Samacheer Kalvi Class 12 Maths Solutions Chapter 6 Applications of Vector Algebra Exercise 6.8

Question 1. Show that the straight lines \( \overline { r } = (5\hat { i } + 7\hat { j } – 3\hat { k }) + s(4\hat { i } + 4\hat { j } – 5\hat { k }) \) and \( \overline { r } = (8\hat { i } + 4\hat { j } + 5\hat { k }) + t(7\hat { i } + \hat { j } + 3\hat { k }) \) are coplanar. Find the vector equation of the plane in which they lie.
Answer:
Given the equations of the straight lines:
Line 1: \( \overline { r } = \overline { a } + s\overline { b } \)
Here, \( \overline { a } = 5\hat { i } + 7\hat { j } - 3\hat { k } \) and \( \overline { b } = 4\hat { i } + 4\hat { j } - 5\hat { k } \)
Line 2: \( \overline { r } = \overline { c } + t\overline { d } \)
Here, \( \overline { c } = 8\hat { i } + 4\hat { j } + 5\hat { k } \) and \( \overline { d } = 7\hat { i } + \hat { j } + 3\hat { k } \)

First, find \( \overline { c } - \overline { a } \):
\( \overline { c } - \overline { a } = (8\hat { i } + 4\hat { j } + 5\hat { k }) - (5\hat { i } + 7\hat { j } - 3\hat { k }) \)
\( \overline { c } - \overline { a } = (8-5)\hat { i } + (4-7)\hat { j } + (5-(-3))\hat { k } \)
\( \overline { c } - \overline { a } = 3\hat { i } - 3\hat { j } + 8\hat { k } \)

Next, find \( \overline { b } \times \overline { d } \):
\( \overline { b } \times \overline { d } = \begin{vmatrix} \hat { i } & \hat { j } & \hat { k } \\ 4 & 4 & -5 \\ 7 & 1 & 3 \end{vmatrix} \)
\( = \hat { i } (4 \times 3 - (-5) \times 1) - \hat { j } (4 \times 3 - (-5) \times 7) + \hat { k } (4 \times 1 - 4 \times 7) \)
\( = \hat { i } (12 + 5) - \hat { j } (12 + 35) + \hat { k } (4 - 28) \)
\( = 17\hat { i } - 47\hat { j } - 24\hat { k } \)

For the lines to be coplanar, the scalar triple product \( (\overline { c } - \overline { a }) \cdot (\overline { b } \times \overline { d }) \) must be zero.
\( (\overline { c } - \overline { a }) \cdot (\overline { b } \times \overline { d }) = (3\hat { i } - 3\hat { j } + 8\hat { k }) \cdot (17\hat { i } - 47\hat { j } - 24\hat { k }) \)
\( = (3 \times 17) + (-3 \times -47) + (8 \times -24) \)
\( = 51 + 141 - 192 \)
\( = 192 - 192 \)
\( = 0 \)
Since the scalar triple product is zero, the lines are coplanar.

Now, find the vector equation of the plane containing these lines. The equation of the plane is given by \( (\overline { r } - \overline { a }) \cdot (\overline { b } \times \overline { d }) = 0 \).
Substitute the values of \( \overline { a } \) and \( \overline { b } \times \overline { d } \):
\( [\overline { r } - (5\hat { i } + 7\hat { j } - 3\hat { k })] \cdot (17\hat { i } - 47\hat { j } - 24\hat { k }) = 0 \)
This can be written as:
\( \overline { r } \cdot (17\hat { i } - 47\hat { j } - 24\hat { k }) - (5\hat { i } + 7\hat { j } - 3\hat { k }) \cdot (17\hat { i } - 47\hat { j } - 24\hat { k }) = 0 \)
Calculate the dot product: \( (5\hat { i } + 7\hat { j } - 3\hat { k }) \cdot (17\hat { i } - 47\hat { j } - 24\hat { k }) \)
\( = (5 \times 17) + (7 \times -47) + (-3 \times -24) \)
\( = 85 - 329 + 72 \)
\( = 157 - 329 \)
\( = -172 \)
So, the vector equation of the plane is:
\( \overline { r } \cdot (17\hat { i } - 47\hat { j } - 24\hat { k }) - (-172) = 0 \)
\( \overline { r } \cdot (17\hat { i } - 47\hat { j } - 24\hat { k }) + 172 = 0 \)
The dot product is a way to check how much two vectors point in the same direction.
In simple words: First, we check if the lines lie on the same flat surface by doing a special calculation with their direction vectors. If the answer is zero, they are on the same plane. Then, we use the points and directions of the lines to find the equation that describes this plane.

🎯 Exam Tip: Remember the condition for coplanarity: \( (\overline { c } - \overline { a }) \cdot (\overline { b } \times \overline { d }) = 0 \). This is a crucial formula to apply correctly.

Question 2. Show that the lines \( \frac { x-2 }{ 1 } = \frac { y-3 }{ 1 } = \frac { z -4}{ 3 } \) and \( \frac { x-1 }{ -3 } = \frac { y-4 }{ 2 } = \frac { z-5 }{ 1 } \) are coplanar. Also, find the plane containing these lines.
Answer:
Given the equations of the straight lines in Cartesian form:
Line 1: \( \frac { x-x_1 }{ l_1 } = \frac { y-y_1 }{ m_1 } = \frac { z-z_1 }{ n_1 } \)
Here, \( (x_1, y_1, z_1) = (2, 3, 4) \) and \( (l_1, m_1, n_1) = (1, 1, 3) \)
Line 2: \( \frac { x-x_2 }{ l_2 } = \frac { y-y_2 }{ m_2 } = \frac { z-z_2 }{ n_2 } \)
Here, \( (x_2, y_2, z_2) = (1, 4, 5) \) and \( (l_2, m_2, n_2) = (-3, 2, 1) \)

For the lines to be coplanar, the determinant of the following matrix must be zero:
\( \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = 0 \)
First, calculate \( (x_2-x_1), (y_2-y_1), (z_2-z_1) \):
\( x_2-x_1 = 1-2 = -1 \)
\( y_2-y_1 = 4-3 = 1 \)
\( z_2-z_1 = 5-4 = 1 \)

Now, form the determinant:
\( \begin{vmatrix} -1 & 1 & 1 \\ 1 & 1 & 3 \\ -3 & 2 & 1 \end{vmatrix} \)
\( = -1((1 \times 1) - (3 \times 2)) - 1((1 \times 1) - (3 \times -3)) + 1((1 \times 2) - (1 \times -3)) \)
\( = -1(1 - 6) - 1(1 - (-9)) + 1(2 - (-3)) \)
\( = -1(-5) - 1(1 + 9) + 1(2 + 3) \)
\( = 5 - 1(10) + 1(5) \)
\( = 5 - 10 + 5 \)
\( = 0 \)
Since the determinant is zero, the lines are coplanar.

Now, find the equation of the plane containing these lines. The equation of the plane is given by:
\( \begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = 0 \)
Substitute \( (x_1, y_1, z_1) = (2, 3, 4) \) and the direction ratios:
\( \begin{vmatrix} x-2 & y-3 & z-4 \\ 1 & 1 & 3 \\ -3 & 2 & 1 \end{vmatrix} = 0 \)
Expand the determinant:
\( (x-2)((1 \times 1) - (3 \times 2)) - (y-3)((1 \times 1) - (3 \times -3)) + (z-4)((1 \times 2) - (1 \times -3)) = 0 \)
\( (x-2)(1 - 6) - (y-3)(1 - (-9)) + (z-4)(2 - (-3)) = 0 \)
\( (x-2)(-5) - (y-3)(10) + (z-4)(5) = 0 \)
Divide the entire equation by -5 to simplify:
\( (x-2) + 2(y-3) - (z-4) = 0 \)
\( x - 2 + 2y - 6 - z + 4 = 0 \)
\( x + 2y - z - 4 = 0 \)
This is the Cartesian equation of the plane. This plane contains all points that make the equation true.
In simple words: We check if two lines lie on the same flat surface by putting their details into a special math table called a determinant. If the answer is zero, they are on the same plane. Then, we use the details of one line and the directions of both lines to write the equation for this plane.

🎯 Exam Tip: When using Cartesian forms, remember to correctly identify \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) from the given equations, as these are the points the lines pass through.

Question 3. If the straight lines \( \frac { x-1 }{ 1 } = \frac { y-2 }{ 2 } = \frac { z-3}{ m^2 } \) and \( \frac { x-3 }{ 1 } = \frac { y-2 }{ m^2 } = \frac { z-1 }{ 2 } \) are coplanar, find the distinct real values of m.
Answer:
Given the equations of the straight lines:
Line 1: \( (x_1, y_1, z_1) = (1, 2, 3) \) and \( (l_1, m_1, n_1) = (1, 2, m^2) \)
Line 2: \( (x_2, y_2, z_2) = (3, 2, 1) \) and \( (l_2, m_2, n_2) = (1, m^2, 2) \)

For the lines to be coplanar, the determinant must be zero:
\( \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = 0 \)
Calculate \( (x_2-x_1), (y_2-y_1), (z_2-z_1) \):
\( x_2-x_1 = 3-1 = 2 \)
\( y_2-y_1 = 2-2 = 0 \)
\( z_2-z_1 = 1-3 = -2 \)

Substitute these values into the determinant:
\( \begin{vmatrix} 2 & 0 & -2 \\ 1 & 2 & m^2 \\ 1 & m^2 & 2 \end{vmatrix} = 0 \)
Expand the determinant:
\( 2((2 \times 2) - (m^2 \times m^2)) - 0((1 \times 2) - (m^2 \times 1)) + (-2)((1 \times m^2) - (2 \times 1)) = 0 \)
\( 2(4 - m^4) - 0(2 - m^2) - 2(m^2 - 2) = 0 \)
\( 8 - 2m^4 - 2m^2 + 4 = 0 \)
\( 12 - 2m^4 - 2m^2 = 0 \)
Divide the entire equation by -2:
\( -6 + m^4 + m^2 = 0 \)
Rearrange the terms:
\( m^4 + m^2 - 6 = 0 \)
This is a quadratic equation in terms of \( m^2 \). Let \( M = m^2 \).
\( M^2 + M - 6 = 0 \)
Factor the quadratic equation:
\( (M+3)(M-2) = 0 \)
So, \( M = -3 \) or \( M = 2 \).
Since \( M = m^2 \), we have:
\( m^2 = -3 \) or \( m^2 = 2 \)
For real values of m, \( m^2 \) cannot be negative.
Therefore, \( m^2 = -3 \) is not possible for real m.
This leaves \( m^2 = 2 \).
So, \( m = \pm \sqrt{2} \).
The distinct real values of m are \( \sqrt{2} \) and \( -\sqrt{2} \). Solving for 'm' involves standard algebraic steps.
In simple words: We are given two lines with a hidden value 'm'. For these lines to lie on the same flat surface, a special math calculation must equal zero. We solve this calculation to find 'm'. Since 'm' squared cannot be a negative number, we only pick the real solutions.

🎯 Exam Tip: Be careful with the algebraic manipulation when expanding the determinant and solving for \( m^2 \). Always check for real vs. imaginary solutions, especially when dealing with squares.

Question 4. If the straight lines \( \frac { x-1 }{ 2 } = \frac { y+1 }{ \lambda } = \frac { z }{ 2 } \) and \( \frac { x+1 }{ 5 } = \frac { y+1 }{ 2 } = \frac { z }{ \lambda } \) are coplanar, find \( \lambda \) and equations of the planes containing these two lines.
Answer:
Given the equations of the straight lines:
Line 1: \( \frac { x-x_1 }{ b_1 } = \frac { y-y_1 }{ b_2 } = \frac { z-z_1 }{ b_3 } \)
Here, \( (x_1, y_1, z_1) = (1, -1, 0) \) and \( (b_1, b_2, b_3) = (2, \lambda, 2) \)
Line 2: \( \frac { x-x_2 }{ d_1 } = \frac { y-y_2 }{ d_2 } = \frac { z-z_2 }{ d_3 } \)
Here, \( (x_2, y_2, z_2) = (-1, -1, 0) \) and \( (d_1, d_2, d_3) = (5, 2, \lambda) \)

For the lines to be coplanar, the determinant of the following matrix must be zero:
\( \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ b_1 & b_2 & b_3 \\ d_1 & d_2 & d_3 \end{vmatrix} = 0 \)
First, calculate \( (x_2-x_1), (y_2-y_1), (z_2-z_1) \):
\( x_2-x_1 = -1-1 = -2 \)
\( y_2-y_1 = -1-(-1) = 0 \)
\( z_2-z_1 = 0-0 = 0 \)

Now, form the determinant:
\( \begin{vmatrix} -2 & 0 & 0 \\ 2 & \lambda & 2 \\ 5 & 2 & \lambda \end{vmatrix} = 0 \)
Expand the determinant along the first row:
\( -2(\lambda \times \lambda - 2 \times 2) - 0(2\lambda - 10) + 0(4 - 5\lambda) = 0 \)
\( -2(\lambda^2 - 4) = 0 \)
\( \lambda^2 - 4 = 0 \)
\( \lambda^2 = 4 \)
\( \lambda = \pm 2 \)
So, the possible values for \( \lambda \) are 2 and -2.

Now we find the equation of the plane for each value of \( \lambda \). The equation of the plane is given by:
\( \begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ b_1 & b_2 & b_3 \\ d_1 & d_2 & d_3 \end{vmatrix} = 0 \)

**Case 1: When \( \lambda = 2 \)**
The points and direction ratios are:
\( (x_1, y_1, z_1) = (1, -1, 0) \)
\( (b_1, b_2, b_3) = (2, 2, 2) \)
\( (d_1, d_2, d_3) = (5, 2, 2) \)
Substitute these into the determinant for the plane equation:
\( \begin{vmatrix} x-1 & y-(-1) & z-0 \\ 2 & 2 & 2 \\ 5 & 2 & 2 \end{vmatrix} = 0 \)
\( \begin{vmatrix} x-1 & y+1 & z \\ 2 & 2 & 2 \\ 5 & 2 & 2 \end{vmatrix} = 0 \)
Notice that the second and third rows are proportional (in fact, identical). If two rows of a determinant are identical, the determinant is zero. So, \( 0 = 0 \). This means that for \( \lambda = 2 \), the lines are coincident or parallel, and any plane containing one also contains the other. However, for distinct lines, this usually indicates an error or a case where the plane is not uniquely determined. Let's recheck the calculation of \( \overline{b} \times \overline{d} \) as this is the normal vector to the plane.
When \( \lambda = 2 \), \( \overline{b} = 2\hat{i} + 2\hat{j} + 2\hat{k} \) and \( \overline{d} = 5\hat{i} + 2\hat{j} + 2\hat{k} \).
\( \overline{b} \times \overline{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 2 \\ 5 & 2 & 2 \end{vmatrix} = \hat{i}(4-4) - \hat{j}(4-10) + \hat{k}(4-10) = 0\hat{i} + 6\hat{j} - 6\hat{k} = 6\hat{j} - 6\hat{k} \)
The plane equation is \( (\overline { r } - \overline { a }) \cdot (\overline { b } \times \overline { d }) = 0 \).
\( [\overline { r } - (\hat { i } - \hat { j } + 0\hat { k })] \cdot (6\hat { j } - 6\hat { k }) = 0 \)
Let \( \overline { r } = x\hat { i } + y\hat { j } + z\hat { k } \).
\( (x\hat { i } + y\hat { j } + z\hat { k } - \hat { i } + \hat { j }) \cdot (6\hat { j } - 6\hat { k }) = 0 \)
\( ((x-1)\hat { i } + (y+1)\hat { j } + z\hat { k }) \cdot (6\hat { j } - 6\hat { k }) = 0 \)
\( (x-1)(0) + (y+1)(6) + z(-6) = 0 \)
\( 6(y+1) - 6z = 0 \)
Divide by 6:
\( y+1-z = 0 \)
\( y - z + 1 = 0 \)

**Case 2: When \( \lambda = -2 \)**
The points and direction ratios are:
\( (x_1, y_1, z_1) = (1, -1, 0) \)
\( (b_1, b_2, b_3) = (2, -2, 2) \)
\( (d_1, d_2, d_3) = (5, 2, -2) \)
Substitute these into the determinant for the plane equation:
\( \begin{vmatrix} x-1 & y-(-1) & z-0 \\ 2 & -2 & 2 \\ 5 & 2 & -2 \end{vmatrix} = 0 \)
\( \begin{vmatrix} x-1 & y+1 & z \\ 2 & -2 & 2 \\ 5 & 2 & -2 \end{vmatrix} = 0 \)
Expand the determinant:
\( (x-1)((-2 \times -2) - (2 \times 2)) - (y+1)((2 \times -2) - (2 \times 5)) + z((2 \times 2) - (-2 \times 5)) = 0 \)
\( (x-1)(4 - 4) - (y+1)(-4 - 10) + z(4 - (-10)) = 0 \)
\( (x-1)(0) - (y+1)(-14) + z(4 + 10) = 0 \)
\( 0 + 14(y+1) + 14z = 0 \)
Divide by 14:
\( y+1+z = 0 \)
\( y + z + 1 = 0 \)
These calculations find the normal vector to the plane, which helps define its orientation in space.
In simple words: We have two lines with an unknown value called lambda. For these lines to lie on the same flat surface, a special math rule must be followed. We solve this rule to find the possible values for lambda. Once we have lambda, we use it to find the equations of the flat surfaces (planes) where these lines can exist.

🎯 Exam Tip: When finding the plane equation, ensure you correctly use the point from one line and the direction vectors of both lines to form the determinant. Double-check calculations when expanding determinants, especially with negative numbers and variables like lambda.