RBSE Solutions Class 12 Maths Chapter 14 Three Dimensional Geometry Exercise 14.7

Question 1. Find the angle between the planes:
(i) \( \vec{r}.(2\hat{i}-\hat{j}+2\hat{k}) = 6 \) and \( \vec{r}.(3\hat{i}+6\hat{j}-2\hat{k}) = 9 \)
(ii) \( \vec{r}.(2\hat{i}+3\hat{j}-6\hat{k}) = 5 \) and \( \vec{r}.(\hat{i}-2\hat{j}+2\hat{k}) = 9 \)
(iii) \( \vec{r}.(\hat{i}+\hat{j}+2\hat{k}) = 5 \) and \( \vec{r}.(2\hat{i}-\hat{j}+2\hat{k}) = 6 \)
Answer:
The angle \( \theta \) between two planes with normal vectors \( \vec{n_1} \) and \( \vec{n_2} \) is given by:
\( \cos \theta = \frac{ | \vec{n_1} \cdot \vec{n_2} | }{ | \vec{n_1} | | \vec{n_2} | } \)
(i) Given planes are \( \vec{r}.(2\hat{i}-\hat{j}+2\hat{k}) = 6 \) and \( \vec{r}.(3\hat{i}+6\hat{j}-2\hat{k}) = 9 \).
The normal vectors are \( \vec{n_1} = 2\hat{i}-\hat{j}+2\hat{k} \) and \( \vec{n_2} = 3\hat{i}+6\hat{j}-2\hat{k} \).
First, calculate the dot product:
\( \vec{n_1} \cdot \vec{n_2} = (2)(3) + (-1)(6) + (2)(-2) \)
\( \implies \vec{n_1} \cdot \vec{n_2} = 6 - 6 - 4 \)
\( \implies \vec{n_1} \cdot \vec{n_2} = -4 \)
Next, calculate the magnitudes of the normal vectors:
\( | \vec{n_1} | = \sqrt{(2)^2 + (-1)^2 + (2)^2} = \sqrt{4+1+4} = \sqrt{9} = 3 \)
\( | \vec{n_2} | = \sqrt{(3)^2 + (6)^2 + (-2)^2} = \sqrt{9+36+4} = \sqrt{49} = 7 \)
Now, find \( \cos \theta \):
\( \cos \theta = \frac{ |-4| }{ (3)(7) } = \frac{ 4 }{ 21 } \)
So, the angle is \( \theta = \cos^{-1}\left(\frac{4}{21}\right) \).
(ii) Given planes are \( \vec{r}.(2\hat{i}+3\hat{j}-6\hat{k}) = 5 \) and \( \vec{r}.(\hat{i}-2\hat{j}+2\hat{k}) = 9 \).
The normal vectors are \( \vec{n_1} = 2\hat{i}+3\hat{j}-6\hat{k} \) and \( \vec{n_2} = \hat{i}-2\hat{j}+2\hat{k} \).
Calculate the dot product:
\( \vec{n_1} \cdot \vec{n_2} = (2)(1) + (3)(-2) + (-6)(2) \)
\( \implies \vec{n_1} \cdot \vec{n_2} = 2 - 6 - 12 \)
\( \implies \vec{n_1} \cdot \vec{n_2} = -16 \)
Calculate the magnitudes:
\( | \vec{n_1} | = \sqrt{(2)^2 + (3)^2 + (-6)^2} = \sqrt{4+9+36} = \sqrt{49} = 7 \)
\( | \vec{n_2} | = \sqrt{(1)^2 + (-2)^2 + (2)^2} = \sqrt{1+4+4} = \sqrt{9} = 3 \)
Find \( \cos \theta \):
\( \cos \theta = \frac{ |-16| }{ (7)(3) } = \frac{ 16 }{ 21 } \)
So, the angle is \( \theta = \cos^{-1}\left(\frac{16}{21}\right) \).
(iii) Given planes are \( \vec{r}.(\hat{i}+\hat{j}+2\hat{k}) = 5 \) and \( \vec{r}.(2\hat{i}-\hat{j}+2\hat{k}) = 6 \).
The normal vectors are \( \vec{n_1} = \hat{i}+\hat{j}+2\hat{k} \) and \( \vec{n_2} = 2\hat{i}-\hat{j}+2\hat{k} \).
Calculate the dot product:
\( \vec{n_1} \cdot \vec{n_2} = (1)(2) + (1)(-1) + (2)(2) \)
\( \implies \vec{n_1} \cdot \vec{n_2} = 2 - 1 + 4 \)
\( \implies \vec{n_1} \cdot \vec{n_2} = 5 \)
Calculate the magnitudes:
\( | \vec{n_1} | = \sqrt{(1)^2 + (1)^2 + (2)^2} = \sqrt{1+1+4} = \sqrt{6} \)
\( | \vec{n_2} | = \sqrt{(2)^2 + (-1)^2 + (2)^2} = \sqrt{4+1+4} = \sqrt{9} = 3 \)
Find \( \cos \theta \):
\( \cos \theta = \frac{ |5| }{ (\sqrt{6})(3) } = \frac{ 5 }{ 3\sqrt{6} } \)
So, the angle is \( \theta = \cos^{-1}\left(\frac{5}{3\sqrt{6}}\right) \). When finding the angle between planes, using their normal vectors is a fundamental concept in 3D geometry.
In simple words: To find the angle between two planes, you look at their "normal" vectors. These vectors are perpendicular to the planes. Then, you use a formula with the dot product and lengths of these normal vectors to calculate the cosine of the angle. Finally, you take the inverse cosine to get the angle itself.

🎯 Exam Tip: Remember to use the absolute value of the dot product \( | \vec{n_1} \cdot \vec{n_2} | \) to ensure the angle \( \theta \) is acute (between 0 and \( \frac{\pi}{2} \) radians or 0 to 90 degrees).

Question 2. Find the angle between the given planes (Cartesian form):
(i) \( x + y + 2z = 9 \) and \( 2x - y - z = 15 \)
(iii) \( x + y - 2z = 3 \) and \( 2x - 2y + z = 5 \)
Answer:
If two planes are given by the equations \( a_1x + b_1y + c_1z + d_1 = 0 \) and \( a_2x + b_2y + c_2z + d_2 = 0 \), the angle \( \theta \) between them is found using the formula:
\( \cos \theta = \frac{ |a_1a_2 + b_1b_2 + c_1c_2| }{ \sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2} } \)
(i) Given planes are \( x + y + 2z = 9 \) and \( 2x - y - z = 15 \).
From the first plane, \( a_1 = 1, b_1 = 1, c_1 = 2 \).
From the second plane, \( a_2 = 2, b_2 = -1, c_2 = -1 \).
Now, substitute these values into the formula:
\( \cos \theta = \frac{ |(1)(2) + (1)(-1) + (2)(-1)| }{ \sqrt{(1)^2 + (1)^2 + (2)^2} \sqrt{(2)^2 + (-1)^2 + (-1)^2} } \)
\( \implies \cos \theta = \frac{ |2 - 1 - 2| }{ \sqrt{1+1+4} \sqrt{4+1+1} } \)
\( \implies \cos \theta = \frac{ |-1| }{ \sqrt{6} \sqrt{6} } \)
\( \implies \cos \theta = \frac{ 1 }{ 6 } \)
So, the angle is \( \theta = \cos^{-1}\left(\frac{1}{6}\right) \). We can also express this as \( \theta = \cos^{-1}\left(\frac{3}{3\sqrt{6}}\right) \), which simplifies the calculation steps.

(iii) Given planes are \( x + y - 2z = 3 \) and \( 2x - 2y + z = 5 \).
From the first plane, \( a_1 = 1, b_1 = 1, c_1 = -2 \).
From the second plane, \( a_2 = 2, b_2 = -2, c_2 = 1 \).
Substitute these values into the formula:
\( \cos \theta = \frac{ |(1)(2) + (1)(-2) + (-2)(1)| }{ \sqrt{(1)^2 + (1)^2 + (-2)^2} \sqrt{(2)^2 + (-2)^2 + (1)^2} } \)
\( \implies \cos \theta = \frac{ |2 - 2 - 2| }{ \sqrt{1+1+4} \sqrt{4+4+1} } \)
\( \implies \cos \theta = \frac{ |-2| }{ \sqrt{6} \sqrt{9} } \)
\( \implies \cos \theta = \frac{ 2 }{ 3\sqrt{6} } \)
So, the angle is \( \theta = \cos^{-1}\left(\frac{2}{3\sqrt{6}}\right) \).
In simple words: When planes are given as equations with x, y, z, you take the numbers in front of x, y, z for each plane. These numbers are like the parts of the normal vector for that plane. Then you use a special formula that combines these numbers to find the cosine of the angle between the planes.

🎯 Exam Tip: Ensure you correctly identify \( a_1, b_1, c_1 \) and \( a_2, b_2, c_2 \) from the plane equations. A common mistake is to miss negative signs or mix up coefficients.

Question 3. Show that following planes are at right angles:
(i) \( x-2y+4z = 10 \) and \( 18x + 17y + 4z = 49 \)
(ii) \( \vec{r}.(2\hat{i}-\hat{j}+\hat{k}) = 4 \) and \( \vec{r}.(-\hat{i}-\hat{j}+\hat{k}) =3 \)
Answer:
Two planes are at right angles (perpendicular) if their normal vectors are perpendicular. This means the dot product of their normal vectors must be zero \( (\vec{n_1} \cdot \vec{n_2} = 0) \).
(i) Given planes are \( x-2y+4z = 10 \) and \( 18x + 17y + 4z = 49 \).
The normal vector for the first plane is \( \vec{n_1} = \hat{i}-2\hat{j}+4\hat{k} \).
The normal vector for the second plane is \( \vec{n_2} = 18\hat{i}+17\hat{j}+4\hat{k} \).
Now, calculate their dot product:
\( \vec{n_1} \cdot \vec{n_2} = (1)(18) + (-2)(17) + (4)(4) \)
\( \implies \vec{n_1} \cdot \vec{n_2} = 18 - 34 + 16 \)
\( \implies \vec{n_1} \cdot \vec{n_2} = 34 - 34 \)
\( \implies \vec{n_1} \cdot \vec{n_2} = 0 \)
Since the dot product is 0, the normal vectors are perpendicular, which means the planes are at right angles.
(ii) Given planes are \( \vec{r}.(2\hat{i}-\hat{j}+\hat{k}) = 4 \) and \( \vec{r}.(-\hat{i}-\hat{j}+\hat{k}) = 3 \).
The normal vector for the first plane is \( \vec{n_1} = 2\hat{i}-\hat{j}+\hat{k} \).
The normal vector for the second plane is \( \vec{n_2} = -\hat{i}-\hat{j}+\hat{k} \).
Now, calculate their dot product:
\( \vec{n_1} \cdot \vec{n_2} = (2)(-1) + (-1)(-1) + (1)(1) \)
\( \implies \vec{n_1} \cdot \vec{n_2} = -2 + 1 + 1 \)
\( \implies \vec{n_1} \cdot \vec{n_2} = 0 \)
Since the dot product is 0, the normal vectors are perpendicular, and thus the planes are at right angles. The concept of perpendicularity is fundamental in geometry, allowing us to determine orthogonal relationships between objects.
In simple words: To check if two planes meet at a perfect right angle, we look at the vectors that stick straight out from each plane (called normal vectors). If these two normal vectors are perpendicular to each other (meaning their dot product is zero), then the planes themselves are also perpendicular.

🎯 Exam Tip: For planes to be perpendicular, the dot product of their normal vectors must be zero. This is a key condition to remember for such problems.

Question 4. Find \( \lambda \), if following planes are perpendicular to each other:
(i) \( \vec{r}.(2\hat{i}-\hat{j}+\lambda\hat{k}) = 5 \) and \( \vec{r}.(3\hat{i}+2\hat{j}+2\hat{k}) = 4 \)
(ii) \( 2x - 4y + 3z = 5 \) and \( x + 2y + \lambda z = 5 \)
Answer:
For two planes to be perpendicular, the dot product of their normal vectors must be zero \( (\vec{n_1} \cdot \vec{n_2} = 0) \).
(i) Given planes are \( \vec{r}.(2\hat{i}-\hat{j}+\lambda\hat{k}) = 5 \) and \( \vec{r}.(3\hat{i}+2\hat{j}+2\hat{k}) = 4 \).
The normal vector for the first plane is \( \vec{n_1} = 2\hat{i}-\hat{j}+\lambda\hat{k} \).
The normal vector for the second plane is \( \vec{n_2} = 3\hat{i}+2\hat{j}+2\hat{k} \).
Since the planes are perpendicular, their dot product must be zero:
\( \vec{n_1} \cdot \vec{n_2} = 0 \)
\( \implies (2)(3) + (-1)(2) + (\lambda)(2) = 0 \)
\( \implies 6 - 2 + 2\lambda = 0 \)
\( \implies 4 + 2\lambda = 0 \)
\( \implies 2\lambda = -4 \)
\( \implies \lambda = -2 \)
So, the value of \( \lambda \) is -2.
(ii) Given planes are \( 2x - 4y + 3z = 5 \) and \( x + 2y + \lambda z = 5 \).
The normal vector for the first plane is \( \vec{n_1} = 2\hat{i}-4\hat{j}+3\hat{k} \).
The normal vector for the second plane is \( \vec{n_2} = \hat{i}+2\hat{j}+\lambda\hat{k} \).
Since the planes are perpendicular, their dot product must be zero:
\( \vec{n_1} \cdot \vec{n_2} = 0 \)
\( \implies (2)(1) + (-4)(2) + (3)(\lambda) = 0 \)
\( \implies 2 - 8 + 3\lambda = 0 \)
\( \implies -6 + 3\lambda = 0 \)
\( \implies 3\lambda = 6 \)
\( \implies \lambda = 2 \)
So, the value of \( \lambda \) is 2. Finding unknown coefficients like \( \lambda \) is a common application of vector properties in geometry.
In simple words: If planes are perpendicular, the multiplication and addition of the matching numbers from their equations must equal zero. You set up this equation with the unknown value (like \( \lambda \)) and then solve for it.

🎯 Exam Tip: Clearly write down the normal vectors \( \vec{n_1} \) and \( \vec{n_2} \) for each plane before applying the dot product condition. This helps avoid calculation errors.

Question 5. Find the angle between the line \( \frac {x+1}{3} = \frac {y-1}{2 } = \frac { z-2 }{4} \) and the plane \( 2x + y - 3z + 4 = 0 \).
Answer:
To find the angle \( \theta \) between a line and a plane, we use the formula involving the direction vector of the line \( \vec{b} \) and the normal vector of the plane \( \vec{n} \):
\( \sin \theta = \frac{ | \vec{n} \cdot \vec{b} | }{ | \vec{n} | | \vec{b} | } \)
Given the line \( \frac {x+1}{3} = \frac {y-1}{2 } = \frac { z-2 }{4} \), its direction vector is \( \vec{b} = 3\hat{i}+2\hat{j}+4\hat{k} \).
Given the plane \( 2x + y - 3z + 4 = 0 \), its normal vector is \( \vec{n} = 2\hat{i}+\hat{j}-3\hat{k} \).
First, calculate the dot product \( \vec{n} \cdot \vec{b} \):
\( \vec{n} \cdot \vec{b} = (2)(3) + (1)(2) + (-3)(4) \)
\( \implies \vec{n} \cdot \vec{b} = 6 + 2 - 12 \)
\( \implies \vec{n} \cdot \vec{b} = -4 \)
Next, calculate the magnitudes of \( \vec{n} \) and \( \vec{b} \):
\( | \vec{n} | = \sqrt{(2)^2 + (1)^2 + (-3)^2} = \sqrt{4+1+9} = \sqrt{14} \)
\( | \vec{b} | = \sqrt{(3)^2 + (2)^2 + (4)^2} = \sqrt{9+4+16} = \sqrt{29} \)
Now, substitute these values into the formula for \( \sin \theta \):
\( \sin \theta = \frac{ |-4| }{ \sqrt{14} \sqrt{29} } \)
\( \implies \sin \theta = \frac{ 4 }{ \sqrt{14 \times 29} } \)
\( \implies \sin \theta = \frac{ 4 }{ \sqrt{406} } \)
So, the angle is \( \theta = \sin^{-1}\left(\frac{4}{\sqrt{406}}\right) \). This angle represents the tilt of the line relative to the plane.
In simple words: To find the angle between a line and a flat surface (plane), we use a special formula. We need the "direction vector" of the line and the "normal vector" of the plane. The formula uses their dot product and lengths to find the sine of the angle, and then we take the inverse sine to get the actual angle.

🎯 Exam Tip: For line-plane angle, use the sine formula, not cosine. Also, ensure you extract the correct direction ratios for the line and normal vector components for the plane.

Question 6. Find the angle between the line \( \frac {x-2}{3} = \frac {y+1 }{-1} = \frac { z-3 }{ 2 } \) and the plane \( 3x + 4y + z + 5 = 0 \).
Answer:
To find the angle \( \theta \) between a line and a plane, we use the formula:
\( \sin \theta = \frac{ | \vec{n} \cdot \vec{b} | }{ | \vec{n} | | \vec{b} | } \)
Given the line \( \frac {x-2}{3} = \frac {y+1 }{-1} = \frac { z-3 }{ 2 } \), its direction vector is \( \vec{b} = 3\hat{i}-1\hat{j}+2\hat{k} \).
Given the plane \( 3x + 4y + z + 5 = 0 \), its normal vector is \( \vec{n} = 3\hat{i}+4\hat{j}+1\hat{k} \).
First, calculate the dot product \( \vec{n} \cdot \vec{b} \):
\( \vec{n} \cdot \vec{b} = (3)(3) + (4)(-1) + (1)(2) \)
\( \implies \vec{n} \cdot \vec{b} = 9 - 4 + 2 \)
\( \implies \vec{n} \cdot \vec{b} = 7 \)
Next, calculate the magnitudes of \( \vec{n} \) and \( \vec{b} \):
\( | \vec{n} | = \sqrt{(3)^2 + (4)^2 + (1)^2} = \sqrt{9+16+1} = \sqrt{26} \)
\( | \vec{b} | = \sqrt{(3)^2 + (-1)^2 + (2)^2} = \sqrt{9+1+4} = \sqrt{14} \)
Now, substitute these values into the formula for \( \sin \theta \):
\( \sin \theta = \frac{ |7| }{ \sqrt{26} \sqrt{14} } \)
\( \implies \sin \theta = \frac{ 7 }{ \sqrt{26 \times 14} } \)
\( \implies \sin \theta = \frac{ 7 }{ \sqrt{364} } \)
\( \implies \sin \theta = \frac{ 7 }{ \sqrt{52 \times 7} } = \frac{ 7 }{ \sqrt{52}\sqrt{7} } = \frac{ \sqrt{7} }{ \sqrt{52} } = \sqrt{\frac{7}{52}} \)
So, the angle is \( \theta = \sin^{-1}\left(\frac{7}{\sqrt{364}}\right) \) or \( \theta = \sin^{-1}\left(\sqrt{\frac{7}{52}}\right) \). Calculating angles between lines and planes is essential for understanding spatial relationships in three dimensions.
In simple words: This is like the last problem. You find the line's direction numbers and the plane's normal numbers. Put them into the sine formula, then solve for the angle using inverse sine. Make sure to simplify the square roots if possible.

🎯 Exam Tip: Simplify the square roots in the denominator if possible, as it often leads to a cleaner final answer. Double-check the calculations for dot products and magnitudes.

Question 7. Find the angle between the line \( \vec{r} = (\hat{i} + 2\hat{j} - \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k}) \) and the plane \( \vec{r}.(2\hat{i} - \hat{j} + \hat{k}) = 4 \).
Answer:
The angle \( \theta \) between a line and a plane is given by:
\( \sin \theta = \frac{ | \vec{n} \cdot \vec{b} | }{ | \vec{n} | | \vec{b} | } \)
Given the line \( \vec{r} = (\hat{i} + 2\hat{j} - \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k}) \), its direction vector is \( \vec{b} = \hat{i}-\hat{j}+\hat{k} \).
Given the plane \( \vec{r}.(2\hat{i} - \hat{j} + \hat{k}) = 4 \), its normal vector is \( \vec{n} = 2\hat{i}-\hat{j}+\hat{k} \).
First, calculate the dot product \( \vec{n} \cdot \vec{b} \):
\( \vec{n} \cdot \vec{b} = (2)(1) + (-1)(-1) + (1)(1) \)
\( \implies \vec{n} \cdot \vec{b} = 2 + 1 + 1 \)
\( \implies \vec{n} \cdot \vec{b} = 4 \)
Next, calculate the magnitudes of \( \vec{n} \) and \( \vec{b} \):
\( | \vec{n} | = \sqrt{(2)^2 + (-1)^2 + (1)^2} = \sqrt{4+1+1} = \sqrt{6} \)
\( | \vec{b} | = \sqrt{(1)^2 + (-1)^2 + (1)^2} = \sqrt{1+1+1} = \sqrt{3} \)
Now, substitute these values into the formula for \( \sin \theta \):
\( \sin \theta = \frac{ |4| }{ \sqrt{6} \sqrt{3} } \)
\( \implies \sin \theta = \frac{ 4 }{ \sqrt{18} } = \frac{ 4 }{ 3\sqrt{2} } \)
\( \implies \sin \theta = \frac{ 4\sqrt{2} }{ 3\sqrt{2}\sqrt{2} } = \frac{ 4\sqrt{2} }{ 6 } = \frac{ 2\sqrt{2} }{ 3 } \)
So, the angle is \( \theta = \sin^{-1}\left(\frac{4}{\sqrt{18}}\right) \) or \( \theta = \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right) \). Understanding how to convert vector equations to extract key information is a crucial skill.
In simple words: This problem asks for the angle between a line and a plane, both given in vector form. You pull out the direction vector from the line and the normal vector from the plane. Then use the sine formula with their dot product and lengths to find the angle.

🎯 Exam Tip: When given vector equations, clearly identify the direction vector \( \vec{b} \) for the line and the normal vector \( \vec{n} \) for the plane before proceeding with calculations.

Question 8. Find the angle between the line \( \vec{r} = (2\hat{i} - 3\hat{j} + \hat{k}) + \lambda(\hat{i} + 2\hat{j} - \hat{k}) \) and the plane \( \vec{r}.(2\hat{i} - \hat{j} + \hat{k}) = 4 \).
Answer:
The angle \( \theta \) between a line and a plane is given by:
\( \sin \theta = \frac{ | \vec{n} \cdot \vec{b} | }{ | \vec{n} | | \vec{b} | } \)
Given the line \( \vec{r} = (2\hat{i} - 3\hat{j} + \hat{k}) + \lambda(\hat{i} + 2\hat{j} - \hat{k}) \), its direction vector is \( \vec{b} = \hat{i}+2\hat{j}-\hat{k} \).
Given the plane \( \vec{r}.(2\hat{i} - \hat{j} + \hat{k}) = 4 \), its normal vector is \( \vec{n} = 2\hat{i}-\hat{j}+\hat{k} \).
First, calculate the dot product \( \vec{n} \cdot \vec{b} \):
\( \vec{n} \cdot \vec{b} = (2)(1) + (-1)(2) + (1)(-1) \)
\( \implies \vec{n} \cdot \vec{b} = 2 - 2 - 1 \)
\( \implies \vec{n} \cdot \vec{b} = -1 \)
Next, calculate the magnitudes of \( \vec{n} \) and \( \vec{b} \):
\( | \vec{n} | = \sqrt{(2)^2 + (-1)^2 + (1)^2} = \sqrt{4+1+1} = \sqrt{6} \)
\( | \vec{b} | = \sqrt{(1)^2 + (2)^2 + (-1)^2} = \sqrt{1+4+1} = \sqrt{6} \)
Now, substitute these values into the formula for \( \sin \theta \):
\( \sin \theta = \frac{ |-1| }{ \sqrt{6} \sqrt{6} } \)
\( \implies \sin \theta = \frac{ 1 }{ 6 } \)
So, the angle is \( \theta = \sin^{-1}\left(\frac{1}{6}\right) \). When the dot product is negative, it simply means the angle between the vectors is obtuse, but we take the absolute value for the angle between geometric objects.
In simple words: Just like before, identify the direction vector of the line and the normal vector of the plane from their equations. Plug these vectors into the sine formula, then solve for the angle using the inverse sine function.

🎯 Exam Tip: The constant term in the line equation (e.g., \( \hat{i} + 2\hat{j} - \hat{k} \) in \( \vec{r} = (\hat{i} + 2\hat{j} - \hat{k}) + \lambda(...) \)) indicates a point on the line but is not used to find the line's direction vector.

Question 9. Determine the value of m, if line \( \vec{r} = (\hat{i} - 2\hat{j} + \hat{k}) + \lambda(2\hat{i} + \hat{j} + 2\hat{k}) \) is parallel to the plane \( \vec{r}.(3\hat{i} - 2\hat{j} + m\hat{k}) = 3 \).
Answer:
For a line to be parallel to a plane, the direction vector of the line \( \vec{b} \) must be perpendicular to the normal vector of the plane \( \vec{n} \). This means their dot product must be zero: \( \vec{b} \cdot \vec{n} = 0 \).
Given the line \( \vec{r} = (\hat{i} - 2\hat{j} + \hat{k}) + \lambda(2\hat{i} + \hat{j} + 2\hat{k}) \), its direction vector is \( \vec{b} = 2\hat{i}+\hat{j}+2\hat{k} \).
Given the plane \( \vec{r}.(3\hat{i} - 2\hat{j} + m\hat{k}) = 3 \), its normal vector is \( \vec{n} = 3\hat{i}-2\hat{j}+m\hat{k} \).
Since the line is parallel to the plane, we set their dot product to zero:
\( \vec{b} \cdot \vec{n} = 0 \)
\( \implies (2)(3) + (1)(-2) + (2)(m) = 0 \)
\( \implies 6 - 2 + 2m = 0 \)
\( \implies 4 + 2m = 0 \)
\( \implies 2m = -4 \)
\( \implies m = -2 \)
So, the value of m is -2. This condition ensures that the line never intersects the plane, maintaining parallelism.
In simple words: When a line runs parallel to a plane, it means the line's direction is at a right angle to the plane's "normal" (outward-pointing) direction. So, you multiply the matching vector components and add them up, setting the total to zero. Then you solve for the unknown 'm'.

🎯 Exam Tip: Remember the critical condition: line parallel to plane means \( \vec{b} \cdot \vec{n} = 0 \). Do not confuse this with the angle formula for line and plane, which uses sine.

Question 10. Find m, if line \( \vec{r} = \hat{i} + \lambda(2\hat{i} - m\hat{j} - 3\hat{k}) \) is parallel to the plane \( \vec{r}.(m\hat{i} + 3\hat{j} + \hat{k}) = 4 \).
Answer:
For a line to be parallel to a plane, the direction vector of the line \( \vec{b} \) must be perpendicular to the normal vector of the plane \( \vec{n} \). This means their dot product must be zero: \( \vec{b} \cdot \vec{n} = 0 \).
Given the line \( \vec{r} = \hat{i} + \lambda(2\hat{i} - m\hat{j} - 3\hat{k}) \), its direction vector is \( \vec{b} = 2\hat{i}-m\hat{j}-3\hat{k} \).
Given the plane \( \vec{r}.(m\hat{i} + 3\hat{j} + \hat{k}) = 4 \), its normal vector is \( \vec{n} = m\hat{i}+3\hat{j}+\hat{k} \).
Since the line is parallel to the plane, we set their dot product to zero:
\( \vec{b} \cdot \vec{n} = 0 \)
\( \implies (2)(m) + (-m)(3) + (-3)(1) = 0 \)
\( \implies 2m - 3m - 3 = 0 \)
\( \implies -m - 3 = 0 \)
\( \implies -m = 3 \)
\( \implies m = -3 \)
So, the value of m is -3. This condition is crucial for determining if a line and plane are coplanar or simply never intersect.
In simple words: If a line is parallel to a plane, the direction of the line is perpendicular to the plane's normal direction. So, the dot product of the line's direction vector and the plane's normal vector must be zero. Solve this equation to find 'm'.

🎯 Exam Tip: Pay close attention to negative signs and the placement of the unknown variable 'm' in both the line's direction vector and the plane's normal vector.