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

Question 1. Find the vector equation of a plane at a distance of 6 units from the origin and has \( \hat{i} \) as the unit vector normal to it.
Answer: The problem gives us the distance of the plane from the origin, which is \( p = 6 \) units. It also states that the unit vector normal to the plane is \( \hat{n} = \hat{i} \). We use the standard form for the vector equation of a plane: \( \vec{r} \cdot \hat{n} = p \).
\( \implies \) Substituting the given values, the required vector equation of the plane is \( \vec{r} \cdot \hat{i} = 6 \). A plane's vector equation is a powerful way to define its orientation and position in space.
In simple words: The plane is 6 units away from the center (origin), and its normal direction is straight along the x-axis. So, its equation is \( \vec{r} \cdot \hat{i} = 6 \).

🎯 Exam Tip: Always remember the standard vector equation of a plane in normal form: \( \vec{r} \cdot \hat{n} = p \), where \( \hat{n} \) is the unit normal vector and \( p \) is the perpendicular distance from the origin.

Question 2. Find the vector equation of a plane which is at a distance of
(i) 8 units from the origin and which is normal to the vector \( \hat{i} + 2 \hat{j} – 2 \hat{k} \).
(ii) 5 units from the origin and which is normal to the vector \( 2 \hat{i} + 6 \hat{j} – 3 \hat{k} \).
Answer:
(i) Here, the distance from the origin is \( p = 8 \) units. The normal vector to the plane is \( \vec{n} = \hat{i} + 2 \hat{j} - 2 \hat{k} \).
First, we find the magnitude of \( \vec{n} \):
\( |\vec{n}| = |\hat{i}+2 \hat{j}-2 \hat{k}| = \sqrt{1^2+2^2+(-2)^2} = \sqrt{1+4+4} = \sqrt{9} = 3 \).
Since \( |\vec{n}| \neq 1 \), \( \vec{n} \) is not a unit vector, so we divide it by its magnitude to get the unit normal vector \( \hat{n} \).
The required equation of the plane is \( \vec{r} \cdot \hat{n} = p \).
\( \implies \vec{r} \cdot \frac{(\hat{i}+2 \hat{j}-2 \hat{k})}{3} = 8 \)
\( \implies \vec{r} \cdot (\hat{i} + 2 \hat{j} - 2 \hat{k}) = 24 \). Normal vectors are crucial because they directly tell us the orientation of the plane in 3D space.
(ii) In this case, the distance from the origin is \( p = 5 \) units. The normal vector is \( \vec{n} = 2 \hat{i} + 6 \hat{j} - 3 \hat{k} \).
Next, we calculate the magnitude of \( \vec{n} \):
\( |\vec{n}| = |2 \hat{i} + 6 \hat{j} - 3 \hat{k}| = \sqrt{2^2+6^2+(-3)^2} = \sqrt{4+36+9} = \sqrt{49} = 7 \).
Since \( |\vec{n}| \neq 1 \), \( \vec{n} \) is not a unit vector, so we convert it to a unit vector \( \hat{n} \).
The required equation of the plane is \( \vec{r} \cdot \hat{n} = p \).
\( \implies \vec{r} \cdot \frac{(2 \hat{i}+6 \hat{j}-3 \hat{k})}{7} = 5 \)
\( \implies \vec{r} \cdot (2 \hat{i} + 6 \hat{j} - 3 \hat{k}) = 35 \).
In simple words: For both parts, first find the length of the normal vector. Then, divide the normal vector by its length to get a unit vector. Finally, use this unit vector and the given distance in the plane equation formula \( \vec{r} \cdot \hat{n} = p \).

🎯 Exam Tip: When given a normal vector \( \vec{n} \) that is not a unit vector, always remember to divide it by its magnitude \( |\vec{n}| \) to get the unit normal vector \( \hat{n} = \frac{\vec{n}}{|\vec{n}|} \) before using the formula \( \vec{r} \cdot \hat{n} = p \).

Question 3. Find the cartesian equation of the following planes
(i) \( \vec{r} \cdot (\hat{i} + 3 \hat{j} – 4 \hat{k}) = 1 \)
(ii) \( \vec{r} \cdot (3 \hat{i} – 7 \hat{j} + \hat{k}) + 3 = 0 \)
(iii) \( \vec{r} = (λ + 3μ) \hat{i} + (2 – μ) \hat{j} + (μ + 2λ) \hat{k} \)
Answer:
(i) The given vector equation of the plane is \( \vec{r} \cdot (2 \hat{i} + 3 \hat{j} – 4 \hat{k}) = 1 \).
To find the Cartesian equation, we replace \( \vec{r} \) with its position vector in Cartesian coordinates, \( x \hat{i} + y \hat{j} + z \hat{k} \).
\( (x \hat{i} + y \hat{j} + z \hat{k}) \cdot (2 \hat{i} + 3 \hat{j} – 4 \hat{k}) = 1 \)
When we perform the dot product, we multiply the corresponding coefficients of \( \hat{i}, \hat{j}, \hat{k} \) and add them.
\( \implies 2x + 3y - 4z = 1 \). This is the required Cartesian equation of the plane.
(ii) The given vector equation of the plane is \( \vec{r} \cdot (3 \hat{i} – 7 \hat{j} + \hat{k}) + 3 = 0 \).
We can rewrite this as \( \vec{r} \cdot (3 \hat{i} – 7 \hat{j} + \hat{k}) = -3 \).
Again, we replace \( \vec{r} \) with \( x \hat{i} + y \hat{j} + z \hat{k} \).
\( (x \hat{i} + y \hat{j} + z \hat{k}) \cdot (3 \hat{i} – 7 \hat{j} + \hat{k}) = -3 \)
Performing the dot product:
\( \implies 3x - 7y + z = -3 \). This is the required Cartesian equation of the plane. The Cartesian form of a plane equation is often easier for visualizing and working with geometric properties in a coordinate system.
(iii) The given vector equation of the plane is \( \vec{r} = (λ + 3μ) \hat{i} + (2 – μ) \hat{j} + (μ + 2λ) \hat{k} \).
We set \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \). By comparing the coefficients of \( \hat{i}, \hat{j}, \hat{k} \) on both sides, we get:
\( x = λ + 3μ \) ......(1)
\( y = 2 - μ \) ......(2)
\( z = μ + 2λ \) ......(3)
From equation (2), we can express \( μ \) in terms of \( y \): \( μ = 2 - y \).
Substitute this value of \( μ \) into equation (3):
\( z = (2 - y) + 2λ \)
\( 2λ = y + z - 2 \implies λ = \frac{y+z-2}{2} \).
Now, substitute the expressions for \( λ \) and \( μ \) into equation (1):
\( x = \frac{y+z-2}{2} + 3(2 - y) \)
Multiply the entire equation by 2 to remove the fraction:
\( 2x = y + z - 2 + 6(2 - y) \)
\( 2x = y + z - 2 + 12 - 6y \)
\( 2x = -5y + z + 10 \)
\( \implies 2x + 5y - z = 10 \). This is the required Cartesian equation of the plane.
In simple words: To get the Cartesian equation from a vector equation like \( \vec{r} \cdot \vec{n} = d \), just replace \( \vec{r} \) with \( x \hat{i} + y \hat{j} + z \hat{k} \). The numbers in \( \vec{n} \) then become the coefficients of \( x, y, z \). For the third part, we compare coefficients and then eliminate \( \lambda \) and \( \mu \) to get an equation with only \( x, y, z \).

🎯 Exam Tip: When converting from vector to Cartesian form, remember that \( \vec{r} \) represents \( x \hat{i} + y \hat{j} + z \hat{k} \). The coefficients of \( \hat{i}, \hat{j}, \hat{k} \) in the normal vector directly correspond to the coefficients of \( x, y, z \) in the Cartesian equation.

Question 4. Find the vector equation of the following planes
(i) \( 5x-7y+2z = 3 \)
(ii) \( x - 2y + 3z+1=0 \)
Answer:
(i) The given Cartesian equation of the plane is \( 5x-7y+2z = 3 \).
To convert this to a vector equation, we can observe that the coefficients of \( x, y, \) and \( z \) form the normal vector to the plane.
So, the normal vector \( \vec{n} = 5 \hat{i} – 7 \hat{j} + 2 \hat{k} \).
We know that the position vector of any point on the plane is \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \).
The equation can be written as the dot product of \( \vec{r} \) and \( \vec{n} \):
\( (x \hat{i} + y \hat{j} + z \hat{k}) \cdot (5 \hat{i} – 7 \hat{j} + 2 \hat{k}) = 3 \)
\( \implies \vec{r} \cdot (5 \hat{i} – 7 \hat{j} + 2 \hat{k}) = 3 \). This is the required vector equation of the plane.
(ii) The given Cartesian equation of the plane is \( x - 2y + 3z + 1 = 0 \).
First, we rewrite it as \( x - 2y + 3z = -1 \).
From the coefficients of \( x, y, \) and \( z \), the normal vector is \( \vec{n} = \hat{i} – 2 \hat{j} + 3 \hat{k} \).
Let \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \).
The equation can be expressed as:
\( (x \hat{i} + y \hat{j} + z \hat{k}) \cdot (\hat{i} – 2 \hat{j} + 3 \hat{k}) = -1 \)
\( \implies \vec{r} \cdot (\hat{i} – 2 \hat{j} + 3 \hat{k}) = -1 \). This is the required vector equation of the plane. Converting between Cartesian and vector forms of plane equations allows for flexibility in solving different types of geometry problems.
In simple words: To change a Cartesian equation \( Ax+By+Cz=D \) into a vector equation, use the numbers \( A, B, C \) to make a normal vector \( \vec{n} = A \hat{i} + B \hat{j} + C \hat{k} \). Then the vector equation is simply \( \vec{r} \cdot \vec{n} = D \).

🎯 Exam Tip: The coefficients of \( x, y, \) and \( z \) in the Cartesian equation \( Ax+By+Cz=D \) directly form the components of the normal vector \( \vec{n} = A \hat{i} + B \hat{j} + C \hat{k} \) for the vector equation \( \vec{r} \cdot \vec{n} = D \).

Question 5. Find the direction cosines of the perpendicular from the origin to the plane
(i) \( \vec{r} \cdot (2 \hat{i} – 3 \hat{j} – 6 \hat{k}) + 5 = 0 \)
(ii) \( \vec{r} \cdot (2 \hat{i} – 2 \hat{j} + \hat{k}) + 2 = 0 \)
Answer:
(i) The given vector equation of the plane is \( \vec{r} \cdot (2 \hat{i} – 3 \hat{j} – 6 \hat{k}) + 5 = 0 \).
First, rewrite it as \( \vec{r} \cdot (2 \hat{i} – 3 \hat{j} – 6 \hat{k}) = -5 \).
For the perpendicular distance \( p \) from the origin to be positive, we multiply the entire equation by -1:
\( \implies \vec{r} \cdot (-2 \hat{i} + 3 \hat{j} + 6 \hat{k}) = 5 \).
Here, the normal vector is \( \vec{n} = -2 \hat{i} + 3 \hat{j} + 6 \hat{k} \).
Next, find the magnitude of \( \vec{n} \): \( |\vec{n}| = \sqrt{(-2)^2+3^2+6^2} = \sqrt{4+9+36} = \sqrt{49} = 7 \).
To get the direction cosines, we form the unit normal vector \( \hat{n} \):
\( \hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{-2 \hat{i} + 3 \hat{j} + 6 \hat{k}}{7} = -\frac{2}{7} \hat{i} + \frac{3}{7} \hat{j} + \frac{6}{7} \hat{k} \).
The direction cosines of the perpendicular from the origin to the plane are \( \langle -\frac{2}{7}, \frac{3}{7}, \frac{6}{7} \rangle \).
(ii) The given vector equation of the plane is \( \vec{r} \cdot (2 \hat{i} – 2 \hat{j} + \hat{k}) + 2 = 0 \).
Rewrite it as \( \vec{r} \cdot (2 \hat{i} – 2 \hat{j} + \hat{k}) = -2 \).
To make the perpendicular distance \( p \) positive, multiply the equation by -1:
\( \implies \vec{r} \cdot (-2 \hat{i} + 2 \hat{j} – \hat{k}) = 2 \).
Here, the normal vector is \( \vec{n} = -2 \hat{i} + 2 \hat{j} – \hat{k} \).
Calculate the magnitude of \( \vec{n} \): \( |\vec{n}| = \sqrt{(-2)^2+2^2+(-1)^2} = \sqrt{4+4+1} = \sqrt{9} = 3 \).
Form the unit normal vector \( \hat{n} \):
\( \hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{-2 \hat{i} + 2 \hat{j} – \hat{k}}{3} = -\frac{2}{3} \hat{i} + \frac{2}{3} \hat{j} – \frac{1}{3} \hat{k} \).
The direction cosines of the perpendicular from the origin to the plane are \( \langle -\frac{2}{3}, \frac{2}{3}, -\frac{1}{3} \rangle \). Direction cosines help define the exact direction of a line in 3D space, which is very useful for perpendicular lines to planes.
In simple words: First, rewrite the plane equation so the right side (distance from origin) is positive. Then, take the vector part, find its length (magnitude), and divide each component of the vector by this length. The resulting numbers are the direction cosines.

🎯 Exam Tip: For the equation \( \vec{r} \cdot \vec{n} = p \), the direction cosines of the perpendicular from the origin are the components of the unit normal vector \( \hat{n} = \frac{\vec{n}}{|\vec{n}|} \). Always ensure that \( p \) is positive before normalizing \( \vec{n} \).

Question 6. Reduce the equation to normal form and hence find length of perpendicular from the origin to the plane.
(i) \( \vec{r} \cdot (3 \hat{i} – 4 \hat{j} + 12 \hat{k}) = 5 \)
(ii) \( \vec{r} \cdot (4 \hat{i} – 2 \hat{j} + 4 \hat{k}) = 18 \)
Answer:
(i) The given vector equation of the plane is \( \vec{r} \cdot (3 \hat{i} – 4 \hat{j} + 12 \hat{k}) = 5 \).
Here, the normal vector is \( \vec{n} = 3 \hat{i} – 4 \hat{j} + 12 \hat{k} \).
First, we calculate the magnitude of the normal vector:
\( |\vec{n}| = \sqrt{3^2+(-4)^2+12^2} = \sqrt{9+16+144} = \sqrt{169} = 13 \).
To reduce the equation to normal form, we divide both sides by \( |\vec{n}| \):
\( \vec{r} \cdot \frac{(3 \hat{i}-4 \hat{j}+12 \hat{k})}{13} = \frac{5}{13} \). This is the normal form of the plane.
The length of the perpendicular from the origin to the plane (which is \( p \)) is \( \frac{5}{13} \) units.
(ii) The given vector equation of the plane is \( \vec{r} \cdot (4 \hat{i} – 2 \hat{j} + 4 \hat{k}) = 18 \).
Here, the normal vector is \( \vec{n} = 4 \hat{i} – 2 \hat{j} + 4 \hat{k} \).
Next, we find the magnitude of this normal vector:
\( |\vec{n}| = \sqrt{4^2+(-2)^2+4^2} = \sqrt{16+4+16} = \sqrt{36} = 6 \).
To convert the equation to normal form, we divide both sides by \( |\vec{n}| \):
\( \vec{r} \cdot \frac{(4 \hat{i}-2 \hat{j}+4 \hat{k})}{6} = \frac{18}{6} = 3 \). This is the normal form of the plane. The normal form of a plane equation, \( \vec{r} \cdot \hat{n} = p \), directly gives both the orientation (unit normal vector \( \hat{n} \)) and its minimum distance from the origin (\( p \)).
The length of the perpendicular from the origin to the plane (which is \( p \)) is \( 3 \) units.
In simple words: To change a plane's equation into its "normal form", take the vector part, find its total length. Then, divide both sides of the original equation by this length. The number on the right side of the new equation is the distance from the origin to the plane.

🎯 Exam Tip: To get the normal form \( \vec{r} \cdot \hat{n} = p \), divide the entire vector equation \( \vec{r} \cdot \vec{n} = d \) by the magnitude of the normal vector \( |\vec{n}| \). The value of \( p \) will be \( \frac{d}{|\vec{n}|} \).

Question 7. The vector equation of a plane is \( \vec{r} \cdot (2 \hat{i} – \hat{j} + 2 \hat{k}) = 9 \), where \( (2 \hat{i} – \hat{j} + 2 \hat{k}) \) is normal to the plane. Find the length of perpendicular from the origin to the plane.
Answer: The given vector equation of the plane is \( \vec{r} \cdot (2 \hat{i} – \hat{j} + 2 \hat{k}) = 9 \).
Here, the normal vector to the plane is \( \vec{n} = 2 \hat{i} – \hat{j} + 2 \hat{k} \).
To find the length of the perpendicular from the origin, we first need to find the magnitude of this normal vector:
\( |\vec{n}| = \sqrt{2^2+(-1)^2+2^2} = \sqrt{4+1+4} = \sqrt{9} = 3 \).
Now, we convert the plane equation into its normal form by dividing both sides by \( |\vec{n}| \):
\( \vec{r} \cdot \frac{(2 \hat{i}-\hat{j}+2 \hat{k})}{3} = \frac{9}{3} \)
\( \implies \vec{r} \cdot \frac{(2 \hat{i}-\hat{j}+2 \hat{k})}{3} = 3 \).
This equation is in the form \( \vec{r} \cdot \hat{n} = p \), where \( p \) is the length of the perpendicular from the origin.
Therefore, the length of the perpendicular from the origin to the given plane is \( 3 \) units. This distance represents the shortest possible distance from the origin to any point on the plane.
In simple words: To find the distance from the origin to a plane, first find the length of the plane's normal vector. Then, divide the constant on the right side of the equation by this length. That result is the perpendicular distance.

🎯 Exam Tip: The length of the perpendicular from the origin to the plane \( \vec{r} \cdot \vec{n} = d \) is given by the formula \( p = \frac{|d|}{|\vec{n}|} \). Make sure to calculate \( |\vec{n}| \) correctly.

Question 8. Find the vector equation of a plane which is at a distance of 5 units from the origin and has -1,2,2 as the direction ratios of a normal to it.
Answer: The plane is at a distance of \( p = 5 \) units from the origin.
The direction ratios of a normal to the plane are \( \langle -1, 2, 2 \rangle \).
From these direction ratios, we can form the normal vector: \( \vec{n} = -\hat{i} + 2 \hat{j} + 2 \hat{k} \).
Next, we find the magnitude of this normal vector:
\( |\vec{n}| = \sqrt{(-1)^2+2^2+2^2} = \sqrt{1+4+4} = \sqrt{9} = 3 \).
Now, we find the unit normal vector \( \hat{n} \) by dividing \( \vec{n} \) by its magnitude:
\( \hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{-\hat{i} + 2 \hat{j} + 2 \hat{k}}{3} = -\frac{1}{3} \hat{i} + \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k} \).
The vector equation of a plane in normal form is \( \vec{r} \cdot \hat{n} = p \).
Substituting the values we found:
\( \implies \vec{r} \cdot (-\frac{1}{3} \hat{i} + \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k}) = 5 \). Direction ratios provide a simple way to express the direction of a vector, and they are proportional to the direction cosines.
In simple words: We are given the distance from the origin and the direction numbers of the plane's normal. First, turn those direction numbers into a normal vector, then find its length. Divide the normal vector by its length to get a "unit" normal vector. Finally, put this unit normal vector and the distance into the plane's standard equation form.

🎯 Exam Tip: When given direction ratios, use them to form the normal vector \( \vec{n} \). Remember to then find the unit normal vector \( \hat{n} = \frac{\vec{n}}{|\vec{n}|} \) before substituting it into the plane equation \( \vec{r} \cdot \hat{n} = p \).

Question 9. Find a unit normal vector to the plane, \( x+2y+3z-5 = 0 \)
Answer: The given Cartesian equation of the plane is \( x+2y+3z-5 = 0 \).
We can rewrite this as \( x+2y+3z = 5 \).
From the coefficients of \( x, y, \) and \( z \), we can identify the normal vector \( \vec{n} = \hat{i} + 2 \hat{j} + 3 \hat{k} \).
To find a unit normal vector, we need to calculate the magnitude of \( \vec{n} \):
\( |\vec{n}| = \sqrt{1^2+2^2+3^2} = \sqrt{1+4+9} = \sqrt{14} \).
The unit normal vector \( \hat{n} \) is found by dividing the normal vector by its magnitude:
\( \hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{\hat{i} + 2 \hat{j} + 3 \hat{k}}{\sqrt{14}} \).
This can also be written as \( \hat{n} = \frac{1}{\sqrt{14}} \hat{i} + \frac{2}{\sqrt{14}} \hat{j} + \frac{3}{\sqrt{14}} \hat{k} \). A unit normal vector is very useful as it directly represents the direction of the normal without being affected by its magnitude.
In simple words: Look at the numbers in front of \( x, y, \) and \( z \) in the plane equation. These form your normal vector. Calculate its length. Then, divide each part of the normal vector by its length to get the unit normal vector.

🎯 Exam Tip: For a Cartesian plane equation \( Ax+By+Cz=D \), the normal vector is \( A \hat{i} + B \hat{j} + C \hat{k} \). To find the unit normal vector, divide this normal vector by its magnitude \( \sqrt{A^2+B^2+C^2} \).

Question 10. Find the vector equation of a plane passing through a point having position vector \( 2 \hat{i} + 3 \hat{j} – 4 \hat{k} \) and perpendicular to the vector \( 2 \hat{i} – \hat{j} + 2 \hat{k} \). Also reduce it to Cartesian form.
Answer: We are given a point on the plane with position vector \( \vec{a} = 2 \hat{i} + 3 \hat{j} – 4 \hat{k} \).
The plane is perpendicular to the vector \( \vec{n} = 2 \hat{i} – \hat{j} + 2 \hat{k} \), which means \( \vec{n} \) is the normal vector to the plane.
The vector equation of a plane passing through a point \( \vec{a} \) and perpendicular to \( \vec{n} \) is given by \( (\vec{r} - \vec{a}) \cdot \vec{n} = 0 \), which simplifies to \( \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \).
First, calculate the dot product \( \vec{a} \cdot \vec{n} \):
\( \vec{a} \cdot \vec{n} = (2 \hat{i} + 3 \hat{j} – 4 \hat{k}) \cdot (2 \hat{i} – \hat{j} + 2 \hat{k}) \)
\( = (2)(2) + (3)(-1) + (-4)(2) = 4 - 3 - 8 = -7 \).
So, the vector equation of the plane is \( \vec{r} \cdot (2 \hat{i} – \hat{j} + 2 \hat{k}) = -7 \). This form of the plane equation is very useful when you know a specific point on the plane and the direction it faces.
To convert this to Cartesian form, we replace \( \vec{r} \) with \( x \hat{i} + y \hat{j} + z \hat{k} \):
\( (x \hat{i} + y \hat{j} + z \hat{k}) \cdot (2 \hat{i} – \hat{j} + 2 \hat{k}) = -7 \)
Performing the dot product:
\( \implies 2x - y + 2z = -7 \).
This can also be written as \( 2x - y + 2z + 7 = 0 \). This is the required Cartesian equation.
In simple words: Use the formula \( \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \), where \( \vec{a} \) is the point and \( \vec{n} \) is the normal vector. Calculate \( \vec{a} \cdot \vec{n} \). Then, to get the Cartesian form, replace \( \vec{r} \) with \( x \hat{i} + y \hat{j} + z \hat{k} \) and perform the dot product.

🎯 Exam Tip: The vector equation of a plane passing through a point \( \vec{a} \) and normal to \( \vec{n} \) is \( \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \). To convert to Cartesian form, substitute \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \) and compute the dot products.

Question 11. Find the equation of the plane through the points \( 2 \hat{i} + 3 \hat{j} – \hat{k} \) and perpendicular to the \( 3 \hat{i} – 2 \hat{j} – 2 \hat{k} \). Determine the perpendicular distance of this plane from the origin.
Answer: The plane passes through the point \( \vec{a} = 2 \hat{i} + 3 \hat{j} – \hat{k} \).
The vector perpendicular to the plane is \( \vec{n} = 3 \hat{i} – 2 \hat{j} – 2 \hat{k} \), so this is the normal vector.
The equation of the plane is \( \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \).
First, calculate \( \vec{a} \cdot \vec{n} \):
\( \vec{a} \cdot \vec{n} = (2 \hat{i} + 3 \hat{j} – \hat{k}) \cdot (3 \hat{i} – 2 \hat{j} – 2 \hat{k}) \)
\( = (2)(3) + (3)(-2) + (-1)(-2) = 6 - 6 + 2 = 2 \).
So, the vector equation of the plane is \( \vec{r} \cdot (3 \hat{i} – 2 \hat{j} – 2 \hat{k}) = 2 \).
Now, we need to determine the perpendicular distance of this plane from the origin. This distance is given by the formula \( p = \frac{|\vec{a} \cdot \vec{n}|}{|\vec{n}|} \).
We already found \( \vec{a} \cdot \vec{n} = 2 \).
Next, calculate the magnitude of the normal vector \( \vec{n} \):
\( |\vec{n}| = \sqrt{3^2+(-2)^2+(-2)^2} = \sqrt{9+4+4} = \sqrt{17} \).
Therefore, the perpendicular distance from the origin is \( p = \frac{|2|}{\sqrt{17}} = \frac{2}{\sqrt{17}} \) units. The absolute value ensures that the distance from the origin is always a positive quantity.
In simple words: First, use the given point and normal vector to write the plane's equation. Then, to find how far the plane is from the origin, divide the constant part of the equation by the length of the normal vector.

🎯 Exam Tip: When finding the perpendicular distance from the origin to a plane \( \vec{r} \cdot \vec{n} = d \), use the formula \( p = \frac{|d|}{|\vec{n}|} \). Ensure that you correctly calculate the dot product \( d \) and the magnitude \( |\vec{n}| \).

Question 12. Find the equation of the plane through the points \( 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \) and perpendicular to the vector \( 6 \hat{i} + 4 \hat{j} + 3 \hat{k} \). Put the above equation in normal form.
Answer: The plane passes through the point \( \vec{a} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \).
The vector perpendicular to the plane (normal vector) is \( \vec{n} = 6 \hat{i} + 4 \hat{j} + 3 \hat{k} \).
The equation of the plane is given by \( \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \).
First, calculate the dot product \( \vec{a} \cdot \vec{n} \):
\( \vec{a} \cdot \vec{n} = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \cdot (6 \hat{i} + 4 \hat{j} + 3 \hat{k}) \)
\( = (2)(6) + (3)(4) + (4)(3) = 12 + 12 + 12 = 36 \).
So, the vector equation of the plane is \( \vec{r} \cdot (6 \hat{i} + 4 \hat{j} + 3 \hat{k}) = 36 \).
Next, we need to put this equation in normal form. For this, we find the magnitude of the normal vector \( \vec{n} \):
\( |\vec{n}| = \sqrt{6^2+4^2+3^2} = \sqrt{36+16+9} = \sqrt{61} \).
Now, divide the entire equation by \( |\vec{n}| \) to get the normal form:
\( \vec{r} \cdot \frac{(6 \hat{i}+4 \hat{j} + 3 \hat{k})}{\sqrt{61}} = \frac{36}{\sqrt{61}} \). This is the required normal form of the plane. Normal form is particularly useful for finding the shortest distance from the origin to the plane and understanding its orientation with respect to the origin.
In simple words: First, find the plane's equation using the point it goes through and its normal vector. Then, to get the "normal form", find the length of the normal vector and divide both sides of the equation by this length.

🎯 Exam Tip: The normal form of the plane equation \( \vec{r} \cdot \hat{n} = p \) explicitly shows the unit normal vector \( \hat{n} \) and the perpendicular distance \( p \) from the origin. Always remember to divide by the magnitude of \( \vec{n} \) to normalize it.

Question 13. Find the vector equations of the coordinate planes.
Answer: The coordinate planes are the XY-plane, YZ-plane, and ZX-plane.
(i) **Equation of the XOY (XY) plane:**
In the XY-plane, any point has its z-coordinate as 0. The normal vector to this plane is along the z-axis, which is \( \hat{k} \). The plane passes through the origin, so the distance \( p=0 \).
Using the normal form \( \vec{r} \cdot \hat{n} = p \), the vector equation is \( \vec{r} \cdot \hat{k} = 0 \).
(ii) **Equation of the YOZ (YZ) plane:**
In the YZ-plane, any point has its x-coordinate as 0. The normal vector to this plane is along the x-axis, which is \( \hat{i} \). The plane passes through the origin, so the distance \( p=0 \).
The vector equation is \( \vec{r} \cdot \hat{i} = 0 \).
(iii) **Equation of the ZOX (ZX) plane:**
In the ZX-plane, any point has its y-coordinate as 0. The normal vector to this plane is along the y-axis, which is \( \hat{j} \). The plane passes through the origin, so the distance \( p=0 \).
The vector equation is \( \vec{r} \cdot \hat{j} = 0 \). These fundamental planes form the basis of the 3D coordinate system and are essential for defining positions and transformations.
In simple words: For the XY-plane, the normal vector points up (along \( \hat{k} \)), and it passes through the origin, so its equation is \( \vec{r} \cdot \hat{k} = 0 \). Similarly for the YZ-plane (\( \vec{r} \cdot \hat{i} = 0 \)) and ZX-plane (\( \vec{r} \cdot \hat{j} = 0 \)).

🎯 Exam Tip: Remember that for any coordinate plane, the normal vector is the unit vector corresponding to the axis that is zero in that plane (e.g., \( z=0 \) for XY-plane, so normal is \( \hat{k} \)). Since they pass through the origin, \( p=0 \).

Question 14. Show that the normals to the following pairs of plants are perpendicular to each other.
(i) \( x-y-z-2 = 0 \) and \( 3x + 2y + z + 4 = 0 \)
(ii) \( \vec{r} \cdot (2 \hat{i} – \hat{j} + 3 \hat{k}) = 5 \) and \( \vec{r} \cdot (2 \hat{i} – 2 \hat{j} – 2 \hat{k}) = 5 \)
Answer: For two planes to be perpendicular, their normal vectors must be perpendicular. This means the dot product of their normal vectors must be zero.
(i) Given planes: \( x-y-z-2 = 0 \) and \( 3x + 2y + z + 4 = 0 \).
From the first plane, the normal vector \( \vec{n}_1 = \hat{i} – \hat{j} – \hat{k} \).
From the second plane, the normal vector \( \vec{n}_2 = 3 \hat{i} + 2 \hat{j} + \hat{k} \).
Now, we calculate their dot product:
\( \vec{n}_1 \cdot \vec{n}_2 = (\hat{i} – \hat{j} – \hat{k}) \cdot (3 \hat{i} + 2 \hat{j} + \hat{k}) \)
\( = (1)(3) + (-1)(2) + (-1)(1) = 3 - 2 - 1 = 0 \).
Since the dot product is 0, the normal vectors \( \vec{n}_1 \) and \( \vec{n}_2 \) are perpendicular to each other.
(ii) Given planes: \( \vec{r} \cdot (2 \hat{i} – \hat{j} + 3 \hat{k}) = 5 \) and \( \vec{r} \cdot (2 \hat{i} – 2 \hat{j} – 2 \hat{k}) = 5 \).
From the first plane, the normal vector \( \vec{n}_1 = 2 \hat{i} – \hat{j} + 3 \hat{k} \).
From the second plane, the normal vector \( \vec{n}_2 = 2 \hat{i} – 2 \hat{j} – 2 \hat{k} \).
Now, we calculate their dot product:
\( \vec{n}_1 \cdot \vec{n}_2 = (2 \hat{i} – \hat{j} + 3 \hat{k}) \cdot (2 \hat{i} – 2 \hat{j} – 2 \hat{k}) \)
\( = (2)(2) + (-1)(-2) + (3)(-2) = 4 + 2 - 6 = 0 \).
Since the dot product is 0, the normal vectors \( \vec{n}_1 \) and \( \vec{n}_2 \) are perpendicular to each other. Perpendicular planes are common in many engineering and architectural designs, like walls meeting at a corner.
In simple words: To check if the normals of two planes are perpendicular, find the normal vector for each plane. Then, calculate the dot product of these two normal vectors. If the dot product is zero, they are perpendicular.

🎯 Exam Tip: Two vectors \( \vec{A} \) and \( \vec{B} \) are perpendicular (orthogonal) if and only if their dot product \( \vec{A} \cdot \vec{B} = 0 \). This condition is key for proving perpendicularity between planes via their normal vectors.

Question 15. Show that the normal vector to the plane \( 2x + 2y + 2z = 3 \) is equally inclined with the coordinate axes.
Answer: The given equation of the plane is \( 2x + 2y + 2z = 3 \).
The normal vector to this plane can be directly taken from the coefficients of \( x, y, \) and \( z \):
\( \vec{n} = 2 \hat{i} + 2 \hat{j} + 2 \hat{k} \).
Next, we find the magnitude of this normal vector:
\( |\vec{n}| = \sqrt{2^2+2^2+2^2} = \sqrt{4+4+4} = \sqrt{12} = 2\sqrt{3} \).
For a vector to be equally inclined with the coordinate axes, the cosines of the angles it makes with each axis (its direction cosines) must be equal. Let \( \alpha, \beta, \gamma \) be the angles \( \vec{n} \) makes with the x, y, and z axes, respectively.
The direction cosine with the x-axis is:
\( \cos \alpha = \frac{\vec{n} \cdot \hat{i}}{|\vec{n}||\hat{i}|} = \frac{(2 \hat{i} + 2 \hat{j} + 2 \hat{k}) \cdot \hat{i}}{2\sqrt{3} \cdot 1} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} \).
The direction cosine with the y-axis is:
\( \cos \beta = \frac{\vec{n} \cdot \hat{j}}{|\vec{n}||\hat{j}|} = \frac{(2 \hat{i} + 2 \hat{j} + 2 \hat{k}) \cdot \hat{j}}{2\sqrt{3} \cdot 1} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} \).
The direction cosine with the z-axis is:
\( \cos \gamma = \frac{\vec{n} \cdot \hat{k}}{|\vec{n}||\hat{k}|} = \frac{(2 \hat{i} + 2 \hat{j} + 2 \hat{k}) \cdot \hat{k}}{2\sqrt{3} \cdot 1} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} \).
Since \( \cos \alpha = \cos \beta = \cos \gamma = \frac{1}{\sqrt{3}} \), it implies that \( \alpha = \beta = \gamma \). A vector equally inclined to the coordinate axes indicates a symmetrical orientation in 3D space.
Therefore, the normal vector to the plane \( 2x + 2y + 2z = 3 \) is equally inclined with the coordinate axes.
In simple words: Take the normal vector of the plane and calculate its direction cosines (cosines of the angles it makes with the x, y, and z axes). If all three cosines are the same, then the vector is equally inclined to the axes.

🎯 Exam Tip: To show a vector is equally inclined to the coordinate axes, calculate its direction cosines \( (\cos \alpha, \cos \beta, \cos \gamma) \). If \( \cos \alpha = \cos \beta = \cos \gamma \), then the vector is equally inclined.