RBSE Solutions Class 12 Maths Chapter 14 Three Dimensional Geometry Exercise 14.6

Question 1. Find the equation of the plane which is perpendicular to x-axis and that passes through the point (2,-1,3).
Answer: Let the equation of a plane passing through the point \( (x_1, y_1, z_1) \) be \( a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \). For the given point \( (2,-1,3) \), the equation is \( a(x - 2) + b(y - (-1)) + c(z - 3) = 0 \), which simplifies to \( a(x - 2) + b(y + 1) + c(z - 3) = 0 \). Since the plane is perpendicular to the x-axis, its normal vector must be parallel to the x-axis. This means the normal vector has direction ratios \( (a, 0, 0) \), so \( b = 0 \) and \( c = 0 \). When we substitute these values into the plane equation, we get:
\( a(x - 2) + 0(y + 1) + 0(z - 3) = 0 \)
\( a(x - 2) = 0 \)
Since \( a \) is a direction ratio of the normal and cannot be zero for a plane to exist, we can divide by \( a \).
\( x - 2 = 0 \) Thus, the equation of the plane is \( x - 2 = 0 \).
In simple words: A plane perpendicular to the x-axis will always have an equation like \( x = \text{constant} \). Since this plane passes through \( (2,-1,3) \), the x-coordinate must be 2, so the equation is \( x - 2 = 0 \).

🎯 Exam Tip: Remember that if a plane is perpendicular to an axis (say, the x-axis), its equation will only involve that coordinate (x in this case) and a constant. The y and z coefficients of the normal vector are zero.

Question 2. Find the equation of the plane that passes through X-axis and point (3,2, 4).
Answer: A plane that passes through the x-axis can be written in the general form \( By + Cz = 0 \), where \( B \) and \( C \) are constants. This is because all points on the x-axis have y and z coordinates equal to zero, which satisfies this equation. The problem states that this plane also passes through the point \( (3,2,4) \). We can substitute these coordinates into the equation \( By + Cz = 0 \):
\( B(2) + C(4) = 0 \)
\( 2B + 4C = 0 \)
Divide by 2:
\( B + 2C = 0 \)
So, \( B = -2C \). Now, substitute the value of \( B \) back into the general equation of the plane \( By + Cz = 0 \):
\( (-2C)y + Cz = 0 \)
\( -2Cy + Cz = 0 \) Assuming \( C \neq 0 \) (otherwise, all coefficients would be zero, which is not a plane), we can divide the entire equation by \( C \):
\( -2y + z = 0 \) This can also be written as \( 2y - z = 0 \).
In simple words: A plane going through the x-axis does not depend on 'x'. Its equation only has 'y' and 'z' terms. We use the given point \( (3,2,4) \) to find the exact relation between 'y' and 'z' parts, which gives us \( 2y - z = 0 \).

🎯 Exam Tip: When a plane passes through a coordinate axis, the coefficient of that axis's variable (and the constant term) in the Cartesian equation becomes zero. For example, if it passes through the x-axis, the equation is of the form \( By + Cz = 0 \).

Question 3. A variable plane passes through the point (p, q, r) and meets the coordinate axis at point A, B and C. Show that the locus of a common point of plane passing through A, B and C and parallel to the coordinate planes, will be \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{p} + \frac{1}{q} + \frac{1}{r} \).
Answer: Let the equation of the variable plane in intercept form be \( \frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 1 \). This plane meets the coordinate axes at points \( A(\alpha, 0, 0) \), \( B(0, \beta, 0) \), and \( C(0, 0, \gamma) \). Since the plane passes through the point \( (p, q, r) \), we can substitute these coordinates into the plane's equation:
\( \frac{p}{\alpha} + \frac{q}{\beta} + \frac{r}{\gamma} = 1 \) ...(1) Now, let \( (x, y, z) \) be a common point of planes passing through \( A, B, C \) and parallel to the coordinate planes. This construction implies that for such a common point, its coordinates are directly related to the intercepts, so \( x = \alpha \), \( y = \beta \), and \( z = \gamma \). This means the point \( (x,y,z) \) is precisely the point whose coordinates are the intercepts of the original variable plane. Therefore, by replacing \( \alpha, \beta, \gamma \) with \( x, y, z \) in equation (1), we get the locus of the common point:
\( \frac{p}{x} + \frac{q}{y} + \frac{r}{z} = 1 \) This equation shows the relationship that the coordinates of the common point must satisfy.
In simple words: We start with a plane that cuts the axes at points (say, at distances alpha, beta, gamma from the origin). This plane also goes through a fixed point \( (p,q,r) \). The special "common point" has coordinates that are the same as these axis-cutting distances \( (\alpha, \beta, \gamma) \). So, we can replace alpha, beta, gamma with \( x, y, z \) in the equation that uses \( (p,q,r) \) to find the path (locus) of this common point.

🎯 Exam Tip: The intercept form of a plane's equation is crucial when dealing with intersections with coordinate axes. When a fixed point lies on a variable plane, it introduces a constraint that helps determine the locus of related points.

Question 4. Find the vector equation of a plane which is at a distance of 7 unit from the origin and has \( \hat{i} \) as the unit vector normal to it.
Answer: The vector equation of a plane in normal form is given by \( \vec{r} \cdot \hat{n} = d \), where \( \vec{r} \) is the position vector of any point on the plane, \( \hat{n} \) is the unit vector normal to the plane, and \( d \) is the perpendicular distance of the plane from the origin. In this question, we are given:
The unit vector normal to the plane, \( \hat{n} = \hat{i} \).
The distance of the plane from the origin, \( d = 7 \) units. Substituting these values directly into the normal form equation:
\( \vec{r} \cdot \hat{i} = 7 \) This is the required vector equation of the plane. This form is very useful for planes with simple orientations.
In simple words: The vector equation of a plane tells us about its position and orientation. When we know the direction it faces (the normal vector) and how far it is from the center point (origin), we just plug those values into the formula \( \vec{r} \cdot \hat{n} = d \). Here, the plane faces along the x-axis and is 7 units away.

🎯 Exam Tip: Remember the standard normal form of a plane's vector equation: \( \vec{r} \cdot \hat{n} = d \). Identify \( \hat{n} \) (unit normal vector) and \( d \) (distance from origin) from the problem statement to quickly form the equation.

Question 5. Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector \( 6\hat{i} + 3\hat{j} - 2\hat{k} \).
Answer: The vector equation of a plane in normal form is \( \vec{r} \cdot \hat{n} = d \). We are given the distance from the origin \( d = 7 \) units. We are also given a vector normal to the plane, \( \vec{N} = 6\hat{i} + 3\hat{j} - 2\hat{k} \). First, we need to find the unit normal vector \( \hat{n} \) from \( \vec{N} \). To do this, we divide \( \vec{N} \) by its magnitude \( |\vec{N}| \).
Calculate the magnitude of \( \vec{N} \):
\( |\vec{N}| = \sqrt{(6)^2 + (3)^2 + (-2)^2} \)
\( |\vec{N}| = \sqrt{36 + 9 + 4} \)
\( |\vec{N}| = \sqrt{49} \)
\( |\vec{N}| = 7 \) Now, find the unit normal vector \( \hat{n} \):
\( \hat{n} = \frac{\vec{N}}{|\vec{N}|} = \frac{6\hat{i} + 3\hat{j} - 2\hat{k}}{7} \) Finally, substitute \( \hat{n} \) and \( d \) into the normal form equation \( \vec{r} \cdot \hat{n} = d \):
\( \vec{r} \cdot \left( \frac{6\hat{i} + 3\hat{j} - 2\hat{k}}{7} \right) = 7 \) To remove the fraction, we can multiply both sides by 7:
\( \vec{r} \cdot (6\hat{i} + 3\hat{j} - 2\hat{k}) = 7 \times 7 \)
\( \vec{r} \cdot (6\hat{i} + 3\hat{j} - 2\hat{k}) = 49 \) This is the required vector equation of the plane.
In simple words: We are given a direction vector that is perpendicular to the plane, but it's not a unit vector. So, first, we make it a unit vector by dividing it by its length. Then, we use this unit vector and the given distance from the origin to write the plane's equation in the form \( \vec{r} \cdot \hat{n} = d \).

🎯 Exam Tip: Always convert the given normal vector to a unit vector before using it in the normal form \( \vec{r} \cdot \hat{n} = d \). Pay attention to signs when calculating the magnitude.

Question 6. Reduce \( 3x - 4y + 12z = 5 \) or \( \vec{r} \cdot (3\hat{i} - 4\hat{j} + 12\hat{k}) = 5 \) to normal form and hence find its direction cosines and perpendicular distance from the origin.
Answer: We are given the Cartesian equation of the plane: \( 3x - 4y + 12z = 5 \). The normal form of a plane's equation is \( lx + my + nz = d \), where \( (l, m, n) \) are the direction cosines of the normal to the plane, and \( d \) is the perpendicular distance from the origin to the plane (and \( d \) must be positive). The coefficients of \( x, y, z \) in the given equation represent the direction ratios of the normal vector, \( \vec{N} = 3\hat{i} - 4\hat{j} + 12\hat{k} \). First, calculate the magnitude of the normal vector \( \vec{N} \):
\( |\vec{N}| = \sqrt{(3)^2 + (-4)^2 + (12)^2} \)
\( |\vec{N}| = \sqrt{9 + 16 + 144} \)
\( |\vec{N}| = \sqrt{169} \)
\( |\vec{N}| = 13 \) Now, to convert the plane equation to normal form, divide the entire equation by the magnitude \( |\vec{N}| = 13 \):
\( \frac{3}{13}x - \frac{4}{13}y + \frac{12}{13}z = \frac{5}{13} \) This is the normal form of the plane's equation. From this normal form, we can identify:
The direction cosines (Dc's) of the normal to the plane are \( \left( \frac{3}{13}, -\frac{4}{13}, \frac{12}{13} \right) \).
The perpendicular distance from the origin to the plane is \( d = \frac{5}{13} \) units. This method helps to normalize the equation.
In simple words: To change a plane's equation into normal form, we divide every term by the length of its normal vector. Once in this special form, the numbers in front of \( x, y, z \) are the direction cosines of the plane's normal, and the number on the right side is how far the plane is from the origin.

🎯 Exam Tip: Always ensure the constant term \( d \) in the normal form \( lx + my + nz = d \) is positive. If it's negative, multiply the entire equation by -1 before dividing by the magnitude of the normal vector.

Question 7. Find the vector equation of a plane which is at a distance of 4 units from the origin and direction cosines of the normal to the plane are 2, -1, 2.
Answer: The vector equation of a plane in normal form is \( \vec{r} \cdot \hat{n} = d \). We are given the distance from the origin \( d = 4 \) units. We are given the direction ratios (DRs) of the normal to the plane as \( (2, -1, 2) \). These are not directly direction cosines (DCs) because their squared sum is not 1. First, we need to convert these direction ratios into a unit normal vector \( \hat{n} \). To do this, we find the magnitude of the vector formed by these direction ratios:
Let \( \vec{N} = 2\hat{i} - \hat{j} + 2\hat{k} \).
\( |\vec{N}| = \sqrt{(2)^2 + (-1)^2 + (2)^2} \)
\( |\vec{N}| = \sqrt{4 + 1 + 4} \)
\( |\vec{N}| = \sqrt{9} \)
\( |\vec{N}| = 3 \) Now, the unit normal vector \( \hat{n} \) is:
\( \hat{n} = \frac{\vec{N}}{|\vec{N}|} = \frac{2\hat{i} - \hat{j} + 2\hat{k}}{3} \) Substitute \( \hat{n} \) and \( d \) into the normal form equation \( \vec{r} \cdot \hat{n} = d \):
\( \vec{r} \cdot \left( \frac{2\hat{i} - \hat{j} + 2\hat{k}}{3} \right) = 4 \) This is the required vector equation of the plane.
In simple words: We are given the length of the perpendicular from the origin and the direction numbers of the line perpendicular to the plane. We first turn these direction numbers into a unit vector. Then we use this unit vector and the distance to write the plane's equation in vector form.

🎯 Exam Tip: Be careful to distinguish between direction ratios and direction cosines. Direction ratios define a direction, while direction cosines are normalized values that represent a unit vector in that direction.

Question 8. Find normal form of the plane \( 2x - 3y + 6z + 14 = 0 \).
Answer: We are given the equation of the plane: \( 2x - 3y + 6z + 14 = 0 \). To convert this to the normal form \( lx + my + nz = d \), where \( d \) is the positive perpendicular distance from the origin, we need to ensure the constant term is positive on the right side. Rearrange the equation to move the constant term to the right side:
\( 2x - 3y + 6z = -14 \) Since \( d \) must be positive, we multiply the entire equation by -1:
\( -2x + 3y - 6z = 14 \) Now, identify the coefficients of \( x, y, z \) as the direction ratios of the normal vector \( \vec{N} = -2\hat{i} + 3\hat{j} - 6\hat{k} \). Calculate the magnitude of this normal vector:
\( |\vec{N}| = \sqrt{(-2)^2 + (3)^2 + (-6)^2} \)
\( |\vec{N}| = \sqrt{4 + 9 + 36} \)
\( |\vec{N}| = \sqrt{49} \)
\( |\vec{N}| = 7 \) Finally, divide the entire modified equation by \( |\vec{N}| = 7 \) to get the normal form:
\( \frac{-2}{7}x + \frac{3}{7}y - \frac{6}{7}z = \frac{14}{7} \)
\( -\frac{2}{7}x + \frac{3}{7}y - \frac{6}{7}z = 2 \) This is the required normal form of the plane. In this form, the direction cosines are \( \left( -\frac{2}{7}, \frac{3}{7}, -\frac{6}{7} \right) \) and the perpendicular distance from the origin is 2 units.
In simple words: To get the normal form, first, make sure the number without \( x, y, z \) is positive and on the right side of the equals sign. Then, find the length of the vector made by the numbers with \( x, y, z \). Divide the whole equation by this length. The new equation is the normal form.

🎯 Exam Tip: Always make the constant term on the right side of the normal form equation positive. If it's negative in the initial equation, multiply the entire equation by -1 before dividing by the magnitude of the normal vector.

Question 9. Find the equation of plane perpendicular of the origin from the plane is 13 and direction ratios of this perpendicular are 4,-3,12.
Answer: We need to find the equation of a plane. We are given the perpendicular distance from the origin to the plane, \( d = 13 \) units. We are also given the direction ratios (DRs) of the perpendicular (normal) to the plane as \( (4, -3, 12) \). The general equation of a plane in normal form is \( lx + my + nz = d \), where \( (l, m, n) \) are the direction cosines of the normal. First, we must convert the given direction ratios into direction cosines. To do this, we find the magnitude of the vector formed by these direction ratios:
Let \( \vec{N} = 4\hat{i} - 3\hat{j} + 12\hat{k} \).
\( |\vec{N}| = \sqrt{(4)^2 + (-3)^2 + (12)^2} \)
\( |\vec{N}| = \sqrt{16 + 9 + 144} \)
\( |\vec{N}| = \sqrt{169} \)
\( |\vec{N}| = 13 \) Now, the direction cosines \( (l, m, n) \) are:
\( l = \frac{4}{13} \), \( m = \frac{-3}{13} \), \( n = \frac{12}{13} \) Substitute these direction cosines and the distance \( d = 13 \) into the normal form equation \( lx + my + nz = d \):
\( \frac{4}{13}x - \frac{3}{13}y + \frac{12}{13}z = 13 \) This is the required equation of the plane.
In simple words: We know how far the plane is from the origin and the direction of the line that is perpendicular to the plane. First, we use the direction numbers to find the actual direction cosines. Then, we use these direction cosines and the distance to write the plane's equation.

🎯 Exam Tip: If the problem gives direction ratios, always calculate the magnitude to find the direction cosines before plugging them into the normal form equation of the plane.

Question 10. Find a unit normal vector to the plane \( x + y + z - 3 = 0 \).
Answer: We are given the Cartesian equation of the plane: \( x + y + z - 3 = 0 \). The general form of a plane's equation is \( Ax + By + Cz + D = 0 \). The normal vector \( \vec{N} \) to this plane is given by \( A\hat{i} + B\hat{j} + C\hat{k} \). From the given equation, the coefficients of \( x, y, z \) are \( A=1, B=1, C=1 \). So, the normal vector to the plane is \( \vec{N} = 1\hat{i} + 1\hat{j} + 1\hat{k} = \hat{i} + \hat{j} + \hat{k} \). To find the unit normal vector \( \hat{n} \), we divide the normal vector \( \vec{N} \) by its magnitude \( |\vec{N}| \). First, calculate the magnitude of \( \vec{N} \):
\( |\vec{N}| = \sqrt{(1)^2 + (1)^2 + (1)^2} \)
\( |\vec{N}| = \sqrt{1 + 1 + 1} \)
\( |\vec{N}| = \sqrt{3} \) Now, find the unit normal vector \( \hat{n} \):
\( \hat{n} = \frac{\vec{N}}{|\vec{N}|} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} \) This can also be written as \( \hat{n} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k} \). This vector points in the direction perpendicular to the plane and has a length of one.
In simple words: For any plane equation like \( ax + by + cz + d = 0 \), the vector \( a\hat{i} + b\hat{j} + c\hat{k} \) is perpendicular to the plane. To make it a "unit" (length one) vector, we divide it by its own length.

🎯 Exam Tip: The coefficients of \( x, y, z \) in the Cartesian equation of a plane directly give you the components of a normal vector. Always remember to divide by the magnitude of this vector to make it a unit normal vector.