Samacheer Kalvi Class 12 Maths Solutions Chapter 6 Applications of Vector Algebra Exercise 6.9

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.9

Question 1. Find the equation of the plane passing through the line of intersection of the planes \( \overline { r } \cdot (2\hat { i } – 7\hat { j } + 4\hat { k }) = 3 \) and \( 3x – 5y + 4z + 11 = 0 \) and the point (- 2, 1, 3).
Answer: The given planes are:
Plane 1: \( \vec{r} \cdot(2 \hat{i}-7 \hat{j}+4 \hat{k}) = 3 \)
This can be written in Cartesian form as \( 2x - 7y + 4z - 3 = 0 \).
Plane 2: \( 3x - 5y + 4z + 11 = 0 \)
The general equation of a plane passing through the line of intersection of two planes \( P_1 = 0 \) and \( P_2 = 0 \) is given by \( P_1 + \lambda P_2 = 0 \). This formula helps us find new planes that share the same intersecting line.
So, the equation of the required plane is:
\( (2x - 7y + 4z - 3) + \lambda (3x - 5y + 4z + 11) = 0 \) ............ (1)
This plane passes through the point (-2, 1, 3). Substitute these coordinates into the equation:
\( (2(-2) - 7(1) + 4(3) - 3) + \lambda (3(-2) - 5(1) + 4(3) + 11) = 0 \)
\( (-4 - 7 + 12 - 3) + \lambda (-6 - 5 + 12 + 11) = 0 \)
\( (-11 + 12 - 3) + \lambda (-11 + 12 + 11) = 0 \)
\( (-2) + \lambda (12) = 0 \)
\( 12\lambda = 2 \)
\( \lambda = \frac{2}{12} \)
\( \lambda = \frac{1}{6} \)
Now, substitute the value of \( \lambda \) back into equation (1) to get the final equation of the plane:
\( (2x - 7y + 4z - 3) + \frac{1}{6} (3x - 5y + 4z + 11) = 0 \)
To remove the fraction, multiply the entire equation by 6:
\( 6(2x - 7y + 4z - 3) + 1(3x - 5y + 4z + 11) = 0 \)
\( 12x - 42y + 24z - 18 + 3x - 5y + 4z + 11 = 0 \)
Combine the like terms:
\( (12x + 3x) + (-42y - 5y) + (24z + 4z) + (-18 + 11) = 0 \)
\( 15x - 47y + 28z - 7 = 0 \)
This is the required equation of the plane.
In simple words: First, we write the equation of a plane that goes through where two other planes cross. Then, we use the extra point given to find a special number called lambda. Finally, we put this number back into our plane equation to get the final answer.

🎯 Exam Tip: Remember to convert vector equations of planes into Cartesian form when mixing with Cartesian equations, and carefully combine like terms after substituting lambda.

Question 2. Find the equation of the plane passing through the line of intersection of the planes \( x + 2y + 3z = 2 \) and \( x - y + z = 3 \) and at a distance \( \frac { 2 }{ \sqrt{3} } \) from the point (3, 1, -1).
Answer: The given planes are:
Plane 1: \( x + 2y + 3z - 2 = 0 \)
Plane 2: \( x - y + z - 3 = 0 \)
The equation of a plane passing through the line of intersection of these two planes is given by \( (x + 2y + 3z - 2) + \lambda (x - y + z - 3) = 0 \) ............ (1)
We need to simplify this equation by grouping terms with x, y, z, and constants:
\( x + 2y + 3z - 2 + \lambda x - \lambda y + \lambda z - 3\lambda = 0 \)
\( (1 + \lambda)x + (2 - \lambda)y + (3 + \lambda)z + (-2 - 3\lambda) = 0 \) ............ (2)
This plane is at a distance of \( \frac { 2 }{ \sqrt{3} } \) from the point (3, 1, -1). We use the distance formula from a point \( (x_1, y_1, z_1) \) to a plane \( Ax + By + Cz + D = 0 \), which is \( \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \). This formula helps us find how far a point is from a flat surface in 3D space.
Here, \( A = (1 + \lambda) \), \( B = (2 - \lambda) \), \( C = (3 + \lambda) \), \( D = (-2 - 3\lambda) \), and the point is \( (3, 1, -1) \). The distance is \( \frac { 2 }{ \sqrt{3} } \).
So, \( \frac{|(1 + \lambda)(3) + (2 - \lambda)(1) + (3 + \lambda)(-1) + (-2 - 3\lambda)|}{\sqrt{(1 + \lambda)^2 + (2 - \lambda)^2 + (3 + \lambda)^2}} = \frac { 2 }{ \sqrt{3} } \)
Let's simplify the numerator:
\( |3 + 3\lambda + 2 - \lambda - 3 - \lambda - 2 - 3\lambda| \)
\( |(3 + 2 - 3 - 2) + (3\lambda - \lambda - \lambda - 3\lambda)| \)
\( |0 - 2\lambda| = |-2\lambda| \)
Now, let's simplify the denominator by squaring both sides to remove the square roots:
\( \frac{(-2\lambda)^2}{(1 + \lambda)^2 + (2 - \lambda)^2 + (3 + \lambda)^2} = \left(\frac { 2 }{ \sqrt{3} }\right)^2 \)
\( \frac{4\lambda^2}{(1 + 2\lambda + \lambda^2) + (4 - 4\lambda + \lambda^2) + (9 + 6\lambda + \lambda^2)} = \frac{4}{3} \)
\( \frac{4\lambda^2}{3\lambda^2 + 4\lambda + 14} = \frac{4}{3} \)
Divide both sides by 4:
\( \frac{\lambda^2}{3\lambda^2 + 4\lambda + 14} = \frac{1}{3} \)
Cross-multiply:
\( 3\lambda^2 = 3\lambda^2 + 4\lambda + 14 \)
\( 0 = 4\lambda + 14 \)
\( 4\lambda = -14 \)
\( \lambda = \frac{-14}{4} = \frac{-7}{2} \)
Substitute this value of \( \lambda \) back into equation (1):
\( (x + 2y + 3z - 2) + \left(\frac{-7}{2}\right) (x - y + z - 3) = 0 \)
Multiply the entire equation by 2 to clear the fraction:
\( 2(x + 2y + 3z - 2) - 7(x - y + z - 3) = 0 \)
\( 2x + 4y + 6z - 4 - 7x + 7y - 7z + 21 = 0 \)
Combine like terms:
\( (2x - 7x) + (4y + 7y) + (6z - 7z) + (-4 + 21) = 0 \)
\( -5x + 11y - z + 17 = 0 \)
We can multiply by -1 to make the first term positive:
\( 5x - 11y + z - 17 = 0 \)
This is the required equation of the plane.
In simple words: We start with a general equation for a plane that cuts across two other planes. Then, we use the fact that this new plane is a certain distance from a given point. By using the distance formula, we find the special number called lambda. Finally, we put lambda back into our equation to get the plane's formula.

🎯 Exam Tip: When using the distance formula, ensure to square both sides correctly and carefully handle the absolute value in the numerator, as it impacts the squaring process.

Question 3. Find the angle between the line \( \overline { r } = (2\hat { i } – \hat {j } + \hat { k }) + t(\hat { i } + 2\hat { j } – 2\hat { k }) \) and the plane \( \overline { r } \cdot (6\hat { i } + 3\hat { j } + 2\hat { k }) = 8 \).
Answer: To find the angle between a line and a plane, we use the formula \( \sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|} \), where \( \vec{b} \) is the direction vector of the line and \( \vec{n} \) is the normal vector to the plane. The sine function is used here because we are finding the angle between the line and the plane itself, not the angle between the line and the normal.
From the given line equation \( \overline { r } = (2\hat { i } – \hat {j } + \hat { k }) + t(\hat { i } + 2\hat { j } – 2\hat { k }) \), the direction vector of the line is \( \vec{b} = \hat { i } + 2\hat { j } – 2\hat { k } \).
From the given plane equation \( \overline { r } \cdot (6\hat { i } + 3\hat { j } + 2\hat { k }) = 8 \), the normal vector to the plane is \( \vec{n} = 6\hat { i } + 3\hat { j } + 2\hat { k } \).
First, calculate the dot product \( \vec{b} \cdot \vec{n} \):
\( \vec{b} \cdot \vec{n} = (1)(6) + (2)(3) + (-2)(2) \)
\( = 6 + 6 - 4 \)
\( = 8 \)
Next, calculate the magnitude of \( \vec{b} \):
\( |\vec{b}| = \sqrt{1^2 + 2^2 + (-2)^2} \)
\( = \sqrt{1 + 4 + 4} \)
\( = \sqrt{9} = 3 \)
Then, calculate the magnitude of \( \vec{n} \):
\( |\vec{n}| = \sqrt{6^2 + 3^2 + 2^2} \)
\( = \sqrt{36 + 9 + 4} \)
\( = \sqrt{49} = 7 \)
Now, substitute these values into the formula for \( \sin \theta \):
\( \sin \theta = \frac{|8|}{(3)(7)} \)
\( \sin \theta = \frac{8}{21} \)
Therefore, the angle \( \theta \) between the line and the plane is:
\( \theta = \sin^{-1} \left( \frac{8}{21} \right) \)
In simple words: To find the angle, we first find the direction the line is going and the direction straight out from the plane. Then, we use a special math formula that involves multiplying these directions and dividing by their lengths. This gives us the sine of the angle, and from that, we can find the angle itself.

🎯 Exam Tip: Be careful to use the sine formula for the angle between a line and a plane, and the cosine formula for the angle between two planes or two lines.

Question 4. Find the angle between the planes \( \overline { r } \cdot (\hat { i } + \hat { j } – 2\hat { k }) = 3 \) and \( 2x – 2y + z = 2 \).
Answer: To find the angle between two planes, we find the angle between their normal vectors. The formula for the angle \( \theta \) between two planes is \( \cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|} \). We use cosine here because the angle between two planes is defined as the angle between their normal vectors, and the dot product directly relates to cosine.
From the first plane \( \overline { r } \cdot (\hat { i } + \hat { j } – 2\hat { k }) = 3 \), the normal vector is \( \vec{n_1} = \hat { i } + \hat { j } – 2\hat { k } \).
The second plane is given as \( 2x – 2y + z = 2 \). The normal vector for this plane is \( \vec{n_2} = 2\hat { i } – 2\hat { j } + \hat { k } \).
First, calculate the dot product \( \vec{n_1} \cdot \vec{n_2} \):
\( \vec{n_1} \cdot \vec{n_2} = (1)(2) + (1)(-2) + (-2)(1) \)
\( = 2 - 2 - 2 \)
\( = -2 \)
Next, calculate the magnitude of \( \vec{n_1} \):
\( |\vec{n_1}| = \sqrt{1^2 + 1^2 + (-2)^2} \)
\( = \sqrt{1 + 1 + 4} \)
\( = \sqrt{6} \)
Then, calculate the magnitude of \( \vec{n_2} \):
\( |\vec{n_2}| = \sqrt{2^2 + (-2)^2 + 1^2} \)
\( = \sqrt{4 + 4 + 1} \)
\( = \sqrt{9} = 3 \)
Now, substitute these values into the formula for \( \cos \theta \):
\( \cos \theta = \frac{|-2|}{\sqrt{6} \cdot 3} \)
\( \cos \theta = \frac{2}{3\sqrt{6}} \)
Therefore, the angle \( \theta \) between the two planes is:
\( \theta = \cos^{-1} \left( \frac{2}{3\sqrt{6}} \right) \)
In simple words: We find the two "normal" vectors, which are like arrows pointing straight out from each plane. Then, we use a special math formula that compares these two arrows to find the angle between them. This angle is the same as the angle between the planes.

🎯 Exam Tip: When using the dot product formula for angles, remember to take the absolute value of the dot product to get the acute angle between the planes.

Question 5. Find the equation of the plane which passes through the point (3, 4, -1) and is parallel to the plane \( 2x – 3y + 5z + 7 = 0 \). Also, find the distance between the two planes.
Answer: We need to find two things: the equation of a new plane and the distance between it and the given plane.
**Part 1: Equation of the parallel plane**
If a plane is parallel to another plane \( Ax + By + Cz + D = 0 \), its equation will have the form \( Ax + By + Cz + \lambda = 0 \). The normal vector remains the same for parallel planes, only the constant term changes. This means the planes face the same direction but are at different positions.
The given plane is \( 2x – 3y + 5z + 7 = 0 \) ............ (1)
So, the equation of the plane parallel to it will be \( 2x – 3y + 5z + \lambda = 0 \) ............ (2)
This new plane passes through the point (3, 4, -1). We can substitute these coordinates into equation (2) to find \( \lambda \):
\( 2(3) - 3(4) + 5(-1) + \lambda = 0 \)
\( 6 - 12 - 5 + \lambda = 0 \)
\( -6 - 5 + \lambda = 0 \)
\( -11 + \lambda = 0 \)
\( \lambda = 11 \)
Substitute \( \lambda = 11 \) back into equation (2):
\( 2x – 3y + 5z + 11 = 0 \) ............ (3)
This is the required equation of the parallel plane.

**Part 2: Distance between the two parallel planes**
The two parallel planes are:
Plane 1: \( 2x – 3y + 5z + 7 = 0 \) (from the question)
Plane 2: \( 2x – 3y + 5z + 11 = 0 \) (the plane we just found)
For two parallel planes \( Ax + By + Cz + D_1 = 0 \) and \( Ax + By + Cz + D_2 = 0 \), the distance between them is given by the formula \( \text{Distance} = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}} \). This formula measures the shortest gap between two parallel flat surfaces.
Here, \( A = 2 \), \( B = -3 \), \( C = 5 \), \( D_1 = 7 \), and \( D_2 = 11 \).
Substitute these values into the distance formula:
\( \text{Distance} = \frac{|7 - 11|}{\sqrt{2^2 + (-3)^2 + 5^2}} \)
\( = \frac{|-4|}{\sqrt{4 + 9 + 25}} \)
\( = \frac{4}{\sqrt{38}} \)
The distance between the two parallel planes is \( \frac{4}{\sqrt{38}} \) units.
In simple words: First, we find the formula for a new plane that runs exactly parallel to the one given. We use the point it goes through to figure out its exact position. Then, we calculate the space between these two parallel planes using a special distance formula.

🎯 Exam Tip: Remember that parallel planes have the same coefficients for x, y, and z, only differing in the constant term. Also, ensure you use the correct distance formula for parallel planes.

Question 6. Find the length of the perpendicular from the point (1, -2, 3) to the plane \( x - y + z = 5 \).
Answer: To find the length of the perpendicular from a point \( (x_1, y_1, z_1) \) to a plane \( Ax + By + Cz + D = 0 \), we use the formula: \( \text{Length} = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \). This formula helps us measure the shortest distance from a single point to a flat plane.
The given point is \( (1, -2, 3) \), so \( x_1 = 1, y_1 = -2, z_1 = 3 \).
The given plane is \( x - y + z = 5 \). To match the formula, we rewrite it as \( x - y + z - 5 = 0 \).
From the plane equation, we have \( A = 1, B = -1, C = 1, D = -5 \).
Now, substitute these values into the formula:
\( \text{Length} = \frac{|(1)(1) + (-1)(-2) + (1)(3) + (-5)|}{\sqrt{1^2 + (-1)^2 + 1^2}} \)
\( = \frac{|1 + 2 + 3 - 5|}{\sqrt{1 + 1 + 1}} \)
\( = \frac{|6 - 5|}{\sqrt{3}} \)
\( = \frac{|1|}{\sqrt{3}} \)
\( = \frac{1}{\sqrt{3}} \)
The length of the perpendicular is \( \frac{1}{\sqrt{3}} \) units.
In simple words: We want to find how far a specific point is from a flat surface. We put the point's numbers and the plane's numbers into a special distance formula. This formula then tells us the shortest distance, which is the length of the perpendicular line.

🎯 Exam Tip: Always ensure the plane equation is in the standard form \( Ax + By + Cz + D = 0 \) before applying the distance formula, correctly identifying the value of D (including its sign).

Question 7. Find the point of intersection of the line with the plane \( (x - 1) = \frac { y }{ 2 } = z + 1 \) with the plane \( 2x – y – 2z = 2 \). Also, find the angle between the line and the plane.
Answer: We need to find two things: where the line crosses the plane and the angle between them.

**Part 1: Point of Intersection**
The equation of the line is given in symmetric form: \( \frac{x - 1}{1} = \frac{y}{2} = \frac{z + 1}{1} \).
We can write any point on this line in parametric form by setting it equal to \( \lambda \):
\( \frac{x - 1}{1} = \lambda \implies x = \lambda + 1 \)
\( \frac{y}{2} = \lambda \implies y = 2\lambda \)
\( \frac{z + 1}{1} = \lambda \implies z = \lambda - 1 \)
So, any point on the line is \( (\lambda + 1, 2\lambda, \lambda - 1) \).
The equation of the plane is \( 2x – y – 2z = 2 \).
To find the point of intersection, substitute the parametric coordinates of the line into the plane equation:
\( 2(\lambda + 1) - (2\lambda) - 2(\lambda - 1) = 2 \)
\( 2\lambda + 2 - 2\lambda - 2\lambda + 2 = 2 \)
Combine the terms:
\( (2\lambda - 2\lambda - 2\lambda) + (2 + 2) = 2 \)
\( -2\lambda + 4 = 2 \)
\( -2\lambda = 2 - 4 \)
\( -2\lambda = -2 \)
\( \lambda = 1 \)
Now, substitute \( \lambda = 1 \) back into the parametric equations of the line to find the coordinates of the intersection point:
\( x = 1 + 1 = 2 \)
\( y = 2(1) = 2 \)
\( z = 1 - 1 = 0 \)
The required point of intersection is (2, 2, 0).

**Part 2: Angle between the line and the plane**
We use the formula \( \sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|} \), where \( \vec{b} \) is the direction vector of the line and \( \vec{n} \) is the normal vector to the plane.
From the line equation \( \frac{x - 1}{1} = \frac{y}{2} = \frac{z + 1}{1} \), the direction vector is \( \vec{b} = 1\hat { i } + 2\hat { j } + 1\hat { k } \).
From the plane equation \( 2x – y – 2z = 2 \), the normal vector is \( \vec{n} = 2\hat { i } – 1\hat { j } – 2\hat { k } \).
First, calculate the dot product \( \vec{b} \cdot \vec{n} \):
\( \vec{b} \cdot \vec{n} = (1)(2) + (2)(-1) + (1)(-2) \)
\( = 2 - 2 - 2 \)
\( = -2 \)
Next, calculate the magnitude of \( \vec{b} \):
\( |\vec{b}| = \sqrt{1^2 + 2^2 + 1^2} \)
\( = \sqrt{1 + 4 + 1} \)
\( = \sqrt{6} \)
Then, calculate the magnitude of \( \vec{n} \):
\( |\vec{n}| = \sqrt{2^2 + (-1)^2 + (-2)^2} \)
\( = \sqrt{4 + 1 + 4} \)
\( = \sqrt{9} = 3 \)
Now, substitute these values into the formula for \( \sin \theta \):
\( \sin \theta = \frac{|-2|}{\sqrt{6} \cdot 3} \)
\( \sin \theta = \frac{2}{3\sqrt{6}} \)
Therefore, the angle \( \theta \) between the line and the plane is:
\( \theta = \sin^{-1} \left( \frac{2}{3\sqrt{6}} \right) \)
In simple words: First, we imagine the line as a moving point with a special number called lambda. We find out what lambda must be for this point to be on the plane. This gives us the point where they cross. Then, we find the line's direction and the plane's straight-out direction. We use a math formula to calculate the angle between them.

🎯 Exam Tip: When finding the point of intersection, convert the line equation to parametric form, substitute it into the plane equation, and solve for the parameter. Remember the correct sine formula for the angle between a line and a plane.

Question 8. Find the co-ordinates of the foot of the perpendicular and length of the perpendicular from the point (4, 3, 2) to the plane \( x + 2y + 3z = 2 \).
Answer: We need to find two things: the exact point on the plane where a perpendicular line from our given point would touch, and the length of that perpendicular line.

**Part 1: Foot of the Perpendicular**
Let the given point be P(4, 3, 2).
The equation of the plane is \( x + 2y + 3z = 2 \).
The direction ratios of the normal to the plane are (1, 2, 3).
Since the perpendicular line (PQ) from P to the plane is parallel to the normal vector of the plane, its direction ratios will also be (1, 2, 3).
The equation of the line PQ passing through P(4, 3, 2) with direction ratios (1, 2, 3) is:
\( \frac{x - 4}{1} = \frac{y - 3}{2} = \frac{z - 2}{3} = \lambda \)
Any point Q on this line can be written in parametric form:
\( x = \lambda + 4 \)
\( y = 2\lambda + 3 \)
\( z = 3\lambda + 2 \)
The foot of the perpendicular, Q, lies on the plane \( x + 2y + 3z = 2 \). So, substitute the coordinates of Q into the plane equation:
\( (\lambda + 4) + 2(2\lambda + 3) + 3(3\lambda + 2) = 2 \)
\( \lambda + 4 + 4\lambda + 6 + 9\lambda + 6 = 2 \)
Combine the like terms:
\( (\lambda + 4\lambda + 9\lambda) + (4 + 6 + 6) = 2 \)
\( 14\lambda + 16 = 2 \)
\( 14\lambda = 2 - 16 \)
\( 14\lambda = -14 \)
\( \lambda = -1 \)
Now, substitute \( \lambda = -1 \) back into the coordinates of Q to find the foot of the perpendicular:
\( x = -1 + 4 = 3 \)
\( y = 2(-1) + 3 = -2 + 3 = 1 \)
\( z = 3(-1) + 2 = -3 + 2 = -1 \)
The coordinates of the foot of the perpendicular Q are (3, 1, -1).

**Part 2: Length of the Perpendicular**
The length of the perpendicular is the distance between point P(4, 3, 2) and the foot of the perpendicular Q(3, 1, -1).
We use the distance formula between two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \):
\( \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \)
\( \text{PQ} = \sqrt{(3 - 4)^2 + (1 - 3)^2 + (-1 - 2)^2} \)
\( = \sqrt{(-1)^2 + (-2)^2 + (-3)^2} \)
\( = \sqrt{1 + 4 + 9} \)
\( = \sqrt{14} \)
The length of the perpendicular is \( \sqrt{14} \) units.
In simple words: First, we imagine a line starting from our given point and going straight down to the plane. We use the plane's normal direction to find this line's path. We then find where this line hits the plane, which is the "foot" of the perpendicular. Finally, we measure the distance between our starting point and this "foot" to get the length of the perpendicular.

🎯 Exam Tip: The line from the point to the foot of the perpendicular is always parallel to the normal vector of the plane. Remember to use the general point on the line to substitute into the plane equation to find the parameter \( \lambda \).