
\( \implies -\frac{9}{11}x - \frac{6}{11}y + \frac{2}{11}z = \frac{7}{11} \)
This is the required normal form of the plane.
Comparing with \( lx + my + nz = p \):
The direction cosines of the normal are \( l = -\frac{9}{11}, m = -\frac{6}{11}, n = \frac{2}{11} \).
The length of the normal from the origin is \( p = \frac{7}{11} \).
In simple words: To change a plane's equation into "normal form," you first need to move any constant numbers to the right side of the equals sign. Make sure the number on the right side is positive. Then, divide every part of the equation by the "length" of the normal vector (which is found by taking the square root of the sum of the squares of the x, y, and z coefficients). Once it's in this form, the new x, y, and z coefficients are your direction cosines, and the number on the right is the length of the normal from the origin.

🎯 Exam Tip: The normal form of a plane \( (lx + my + nz = p) \) requires \( p \) to be positive, representing the distance from the origin. If the constant term on the right side is initially negative, multiply the entire equation by -1 before dividing by the magnitude of the normal vector.

Question 10. For the following planes, find the direction ratios and the direction cosines of the normal from the origin.
(i) \( 3x + 2y - 6z + 22 = 0 \)
(ii) \( 4x - 3y + 5z - 30 = 0 \)
(iii) \( 5y - 12z + 20 = 0 \)
Answer:
(i) Given equation of the plane is \( 3x + 2y - 6z + 22 = 0 \).
Rewrite it as \( 3x + 2y - 6z = -22 \).
To make the constant term positive, multiply by -1:
\( -3x - 2y + 6z = 22 \).
The direction ratios of the normal are \( <-3, -2, 6> \).
Now, find the magnitude of this normal vector: \( \sqrt{(-3)^2 + (-2)^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 \).
Divide the direction ratios by the magnitude to get the direction cosines:
\( \left< \frac{-3}{7}, \frac{-2}{7}, \frac{6}{7} \right> \).

(ii) Given equation of the plane is \( 4x - 3y + 5z - 30 = 0 \).
Rewrite it as \( 4x - 3y + 5z = 30 \).
The constant term is already positive.
The direction ratios of the normal are \( <4, -3, 5> \).
Now, find the magnitude of this normal vector: \( \sqrt{4^2 + (-3)^2 + 5^2} = \sqrt{16 + 9 + 25} = \sqrt{50} = 5\sqrt{2} \).
Divide the direction ratios by the magnitude to get the direction cosines:
\( \left< \frac{4}{5\sqrt{2}}, \frac{-3}{5\sqrt{2}}, \frac{5}{5\sqrt{2}} \right> \)
\( \implies \left< \frac{4}{5\sqrt{2}}, \frac{-3}{5\sqrt{2}}, \frac{1}{\sqrt{2}} \right> \).

(iii) Given equation of the plane is \( 5y - 12z + 20 = 0 \).
Rewrite it as \( 5y - 12z = -20 \).
To make the constant term positive, multiply by -1:
\( -5y + 12z = 20 \).
Note that the coefficient of x is 0. So the direction ratios of the normal are \( <0, -5, 12> \).
Now, find the magnitude of this normal vector: \( \sqrt{0^2 + (-5)^2 + 12^2} = \sqrt{0 + 25 + 144} = \sqrt{169} = 13 \).
Divide the direction ratios by the magnitude to get the direction cosines:
\( \left< \frac{0}{13}, \frac{-5}{13}, \frac{12}{13} \right> \)
\( \implies \left< 0, -\frac{5}{13}, \frac{12}{13} \right> \).
In simple words: For each plane equation, first rearrange it so the constant number is by itself on the right side and is positive. The numbers attached to x, y, and z are the "direction ratios" of the line that is normal (perpendicular) to the plane. To get the "direction cosines", you need to divide each of these direction ratios by the total length of the normal vector. This length is found by squaring each direction ratio, adding them up, and then taking the square root.

🎯 Exam Tip: The coefficients of x, y, and z in the plane equation \( Ax + By + Cz + D = 0 \) directly give the direction ratios \( \) of the normal. Always ensure the constant term is positive before forming the normal vector for length calculations if you intend to directly use the formula for distance from the origin.

Question 11.
(i) Find the coordinates of the point, where the line \( \frac{x+1}{2} = \frac{y+2}{3} = \frac{z+3}{4} \) meets the plane \( x + y + 4z = 6 \).
(ii) Find the coordinates of the point where the line joining the points (1, -2, 3) and (2, -1, 5) cuts the plane \( x - 2y + 3z = 1 \). Hence, find the distance of this point from the point (5, 4, 1).
(iii) Find the coordinates of the point where the line through the points A(3, 4, 1) and B(5, 1, 6) crosses the XZ plane: Also find the angle which the line makes with the XZ plane.
Answer:
(i) Given equation of the line is \( \frac{x+1}{2} = \frac{y+2}{3} = \frac{z+3}{4} \).
Let this equal \( t \), so \( \frac{x+1}{2} = t \implies x = 2t - 1 \).
\( \frac{y+2}{3} = t \implies y = 3t - 2 \).
\( \frac{z+3}{4} = t \implies z = 4t - 3 \).
So, any point P on the line can be represented as \( (2t - 1, 3t - 2, 4t - 3) \).
The given equation of the plane is \( x + y + 4z = 6 \).
For the point of intersection, point P must lie on the plane. Substitute the coordinates of P into the plane equation:
\( (2t - 1) + (3t - 2) + 4(4t - 3) = 6 \)
\( \implies 2t - 1 + 3t - 2 + 16t - 12 = 6 \)
\( \implies 21t - 15 = 6 \)
\( \implies 21t = 21 \)
\( \implies t = 1 \)
Now, substitute \( t = 1 \) back into the coordinates of P to find the intersection point:
\( x = 2(1) - 1 = 1 \)
\( y = 3(1) - 2 = 1 \)
\( z = 4(1) - 3 = 1 \)
So, the required point of intersection is \( P(1, 1, 1) \).

(ii) First, find the equation of the line joining points \( (1, -2, 3) \) and \( (2, -1, 5) \).
The equation of a line passing through \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) 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 the given points:
\( \frac{x - 1}{2 - 1} = \frac{y - (-2)}{-1 - (-2)} = \frac{z - 3}{5 - 3} \)
\( \implies \frac{x - 1}{1} = \frac{y + 2}{1} = \frac{z - 3}{2} \).
Let this equal \( t \):
\( x = t + 1 \)
\( y = t - 2 \)
\( z = 2t + 3 \)
So, any point P on the line is \( (t + 1, t - 2, 2t + 3) \).
The given equation of the plane is \( x - 2y + 3z = 19 \).
For the point of intersection, P lies on the plane. Substitute the coordinates of P into the plane equation:
\( (t + 1) - 2(t - 2) + 3(2t + 3) = 19 \)
\( \implies t + 1 - 2t + 4 + 6t + 9 = 19 \)
\( \implies 5t + 14 = 19 \)
\( \implies 5t = 5 \)
\( \implies t = 1 \)
Substitute \( t = 1 \) back into the coordinates of P to find the intersection point:
\( x = 1 + 1 = 2 \)
\( y = 1 - 2 = -1 \)
\( z = 2(1) + 3 = 5 \)
So, the point of intersection is \( P(2, -1, 5) \).
Now, find the distance of this point \( P(2, -1, 5) \) from the given point \( Q(5, 4, 1) \).
The distance formula between two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) is \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \).
Distance \( PQ = \sqrt{(5 - 2)^2 + (4 - (-1))^2 + (1 - 5)^2} \)
\( = \sqrt{3^2 + (4 + 1)^2 + (-4)^2} \)
\( = \sqrt{9 + 5^2 + 16} \)
\( = \sqrt{9 + 25 + 16} \)
\( = \sqrt{50} \)
\( = \sqrt{25 \times 2} \)
\( = 5\sqrt{2} \) units.

(iii) First, find the equation of the line passing through points \( A(3, 4, 1) \) and \( B(5, 1, 6) \).
\( \frac{x - 3}{5 - 3} = \frac{y - 4}{1 - 4} = \frac{z - 1}{6 - 1} \)
\( \implies \frac{x - 3}{2} = \frac{y - 4}{-3} = \frac{z - 1}{5} \).
Let this equal \( t \):
\( x = 2t + 3 \)
\( y = -3t + 4 \)
\( z = 5t + 1 \)
So, any point P on the line is \( (2t + 3, -3t + 4, 5t + 1) \).
The XZ plane is defined by the equation \( y = 0 \).
When the line crosses the XZ plane, the y-coordinate of the point of intersection is 0.
So, set the y-coordinate of P to 0:
\( -3t + 4 = 0 \)
\( \implies 3t = 4 \)
\( \implies t = \frac{4}{3} \)
Substitute \( t = \frac{4}{3} \) back into the coordinates of P to find the intersection point:
\( x = 2\left(\frac{4}{3}\right) + 3 = \frac{8}{3} + 3 = \frac{8 + 9}{3} = \frac{17}{3} \)
\( y = -3\left(\frac{4}{3}\right) + 4 = -4 + 4 = 0 \)
\( z = 5\left(\frac{4}{3}\right) + 1 = \frac{20}{3} + 1 = \frac{20 + 3}{3} = \frac{23}{3} \)
The coordinates of the point where the line crosses the XZ plane are \( \left(\frac{17}{3}, 0, \frac{23}{3}\right) \).

Now, find the angle the line makes with the XZ plane.
Let \( \theta \) be the angle between the line and the XZ plane. The XZ plane has a normal vector along the y-axis. The direction ratios of the normal to the XZ plane are \( <0, 1, 0> \).
The direction ratios of the line are \( <2, -3, 5> \).
The angle \( \phi \) between the line (with direction ratios \( \)) and the normal to the plane (with direction ratios \( \)) is given by:
\( \cos \phi = \frac{|l_1l_2 + m_1m_2 + n_1n_2|}{\sqrt{l_1^2 + m_1^2 + n_1^2} \sqrt{l_2^2 + m_2^2 + n_2^2}} \)
Here, line direction ratios \( <2, -3, 5> \) and normal direction ratios \( <0, 1, 0> \).
\( \cos \phi = \frac{|(2)(0) + (-3)(1) + (5)(0)|}{\sqrt{2^2 + (-3)^2 + 5^2} \sqrt{0^2 + 1^2 + 0^2}} \)
\( = \frac{|0 - 3 + 0|}{\sqrt{4 + 9 + 25} \sqrt{1}} \)
\( = \frac{|-3|}{\sqrt{38}} = \frac{3}{\sqrt{38}} \)
The angle \( \theta \) between the line and the plane is related to \( \phi \) by \( \theta = 90^\circ - \phi \) or \( \sin \theta = \cos \phi \).
So, \( \sin \theta = \frac{3}{\sqrt{38}} \).
Therefore, the angle the line makes with the XZ plane is \( \theta = \arcsin\left(\frac{3}{\sqrt{38}}\right) \).
In simple words: (i) To find where a line hits a plane, first write down any point on the line using a variable (like 't'). Then, put these variable coordinates into the plane's equation. Solve for 't' and then put 't' back into the point's coordinates to get the exact spot. (ii) For a line joining two points, first find its equation using those points. Then, do the same steps as in (i) to find the intersection point. After that, use the distance formula to find how far this intersection point is from another given point. (iii) To find where a line crosses the XZ plane, set the y-coordinate of any point on the line to zero (because on the XZ plane, y is always zero). Solve for 't' and find the intersection point. To find the angle the line makes with the XZ plane, find the direction numbers of the line and the direction numbers of the XZ plane's "normal" (which is the y-axis, or <0,1,0>). Then use the sine formula: \( \sin \theta = \frac{|l_1l_2 + m_1m_2 + n_1n_2|}{\text{magnitude of line} \times \text{magnitude of normal}} \).

🎯 Exam Tip: When finding the intersection of a line and a plane, parameterize the line (e.g., using 't') and substitute the parametric equations into the plane's equation. Solving for 't' gives the specific point of intersection. For angle with a plane, remember to use the sine of the angle between the line and the plane, which is equal to the cosine of the angle between the line and the plane's normal.

Question 12. Foot of the perpendicular from the origin to the plane is (2, 3, 4). Find the equation of the Plane.
Answer:
The foot of the perpendicular from the origin \( O(0,0,0) \) to the plane is the point \( N(2, 3, 4) \).
The line segment ON (connecting the origin to N) is the normal to the plane.
The direction ratios of the normal to the plane are given by the coordinates of N (since O is the origin): \( <2-0, 3-0, 4-0> = <2, 3, 4> \).
The plane passes through the point N(2, 3, 4).
The equation of a plane passing through a point \( (x_1, y_1, z_1) \) with normal direction ratios \( \) is given by:
\( a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \).
Here, \( (x_1, y_1, z_1) = (2, 3, 4) \) and \( = <2, 3, 4> \).
Substitute these values:
\( 2(x - 2) + 3(y - 3) + 4(z - 4) = 0 \)
\( \implies 2x - 4 + 3y - 9 + 4z - 16 = 0 \)
\( \implies 2x + 3y + 4z - 29 = 0 \)
This is the required equation of the plane. The plane is always perpendicular to the line from the origin to the foot of the perpendicular.
In simple words: The line from the center (origin) to the "foot of the perpendicular" on the plane is the normal line to that plane. This means the coordinates of the foot of the perpendicular are the direction numbers for the plane's normal. Since the plane also passes through this "foot" point, you can use the formula for a plane passing through a point with known normal direction to find its equation.

🎯 Exam Tip: The line connecting the origin to the foot of the perpendicular on a plane is always normal to that plane. Hence, the coordinates of the foot directly provide the direction ratios of the normal, and the foot itself is a point on the plane.

Question 13. Find the distance of the point (2, 3, 4) from the plane \( 3x + 2y + 2z + 5 = 0 \), measured parallel to the line \( \frac{x+3}{3} = \frac{y-2}{6} = \frac{z}{2} \).
Answer:
Let the given point be \( A(2, 3, 4) \).
The equation of the line is \( \frac{x+3}{3} = \frac{y-2}{6} = \frac{z}{2} \).
We need to find the distance from point A to the plane \( 3x + 2y + 2z + 5 = 0 \), measured parallel to this line.
First, write the equation of a line passing through point A(2, 3, 4) and parallel to the given line. Since it's parallel, it will have the same direction ratios \( <3, 6, 2> \).
The equation of the line passing through A(2, 3, 4) parallel to the given line is:
\( \frac{x - 2}{3} = \frac{y - 3}{6} = \frac{z - 4}{2} \).
Let this equal \( t \). So, any point P on this line can be represented as:
\( x = 3t + 2 \)
\( y = 6t + 3 \)
\( z = 2t + 4 \)

This point P is the intersection of the line from A parallel to the given line and the plane.
Substitute the coordinates of P into the plane equation \( 3x + 2y + 2z + 5 = 0 \):
\( 3(3t + 2) + 2(6t + 3) + 2(2t + 4) + 5 = 0 \)
\( \implies 9t + 6 + 12t + 6 + 4t + 8 + 5 = 0 \)
\( \implies 25t + 25 = 0 \)
\( \implies 25t = -25 \)
\( \implies t = -1 \)
Now, substitute \( t = -1 \) back into the coordinates of P to find the point of intersection on the plane:
\( x = 3(-1) + 2 = -3 + 2 = -1 \)
\( y = 6(-1) + 3 = -6 + 3 = -3 \)
\( z = 2(-1) + 4 = -2 + 4 = 2 \)
So, the point of intersection P is \( (-1, -3, 2) \).
The required distance is the distance between point A\( (2, 3, 4) \) and point P\( (-1, -3, 2) \).
Using the distance formula \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \):
Distance \( AP = \sqrt{(-1 - 2)^2 + (-3 - 3)^2 + (2 - 4)^2} \)
\( = \sqrt{(-3)^2 + (-6)^2 + (-2)^2} \)
\( = \sqrt{9 + 36 + 4} \)
\( = \sqrt{49} \)
\( = 7 \) units.
In simple words: First, find the equation of a new line that starts from the given point and runs parallel to the given line. Next, find where this new line hits the plane; this is your point P. Finally, calculate the distance between the original point and this new point P. This distance is the answer.

🎯 Exam Tip: When measuring distance parallel to a line, always construct a new line passing through the given point with the same direction ratios as the parallel line. Then, find the intersection of this new line with the plane and calculate the distance between the two points.

Question 14. Find the equation in Cartesian form of the plane passing through the point (3, -3, 1) and normal to the line joining the points (3, 4, -1) and (2, -1, 5).
Answer:
The plane passes through the point \( (3, -3, 1) \). Let this be \( (x_1, y_1, z_1) \).
The plane is normal (perpendicular) to the line joining points \( P(3, 4, -1) \) and \( Q(2, -1, 5) \).
The direction ratios of this line PQ will be the direction ratios of the normal to the plane. Let these be \( \).
\( a = x_2 - x_1 = 2 - 3 = -1 \)
\( b = y_2 - y_1 = -1 - 4 = -5 \)
\( c = z_2 - z_1 = 5 - (-1) = 5 + 1 = 6 \)
So, the direction ratios of the normal to the plane are \( <-1, -5, 6> \).
The equation of a plane passing through a point \( (x_1, y_1, z_1) \) with normal direction ratios \( \) is given by:
\( a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \).
Substitute the point \( (3, -3, 1) \) and normal direction ratios \( <-1, -5, 6> \):
\( -1(x - 3) + (-5)(y - (-3)) + 6(z - 1) = 0 \)
\( \implies -1(x - 3) - 5(y + 3) + 6(z - 1) = 0 \)
\( \implies -x + 3 - 5y - 15 + 6z - 6 = 0 \)
\( \implies -x - 5y + 6z - 18 = 0 \)
To make the leading coefficient positive, multiply by -1:
\( \implies x + 5y - 6z + 18 = 0 \)
This is the required equation of the plane in Cartesian form.
In simple words: First, find the direction numbers of the line that is "normal" (at 90 degrees) to the plane. You do this by subtracting the coordinates of the two points given for that line. These direction numbers become the 'a', 'b', and 'c' in your plane equation. Then, use the given point that the plane passes through and these 'a', 'b', 'c' values in the plane formula \( a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \) to get the final equation.

🎯 Exam Tip: The line normal to a plane directly provides the direction ratios \( \) for the plane's equation \( A(x-x_1) + B(y-y_1) + C(z-z_1) = 0 \). Ensure proper calculation of direction ratios from the two given points, and careful substitution of the passing point's coordinates.

Question 15. Find the image of the point
(i) (3, -2, 1) in the plane \( 3x - y + 4z = 2 \)
(ii) (0,0,0) in the plane \( 3x + 4y - 6z + 1 = 0 \)
Answer:
(i) Let the given point be \( P(3, -2, 1) \). Let the image of P in the plane be \( Q(x', y', z') \).
The given plane is \( 3x - y + 4z = 2 \) (Equation 1).
The line PM (where M is the foot of the perpendicular from P to the plane) is normal to the plane.
The direction ratios of the normal to the plane are \( <3, -1, 4> \).
The equation of the line PM passing through \( P(3, -2, 1) \) with direction ratios \( <3, -1, 4> \) is:
\( \frac{x - 3}{3} = \frac{y - (-2)}{-1} = \frac{z - 1}{4} \).
Let this equal \( t \). So any point M on the line PM can be written as:
\( x = 3t + 3 \)
\( y = -t - 2 \)
\( z = 4t + 1 \)
Since M is the foot of the perpendicular, it lies on the plane. Substitute the coordinates of M into the plane equation:
\( 3(3t + 3) - (-t - 2) + 4(4t + 1) = 2 \)
\( \implies 9t + 9 + t + 2 + 16t + 4 = 2 \)
\( \implies 26t + 15 = 2 \)
\( \implies 26t = -13 \)
\( \implies t = -\frac{13}{26} = -\frac{1}{2} \)
Now, find the coordinates of M by substituting \( t = -\frac{1}{2} \):
\( x_M = 3\left(-\frac{1}{2}\right) + 3 = -\frac{3}{2} + 3 = \frac{3}{2} \)
\( y_M = -\left(-\frac{1}{2}\right) - 2 = \frac{1}{2} - 2 = -\frac{3}{2} \)
\( z_M = 4\left(-\frac{1}{2}\right) + 1 = -2 + 1 = -1 \)
So, the foot of the perpendicular is \( M\left(\frac{3}{2}, -\frac{3}{2}, -1\right) \).
M is the midpoint of the line segment PQ, where Q is the image of P. Let \( Q = (x', y', z') \).
Using the midpoint formula:
\( \frac{3 + x'}{2} = \frac{3}{2} \implies 3 + x' = 3 \implies x' = 0 \)
\( \frac{-2 + y'}{2} = -\frac{3}{2} \implies -2 + y' = -3 \implies y' = -1 \)
\( \frac{1 + z'}{2} = -1 \implies 1 + z' = -2 \implies z' = -3 \)
So, the image of the point \( (3, -2, 1) \) in the plane is \( Q(0, -1, -3) \).

(ii) Let the given point be the origin \( O(0, 0, 0) \). Let the image of O in the plane be \( Q(x', y', z') \).
The given plane is \( 3x + 4y - 6z + 1 = 0 \) (Equation 1).
The line OM (where M is the foot of the perpendicular from O to the plane) is normal to the plane.
The direction ratios of the normal to the plane are \( <3, 4, -6> \).
The equation of the line OM passing through \( O(0, 0, 0) \) with direction ratios \( <3, 4, -6> \) is:
\( \frac{x - 0}{3} = \frac{y - 0}{4} = \frac{z - 0}{-6} \).
Let this equal \( t \). So any point M on the line OM can be written as:
\( x = 3t \)
\( y = 4t \)
\( z = -6t \)
Since M is the foot of the perpendicular, it lies on the plane. Substitute the coordinates of M into the plane equation:
\( 3(3t) + 4(4t) - 6(-6t) + 1 = 0 \)
\( \implies 9t + 16t + 36t + 1 = 0 \)
\( \implies 61t + 1 = 0 \)
\( \implies 61t = -1 \)
\( \implies t = -\frac{1}{61} \)
Now, find the coordinates of M by substituting \( t = -\frac{1}{61} \):
\( x_M = 3\left(-\frac{1}{61}\right) = -\frac{3}{61} \)
\( y_M = 4\left(-\frac{1}{61}\right) = -\frac{4}{61} \)
\( z_M = -6\left(-\frac{1}{61}\right) = \frac{6}{61} \)
So, the foot of the perpendicular is \( M\left(-\frac{3}{61}, -\frac{4}{61}, \frac{6}{61}\right) \).
M is the midpoint of the line segment OQ, where Q is the image of O. Let \( Q = (x', y', z') \).
Using the midpoint formula:
\( \frac{0 + x'}{2} = -\frac{3}{61} \implies x' = -\frac{6}{61} \)
\( \frac{0 + y'}{2} = -\frac{4}{61} \implies y' = -\frac{8}{61} \)
\( \frac{0 + z'}{2} = \frac{6}{61} \implies z' = \frac{12}{61} \)
So, the image of the origin \( (0, 0, 0) \) in the plane is \( Q\left(-\frac{6}{61}, -\frac{8}{61}, \frac{12}{61}\right) \).
In simple words: To find a point's "image" in a plane, imagine the plane acts like a mirror. First, find the line that goes straight from the point and hits the plane at 90 degrees. Any point on this line can be written using a variable 't'. Find where this line hits the plane; this is the "foot of the perpendicular." This foot is the middle point between your original point and its image. Use the midpoint formula to find the coordinates of the image.

🎯 Exam Tip: The foot of the perpendicular from a point to a plane is the midpoint of the line segment connecting the original point and its image. This relationship is crucial for finding the image coordinates after determining the foot of the perpendicular.

Question 15. Find the image of the point
(i) (3, -2, 1) in the plane 3x-y+4z= 2
(ii) (0,0,0) in the plane 3x+4y-6z+1=0
Answer:
(i) Let the given point be \( P(3, -2, 1) \) and the plane be \( 3x - y + 4z = 2 \).
We consider a line PM that passes through \( P \) and is perpendicular to the plane.
The direction ratios of the normal to the plane are \( (3, -1, 4) \).
So, the equation of the line PM is \( \frac{x-3}{3} = \frac{y-(-2)}{-1} = \frac{z-1}{4} = t \) (say).
Any point M on this line can be written as \( M(3t+3, -t-2, 4t+1) \). This point M is the foot of the perpendicular.
Since M lies on the plane \( 3x - y + 4z = 2 \), we substitute its coordinates into the plane equation:
\( 3(3t+3) - (-t-2) + 4(4t+1) = 2 \)
\( 9t + 9 + t + 2 + 16t + 4 = 2 \)
\( 26t + 15 = 2 \)
\( 26t = -13 \)
\( t = -\frac{13}{26} = -\frac{1}{2} \)
Now, we find the coordinates of M by substituting \( t = -\frac{1}{2} \):
\( M \left( 3\left(-\frac{1}{2}\right)+3, -\left(-\frac{1}{2}\right)-2, 4\left(-\frac{1}{2}\right)+1 \right) \)
\( M \left( -\frac{3}{2}+3, \frac{1}{2}-2, -2+1 \right) \)
\( M \left( \frac{-3+6}{2}, \frac{1-4}{2}, -1 \right) \)
\( M \left( \frac{3}{2}, -\frac{3}{2}, -1 \right) \)
Let \( Q(x', y', z') \) be the image of point P. M is the midpoint of the line segment PQ.
\( \frac{3}{2} = \frac{3+x'}{2} \implies 3 = 3+x' \implies x' = 0 \)
\( -\frac{3}{2} = \frac{-2+y'}{2} \implies -3 = -2+y' \implies y' = -1 \)
\( -1 = \frac{1+z'}{2} \implies -2 = 1+z' \implies z' = -3 \)
So, the image of the point \( (3, -2, 1) \) in the plane is \( (0, -1, -3) \). When finding an image, we essentially reflect the point across the plane, with the foot of the perpendicular being the exact halfway point.
(ii) Let the given point be \( O(0, 0, 0) \) and the plane be \( 3x + 4y - 6z + 1 = 0 \).
We consider a line OM that passes through \( O \) and is perpendicular to the plane.
The direction ratios of the normal to the plane are \( (3, 4, -6) \).
So, the equation of the line OM is \( \frac{x-0}{3} = \frac{y-0}{4} = \frac{z-0}{-6} = t \) (say).
Any point M on this line can be written as \( M(3t, 4t, -6t) \). This M is the foot of the perpendicular from the origin.
Since M lies on the plane \( 3x + 4y - 6z + 1 = 0 \), we substitute its coordinates:
\( 3(3t) + 4(4t) - 6(-6t) + 1 = 0 \)
\( 9t + 16t + 36t + 1 = 0 \)
\( 61t + 1 = 0 \)
\( t = -\frac{1}{61} \)
Now, we find the coordinates of M:
\( M \left( 3\left(-\frac{1}{61}\right), 4\left(-\frac{1}{61}\right), -6\left(-\frac{1}{61}\right) \right) \)
\( M \left( -\frac{3}{61}, -\frac{4}{61}, \frac{6}{61} \right) \)
Let \( Q(x', y', z') \) be the image of the origin \( O(0, 0, 0) \). M is the midpoint of OQ.
\( -\frac{3}{61} = \frac{0+x'}{2} \implies x' = -\frac{6}{61} \)
\( -\frac{4}{61} = \frac{0+y'}{2} \implies y' = -\frac{8}{61} \)
\( \frac{6}{61} = \frac{0+z'}{2} \implies z' = \frac{12}{61} \)
So, the image of the point \( (0,0,0) \) in the plane is \( \left( -\frac{6}{61}, -\frac{8}{61}, \frac{12}{61} \right) \). This reflection helps in understanding how points behave relative to a given plane.
In simple words: To find the image of a point in a plane, imagine the plane is a mirror. First, find the point on the mirror (the plane) that is directly below or above your original point. This is called the 'foot of the perpendicular'. Then, move the same distance from this mirror point in the same direction, and you will find the reflected image.

🎯 Exam Tip: When finding the image of a point in a plane, remember that the foot of the perpendicular from the point to the plane is the midpoint between the original point and its image. Make sure to solve for 't' accurately.

Question 16. Find the reflection of the point (1, 2, -1) in the plane 3x – 5y + 4z = 5.
Answer:
Let the given point be \( P(1, 2, -1) \) and the plane be \( 3x - 5y + 4z = 5 \).
We need to find the image of P in this plane.
First, find the equation of the line PM passing through P and perpendicular to the plane.
The direction ratios of the normal to the plane are \( (3, -5, 4) \).
So, the equation of the line PM is \( \frac{x-1}{3} = \frac{y-2}{-5} = \frac{z-(-1)}{4} = t \) (say).
Any point M on this line can be written as \( M(3t+1, -5t+2, 4t-1) \). This point M is the foot of the perpendicular.
Since M lies on the plane \( 3x - 5y + 4z = 5 \), we substitute its coordinates:
\( 3(3t+1) - 5(-5t+2) + 4(4t-1) = 5 \)
\( 9t + 3 + 25t - 10 + 16t - 4 = 5 \)
\( 50t - 11 = 5 \)
\( 50t = 16 \)
\( t = \frac{16}{50} = \frac{8}{25} \)
Now, we find the coordinates of M by substituting \( t = \frac{8}{25} \):
\( M \left( 3\left(\frac{8}{25}\right)+1, -5\left(\frac{8}{25}\right)+2, 4\left(\frac{8}{25}\right)-1 \right) \)
\( M \left( \frac{24}{25}+1, -\frac{40}{25}+2, \frac{32}{25}-1 \right) \)
\( M \left( \frac{24+25}{25}, \frac{-40+50}{25}, \frac{32-25}{25} \right) \)
\( M \left( \frac{49}{25}, \frac{10}{25}, \frac{7}{25} \right) = M \left( \frac{49}{25}, \frac{2}{5}, \frac{7}{25} \right) \)
Let \( Q(x', y', z') \) be the reflection (image) of point P. M is the midpoint of the line segment PQ.
\( \frac{49}{25} = \frac{1+x'}{2} \implies 98 = 25(1+x') \implies 98 = 25+25x' \implies 25x' = 73 \implies x' = \frac{73}{25} \)
\( \frac{2}{5} = \frac{2+y'}{2} \implies 4 = 5(2+y') \implies 4 = 10+5y' \implies 5y' = -6 \implies y' = -\frac{6}{5} \)
\( \frac{7}{25} = \frac{-1+z'}{2} \implies 14 = 25(-1+z') \implies 14 = -25+25z' \implies 25z' = 39 \implies z' = \frac{39}{25} \)
Thus, the reflection of the point \( (1, 2, -1) \) in the plane is \( \left( \frac{73}{25}, -\frac{6}{5}, \frac{39}{25} \right) \). The process of reflection involves finding a symmetrical point across the given plane.
In simple words: To find the reflected point, imagine the plane is a perfect mirror. First, draw a line from your point straight to the mirror, touching it at one spot. This spot is the middle point between your original point and its reflection. Then, extend the line from the mirror spot, moving the same distance away. That new spot is the reflection.

🎯 Exam Tip: Be careful with signs when calculating coordinates and ensure you substitute the 't' value back correctly into the point formula. Double-check your midpoint calculations.

Question 17. From the point P(1, 2, 4), a perpendicular is drawn on the plane \( 2x + y – 2z + 3 = 0 \). Find the equation, the length and the coordinates of the foot of the perpendicular.
Answer:
Let the given point be \( P(1, 2, 4) \) and the plane be \( 2x + y - 2z + 3 = 0 \).
First, we find the equation of the line PM, which is perpendicular to the plane and passes through P.
The direction ratios of the normal to the plane are \( (2, 1, -2) \).
So, the equation of the perpendicular line PM is \( \frac{x-1}{2} = \frac{y-2}{1} = \frac{z-4}{-2} = t \) (say). This equation describes all points on the line that is normal to the plane.
Any point M on this line can be written as \( M(2t+1, t+2, -2t+4) \). This point M is the foot of the perpendicular.
Since M lies on the plane \( 2x + y - 2z + 3 = 0 \), we substitute its coordinates:
\( 2(2t+1) + (t+2) - 2(-2t+4) + 3 = 0 \)
\( 4t + 2 + t + 2 + 4t - 8 + 3 = 0 \)
\( 9t - 1 = 0 \)
\( 9t = 1 \)
\( t = \frac{1}{9} \)
Now, we find the coordinates of the foot of the perpendicular M by substituting \( t = \frac{1}{9} \):
\( M \left( 2\left(\frac{1}{9}\right)+1, \frac{1}{9}+2, -2\left(\frac{1}{9}\right)+4 \right) \)
\( M \left( \frac{2}{9}+1, \frac{1}{9}+2, -\frac{2}{9}+4 \right) \)
\( M \left( \frac{2+9}{9}, \frac{1+18}{9}, \frac{-2+36}{9} \right) \)
\( M \left( \frac{11}{9}, \frac{19}{9}, \frac{34}{9} \right) \). These coordinates show the exact point where the perpendicular meets the plane.
Finally, we calculate the length of the perpendicular PM, which is the distance between \( P(1, 2, 4) \) and \( M \left( \frac{11}{9}, \frac{19}{9}, \frac{34}{9} \right) \).
\( PM = \sqrt{\left(\frac{11}{9}-1\right)^2 + \left(\frac{19}{9}-2\right)^2 + \left(\frac{34}{9}-4\right)^2} \)
\( PM = \sqrt{\left(\frac{11-9}{9}\right)^2 + \left(\frac{19-18}{9}\right)^2 + \left(\frac{34-36}{9}\right)^2} \)
\( PM = \sqrt{\left(\frac{2}{9}\right)^2 + \left(\frac{1}{9}\right)^2 + \left(-\frac{2}{9}\right)^2} \)
\( PM = \sqrt{\frac{4}{81} + \frac{1}{81} + \frac{4}{81}} = \sqrt{\frac{9}{81}} = \sqrt{\frac{1}{9}} = \frac{1}{3} \) unit.
In simple words: We find a line from the given point that goes straight to the plane. The equation of this line helps us find where it touches the plane (the foot). Then, we measure how long this line is, which is the shortest distance from the point to the plane.

🎯 Exam Tip: When calculating the length of the perpendicular, simplify the fractions inside the square root before squaring to avoid larger numbers and potential errors.

Question 18. Find the distance of the point (1, -2, 3) from the plane \( x – y + z = 5 \) measured along a line parallel to \( \frac{x}{2} = \frac{y}{3} = \frac{z}{-6} \).
Answer:
Let the given point be \( A(1, -2, 3) \) and the plane be \( x - y + z = 5 \).
We need to find the distance along a line parallel to \( \frac{x}{2} = \frac{y}{3} = \frac{z}{-6} \).
First, write the equation of the line passing through point A and parallel to the given line:
\( \frac{x-1}{2} = \frac{y-(-2)}{3} = \frac{z-3}{-6} = t \) (say). This line represents the path along which the distance is measured.
Any point P on this line can be written as \( P(2t+1, 3t-2, -6t+3) \).
Since this point P also lies on the plane \( x - y + z = 5 \), we substitute its coordinates into the plane equation:
\( (2t+1) - (3t-2) + (-6t+3) = 5 \)
\( 2t + 1 - 3t + 2 - 6t + 3 = 5 \)
\( -7t + 6 = 5 \)
\( -7t = -1 \)
\( t = \frac{1}{7} \)
Now, we find the coordinates of point P by substituting \( t = \frac{1}{7} \):
\( P \left( 2\left(\frac{1}{7}\right)+1, 3\left(\frac{1}{7}\right)-2, -6\left(\frac{1}{7}\right)+3 \right) \)
\( P \left( \frac{2}{7}+1, \frac{3}{7}-2, -\frac{6}{7}+3 \right) \)
\( P \left( \frac{2+7}{7}, \frac{3-14}{7}, \frac{-6+21}{7} \right) \)
\( P \left( \frac{9}{7}, -\frac{11}{7}, \frac{15}{7} \right) \). This is the point where the line intersects the plane.
The required distance is the distance between point A(1, -2, 3) and point P \( \left( \frac{9}{7}, -\frac{11}{7}, \frac{15}{7} \right) \).
\( AP = \sqrt{\left(\frac{9}{7}-1\right)^2 + \left(-\frac{11}{7}-(-2)\right)^2 + \left(\frac{15}{7}-3\right)^2} \)
\( AP = \sqrt{\left(\frac{9-7}{7}\right)^2 + \left(\frac{-11+14}{7}\right)^2 + \left(\frac{15-21}{7}\right)^2} \)
\( AP = \sqrt{\left(\frac{2}{7}\right)^2 + \left(\frac{3}{7}\right)^2 + \left(-\frac{6}{7}\right)^2} \)
\( AP = \sqrt{\frac{4}{49} + \frac{9}{49} + \frac{36}{49}} = \sqrt{\frac{49}{49}} = \sqrt{1} = 1 \) unit.
This distance represents the length along the specified line from the point to the plane.
In simple words: We are looking for the distance from a point to a plane, but not the shortest distance. Instead, we want the distance measured along a specific direction. So, we draw a line from the point in that direction until it hits the plane, and then we measure the length of that line.

🎯 Exam Tip: Distinguish between finding the shortest distance (perpendicular) and distance along a parallel line. For parallel lines, the direction vector of the line is used directly in the parametric equation.

Question 19. Prove that a variable plane which moves so that the sum of reciprocals of its intercepts on the three coordinate axes is constant, passes through a fixed point.
Answer:
Let the equation of the variable plane in intercept form be \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \) (1).
Here, \( a, b, c \) are the intercepts the plane makes on the x, y, and z axes, respectively.
We are given that the sum of the reciprocals of its intercepts is constant. Let this constant be \( k \).
So, \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = k \) (2).
We can rewrite equation (2) by dividing both sides by \( k \):
\( \frac{1}{k} \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) = \frac{k}{k} \)
\( \frac{1}{ka} + \frac{1}{kb} + \frac{1}{kc} = 1 \)
This can also be written as: \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \) where \( x = \frac{1}{k}, y = \frac{1}{k}, z = \frac{1}{k} \).
Comparing this with equation (1), if we substitute \( x = \frac{1}{k}, y = \frac{1}{k}, z = \frac{1}{k} \), then the plane equation becomes \( \frac{1/k}{a} + \frac{1/k}{b} + \frac{1/k}{c} = 1 \), which simplifies to \( \frac{1}{k} \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) = 1 \).
From our given condition (2), we know \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = k \).
Substituting this into the equation, we get \( \frac{1}{k} (k) = 1 \), which means \( 1 = 1 \).
This shows that the point \( \left( \frac{1}{k}, \frac{1}{k}, \frac{1}{k} \right) \) always satisfies the equation of the plane (1), regardless of the specific values of \( a, b, c \) (as long as they maintain the sum of reciprocals as \( k \)).
Since \( k \) is a constant, the point \( \left( \frac{1}{k}, \frac{1}{k}, \frac{1}{k} \right) \) is a fixed point. Therefore, the variable plane always passes through this fixed point.
In simple words: Imagine a plane that moves but always keeps a special rule: if you take 1 divided by where it cuts each axis and add them up, the total is always the same number. We can show that because of this rule, the plane must always pass through one specific point in space, no matter how it moves.

🎯 Exam Tip: To prove a plane passes through a fixed point, you often need to manipulate the given condition to match the intercept form of the plane equation, showing that a constant coordinate triplet always satisfies it.

Question 20.
(i) A variable plane is at a constant distance p from the origin and meets the axes in A, B, C. Show that the locus of the centroid of triangle A B C is \( x^{-2} + y^{-2} + z^{-2} = 9 p^{-2} \).
(ii) A variable plane, which remains at a constant distance of 9 units from the origin, cuts the coordinates axes at the points A, B and C. Show that the locus of the centroid of \( \triangle ABC \) is \( \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{1}{9} \).
Answer:
(i) Let the equation of the variable plane in intercept form be \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \) (1).
The plane meets the coordinate axes at \( A(a,0,0) \), \( B(0,b,0) \), and \( C(0,0,c) \).
The distance \( p \) from the origin \( (0,0,0) \) to the plane \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} - 1 = 0 \) is given by the formula:
\( p = \frac{| \frac{0}{a} + \frac{0}{b} + \frac{0}{c} - 1 |}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} \)
Squaring both sides, we get: \( p^2 = \frac{1}{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}} \)
This implies \( \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{p^2} \) (2). This equation relates the intercepts to the distance from the origin.
The centroid \( G \) of the triangle formed by the intercepts \( A, B, C \) is given by:
\( G \left( \frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3} \right) = \left( \frac{a}{3}, \frac{b}{3}, \frac{c}{3} \right) \)
Let the coordinates of the centroid be \( (x_G, y_G, z_G) \). So, \( x_G = \frac{a}{3}, y_G = \frac{b}{3}, z_G = \frac{c}{3} \).
From these relations, we can express \( a, b, c \) in terms of the centroid coordinates: \( a = 3x_G, b = 3y_G, c = 3z_G \).
Substitute these expressions for \( a, b, c \) into equation (2):
\( \frac{1}{(3x_G)^2} + \frac{1}{(3y_G)^2} + \frac{1}{(3z_G)^2} = \frac{1}{p^2} \)
\( \frac{1}{9x_G^2} + \frac{1}{9y_G^2} + \frac{1}{9z_G^2} = \frac{1}{p^2} \)
Multiplying the entire equation by 9, we get:
\( \frac{1}{x_G^2} + \frac{1}{y_G^2} + \frac{1}{z_G^2} = \frac{9}{p^2} \)
Replacing \( x_G, y_G, z_G \) with \( x, y, z \) to represent the locus, we have:
\( x^{-2} + y^{-2} + z^{-2} = 9 p^{-2} \). This equation defines the path of the centroid.
(ii) For this part, the variable plane remains at a constant distance of 9 units from the origin, meaning \( p=9 \).
We use the result from part (i) for the locus of the centroid of \( \triangle ABC \):
\( \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{9}{p^2} \)
Substitute \( p=9 \) into this equation:
\( \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{9}{9^2} \)
\( \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{9}{81} \)
\( \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{1}{9} \). This demonstrates the specific locus when the distance is 9 units.
In simple words: When a plane is always the same distance from the center point (origin) and cuts the axes, the midpoint of the triangle formed by these cuts (the centroid) follows a specific path. We showed that this path is described by a formula involving the reciprocals of its coordinates squared. If the distance from the origin is exactly 9, this formula simplifies further.

🎯 Exam Tip: Remember the formula for the distance of a plane from the origin and the coordinates of the centroid of a triangle formed by intercepts. These are key for solving locus problems involving planes.

Question 21. Find the equation of the plane passing through the following points :
(i) A(2, 2, -1), B(3, 4, 2), C(7, 0, 6)
(ii) A(2, 1, 0), B(3, -2, -2), C(3, 1, 7).
Answer:
(i) To find the equation of the plane passing through \( A(2, 2, -1) \), \( B(3, 4, 2) \), and \( C(7, 0, 6) \).
Let the equation of the plane passing through \( A(2, 2, -1) \) be \( a(x-2) + b(y-2) + c(z-(-1)) = 0 \)
\( a(x-2) + b(y-2) + c(z+1) = 0 \) (1).
Since the plane also passes through \( B(3, 4, 2) \), we substitute its coordinates into (1):
\( a(3-2) + b(4-2) + c(2+1) = 0 \)
\( a(1) + b(2) + c(3) = 0 \)
\( a + 2b + 3c = 0 \) (2).
Since the plane also passes through \( C(7, 0, 6) \), we substitute its coordinates into (1):
\( a(7-2) + b(0-2) + c(6+1) = 0 \)
\( a(5) + b(-2) + c(7) = 0 \)
\( 5a - 2b + 7c = 0 \) (3).
Now, we solve equations (2) and (3) for \( a, b, c \) using the cross-multiplication method:
\( \frac{a}{(2)(7) - (3)(-2)} = \frac{b}{(3)(5) - (1)(7)} = \frac{c}{(1)(-2) - (2)(5)} \)
\( \frac{a}{14 - (-6)} = \frac{b}{15 - 7} = \frac{c}{-2 - 10} \)
\( \frac{a}{20} = \frac{b}{8} = \frac{c}{-12} = k \) (say, where \( k \) is a non-zero constant).
So, \( a = 20k, b = 8k, c = -12k \). These are the direction ratios of the normal to the plane.
Substitute these values of \( a, b, c \) back into equation (1):
\( 20k(x-2) + 8k(y-2) - 12k(z+1) = 0 \)
Divide the entire equation by \( 4k \) (assuming \( k \ne 0 \)):
\( 5(x-2) + 2(y-2) - 3(z+1) = 0 \)
\( 5x - 10 + 2y - 4 - 3z - 3 = 0 \)
\( 5x + 2y - 3z - 17 = 0 \). This is the equation of the plane passing through the three given points.
(ii) To find the equation of the plane passing through \( A(2, 1, 0) \), \( B(3, -2, -2) \), and \( C(3, 1, 7) \).
Let the equation of the plane passing through \( A(2, 1, 0) \) be \( a(x-2) + b(y-1) + c(z-0) = 0 \)
\( a(x-2) + b(y-1) + cz = 0 \) (1).
Since the plane also passes through \( B(3, -2, -2) \), we substitute its coordinates into (1):
\( a(3-2) + b(-2-1) + c(-2) = 0 \)
\( a(1) + b(-3) + c(-2) = 0 \)
\( a - 3b - 2c = 0 \) (2).
Since the plane also passes through \( C(3, 1, 7) \), we substitute its coordinates into (1):
\( a(3-2) + b(1-1) + c(7) = 0 \)
\( a(1) + b(0) + c(7) = 0 \)
\( a + 7c = 0 \) (3).
Now, we solve equations (2) and (3) for \( a, b, c \) using the cross-multiplication method:
\( \frac{a}{(-3)(7) - (-2)(0)} = \frac{b}{(-2)(1) - (1)(7)} = \frac{c}{(1)(0) - (-3)(1)} \)
\( \frac{a}{-21 - 0} = \frac{b}{-2 - 7} = \frac{c}{0 - (-3)} \)
\( \frac{a}{-21} = \frac{b}{-9} = \frac{c}{3} = k \) (say, where \( k \) is a non-zero constant).
So, \( a = -21k, b = -9k, c = 3k \). These represent the direction ratios of the plane's normal.
Substitute these values of \( a, b, c \) back into equation (1):
\( -21k(x-2) - 9k(y-1) + 3k(z) = 0 \)
Divide the entire equation by \( 3k \) (assuming \( k \ne 0 \)):
\( -7(x-2) - 3(y-1) + z = 0 \)
\( -7x + 14 - 3y + 3 + z = 0 \)
\( -7x - 3y + z + 17 = 0 \)
Multiplying by \( -1 \): \( 7x + 3y - z - 17 = 0 \). This is the equation of the plane passing through the three specified points. Finding a plane from three points is like defining a flat surface that touches all three points.
In simple words: To find the equation of a flat surface (plane) that passes through three points, we first write a general equation for a plane using one point. Then, we use the other two points to find the numbers that make the equation true. After some calculations, we get the final equation that defines that specific plane.

🎯 Exam Tip: When finding the equation of a plane through three points, ensure your first equation uses one of the points correctly. Be very careful with cross-multiplication steps and signs, as small errors can lead to incorrect plane equations.

Question 22. Show that the following points are coplanar: (-6, 3, 2), (3, -2, 4), (5, 7, 3) and (-13, 17, -1).
Answer:
Let the given points be \( A(-6, 3, 2) \), \( B(3, -2, 4) \), \( C(5, 7, 3) \), and \( D(-13, 17, -1) \).
To show that these four points are coplanar, we first find the equation of the plane passing through any three of them (say, A, B, and C). Then, we check if the fourth point (D) lies on this plane.
Let the equation of the plane passing through \( A(-6, 3, 2) \) be \( a(x-(-6)) + b(y-3) + c(z-2) = 0 \)
\( a(x+6) + b(y-3) + c(z-2) = 0 \) (1).
Since the plane passes through \( B(3, -2, 4) \), substitute its coordinates into (1):
\( a(3+6) + b(-2-3) + c(4-2) = 0 \)
\( 9a - 5b + 2c = 0 \) (2).
Since the plane passes through \( C(5, 7, 3) \), substitute its coordinates into (1):
\( a(5+6) + b(7-3) + c(3-2) = 0 \)
\( 11a + 4b + c = 0 \) (3).
Now, we solve equations (2) and (3) for \( a, b, c \) using the cross-multiplication method:
\( \frac{a}{(-5)(1) - (2)(4)} = \frac{b}{(2)(11) - (9)(1)} = \frac{c}{(9)(4) - (-5)(11)} \)
\( \frac{a}{-5 - 8} = \frac{b}{22 - 9} = \frac{c}{36 - (-55)} \)
\( \frac{a}{-13} = \frac{b}{13} = \frac{c}{91} = k \) (say).
So, \( a = -13k, b = 13k, c = 91k \).
Substitute these values of \( a, b, c \) back into equation (1):
\( -13k(x+6) + 13k(y-3) + 91k(z-2) = 0 \)
Divide the entire equation by \( 13k \) (assuming \( k \ne 0 \)):
\( -(x+6) + (y-3) + 7(z-2) = 0 \)
\( -x - 6 + y - 3 + 7z - 14 = 0 \)
\( -x + y + 7z - 23 = 0 \)
Multiplying by \( -1 \) to make the x-coefficient positive:
\( x - y - 7z + 23 = 0 \). This is the equation of the plane containing points A, B, and C.
Now, we check if the fourth point \( D(-13, 17, -1) \) lies on this plane. Substitute its coordinates into the plane equation:
\( (-13) - (17) - 7(-1) + 23 = 0 \)
\( -13 - 17 + 7 + 23 = 0 \)
\( -30 + 30 = 0 \)
\( 0 = 0 \).
Since point D satisfies the equation of the plane, it lies on the plane. Therefore, all four given points are coplanar. Coplanar means all points lie on the same flat surface.
In simple words: To prove that four points lie on the same flat surface (plane), we first find the equation for a plane using any three of the points. Then, we check if the fourth point also fits into that plane's equation. If it does, all four points are on the same plane.

🎯 Exam Tip: This method of checking coplanarity is robust. Always verify the fourth point by substituting its coordinates into the derived plane equation; if the equation holds true, the points are coplanar.

Question 23. Show that the line joining the points (0, -1, 2) and (2, -1, -1) is coplanar with the line joining the points (1, 1, 1) and (0, 3, 3).
Answer:
Let the first line \( L_1 \) join points \( A(0, -1, 2) \) and \( B(2, -1, -1) \).
Let the second line \( L_2 \) join points \( C(1, 1, 1) \) and \( D(0, 3, 3) \).
To show that two lines are coplanar, we can find the equation of a plane containing the first line and then check if the second line also lies entirely within this plane. This means both points of the second line must satisfy the plane's equation.
First, find the equation of the plane containing line \( L_1 \) (points A and B).
Let the equation of the plane passing through \( A(0, -1, 2) \) be \( a(x-0) + b(y-(-1)) + c(z-2) = 0 \)
\( ax + b(y+1) + c(z-2) = 0 \) (1).
Since point \( B(2, -1, -1) \) lies on this plane, substitute its coordinates into (1):
\( a(2) + b(-1+1) + c(-1-2) = 0 \)
\( 2a + 0b - 3c = 0 \)
\( 2a - 3c = 0 \) (2).
Now, to define the plane, we need a third point that is not on line \( L_1 \). We can pick one of the points from \( L_2 \), say \( C(1, 1, 1) \), and assume the plane contains it.
Substitute \( C(1, 1, 1) \) into equation (1):
\( a(1) + b(1+1) + c(1-2) = 0 \)
\( a + 2b - c = 0 \) (3).
Now, solve equations (2) and (3) for \( a, b, c \) using the cross-multiplication method:
\( \frac{a}{(0)(-1) - (-3)(2)} = \frac{b}{(-3)(1) - (2)(-1)} = \frac{c}{(2)(2) - (0)(1)} \)
\( \frac{a}{0 - (-6)} = \frac{b}{-3 - (-2)} = \frac{c}{4 - 0} \)
\( \frac{a}{6} = \frac{b}{-1} = \frac{c}{4} = k \) (say).
So, \( a = 6k, b = -k, c = 4k \).
Substitute these values of \( a, b, c \) back into equation (1):
\( 6k(x) - k(y+1) + 4k(z-2) = 0 \)
Divide the entire equation by \( k \) (assuming \( k \ne 0 \)):
\( 6x - (y+1) + 4(z-2) = 0 \)
\( 6x - y - 1 + 4z - 8 = 0 \)
\( 6x - y + 4z - 9 = 0 \) (4). This is the equation of the plane that contains line \( L_1 \) and point C.
Finally, we need to check if the other point of line \( L_2 \), which is \( D(0, 3, 3) \), also lies on this plane. Substitute its coordinates into equation (4):
\( 6(0) - (3) + 4(3) - 9 = 0 \)
\( 0 - 3 + 12 - 9 = 0 \)
\( 9 - 9 = 0 \)
\( 0 = 0 \).
Since point D satisfies the plane equation, it means point D also lies on the plane. Because both points C and D of line \( L_2 \) lie on the plane that contains line \( L_1 \), the two lines \( L_1 \) and \( L_2 \) are coplanar.
In simple words: To see if two lines are on the same flat surface, we make a surface using the first line and one point from the second line. Then, we check if the other point from the second line also sits on this same surface. If it does, both lines are on that surface together.

🎯 Exam Tip: Two lines are coplanar if they are parallel or if they intersect. An effective way to prove coplanarity is to show that both lines lie on the same plane, which means all four points (two from each line) satisfy the equation of that plane.