Samacheer Kalvi Class 12 Maths Solutions Chapter 6 Applications of Vector Algebra Exercise 6.7

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.7

Question 1. Find the non-parametric form of vector equation and Cartesian equation of the plane passing through the point (2, 3, 6) and parallel to the straight lines: \( \frac {x-1 }{ 2 } = \frac { y+1 }{ 3 } = \frac { z-3 }{ 1 } \) and \( \frac { x+3 }{ 2 } = \frac { y-3 }{ -5 } = \frac { z+1 }{ -3 } \)
Answer:
Let the given point be \( \vec{a} = 2\hat{i} + 3\hat{j} + 6\hat{k} \).
The direction vectors of the two parallel lines are:
\( \vec{b} = 2\hat{i} + 3\hat{j} + \hat{k} \)
\( \vec{c} = 2\hat{i} - 5\hat{j} - 3\hat{k} \)
The normal vector to the plane is perpendicular to both \( \vec{b} \) and \( \vec{c} \). So, we find the cross product \( \vec{b} \times \vec{c} \).
\( \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 2 & -5 & -3 \end{vmatrix} \)
\( \implies \hat{i}((3)(-3) - (1)(-5)) - \hat{j}((2)(-3) - (1)(2)) + \hat{k}((2)(-5) - (3)(2)) \)
\( \implies \hat{i}(-9 + 5) - \hat{j}(-6 - 2) + \hat{k}(-10 - 6) \)
\( \implies -4\hat{i} + 8\hat{j} - 16\hat{k} \)
This can be written as \( -4(\hat{i} - 2\hat{j} + 4\hat{k}) \). Let \( \vec{n} = \hat{i} - 2\hat{j} + 4\hat{k} \).
The non-parametric vector equation of the plane passing through \( \vec{a} \) with normal \( \vec{n} \) is \( (\vec{r} - \vec{a}) \cdot \vec{n} = 0 \).
\( [\vec{r} - (2\hat{i} + 3\hat{j} + 6\hat{k})] \cdot (\hat{i} - 2\hat{j} + 4\hat{k}) = 0 \)
\( \implies \vec{r} \cdot (\hat{i} - 2\hat{j} + 4\hat{k}) - (2\hat{i} + 3\hat{j} + 6\hat{k}) \cdot (\hat{i} - 2\hat{j} + 4\hat{k}) = 0 \)
\( \implies \vec{r} \cdot (\hat{i} - 2\hat{j} + 4\hat{k}) - (2(1) + 3(-2) + 6(4)) = 0 \)
\( \implies \vec{r} \cdot (\hat{i} - 2\hat{j} + 4\hat{k}) - (2 - 6 + 24) = 0 \)
\( \implies \vec{r} \cdot (\hat{i} - 2\hat{j} + 4\hat{k}) - 20 = 0 \)
Thus, the non-parametric vector equation is \( \vec{r} \cdot (\hat{i} - 2\hat{j} + 4\hat{k}) = 20 \).
To find the Cartesian equation, let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
\( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} - 2\hat{j} + 4\hat{k}) = 20 \)
\( \implies x(1) + y(-2) + z(4) = 20 \)
\( \implies x - 2y + 4z = 20 \)
Therefore, the Cartesian equation of the plane is \( x - 2y + 4z - 20 = 0 \). This method clearly shows how the direction of the lines defines the orientation of the plane.
In simple words: First, find a point on the plane and two vectors parallel to it. Then, calculate the cross product of these parallel vectors to get the plane's normal vector. Use this normal vector with the point to write the non-parametric vector equation. Finally, replace the position vector with \( x\hat{i} + y\hat{j} + z\hat{k} \) to get the Cartesian equation.

🎯 Exam Tip: Remember that a plane parallel to two given lines means its normal vector is perpendicular to the direction vectors of both lines. This is why a cross product is used.

Question 2. Find the non-parametric form of vector equation and Cartesian equations of the plane passing through the points (2, 2, 1), (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z = 9.
Answer:
Let the two given points be \( \vec{a} = 2\hat{i} + 2\hat{j} + \hat{k} \) and \( \vec{b} = 9\hat{i} + 3\hat{j} + 6\hat{k} \).
The vector connecting these two points is \( \vec{AB} = \vec{b} - \vec{a} \).
\( \vec{AB} = (9\hat{i} + 3\hat{j} + 6\hat{k}) - (2\hat{i} + 2\hat{j} + \hat{k}) \)
\( \implies \vec{AB} = (9-2)\hat{i} + (3-2)\hat{j} + (6-1)\hat{k} \)
\( \implies \vec{AB} = 7\hat{i} + \hat{j} + 5\hat{k} \)
The plane we are looking for is perpendicular to the plane \( 2x + 6y + 6z = 9 \).
The normal vector of this given perpendicular plane is \( \vec{c} = 2\hat{i} + 6\hat{j} + 6\hat{k} \).
Since the required plane passes through points \( \vec{a} \) and \( \vec{b} \), the vector \( \vec{AB} \) lies in the plane. Also, the required plane is perpendicular to the plane with normal \( \vec{c} \), so its normal vector must be perpendicular to \( \vec{c} \).
Therefore, the normal vector to our required plane is parallel to \( \vec{AB} \times \vec{c} \).
\( \vec{n} = \vec{AB} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 1 & 5 \\ 2 & 6 & 6 \end{vmatrix} \)
\( \implies \hat{i}((1)(6) - (5)(6)) - \hat{j}((7)(6) - (5)(2)) + \hat{k}((7)(6) - (1)(2)) \)
\( \implies \hat{i}(6 - 30) - \hat{j}(42 - 10) + \hat{k}(42 - 2) \)
\( \implies -24\hat{i} - 32\hat{j} + 40\hat{k} \)
This vector can be simplified by taking out a common factor of -8: \( -8(3\hat{i} + 4\hat{j} - 5\hat{k}) \).
Let's use \( \vec{n} = 3\hat{i} + 4\hat{j} - 5\hat{k} \) as the normal vector for our plane.
The non-parametric vector equation of the plane passing through \( \vec{a} \) with normal \( \vec{n} \) is \( (\vec{r} - \vec{a}) \cdot \vec{n} = 0 \).
\( [\vec{r} - (2\hat{i} + 2\hat{j} + \hat{k})] \cdot (3\hat{i} + 4\hat{j} - 5\hat{k}) = 0 \)
\( \implies \vec{r} \cdot (3\hat{i} + 4\hat{j} - 5\hat{k}) - (2\hat{i} + 2\hat{j} + \hat{k}) \cdot (3\hat{i} + 4\hat{j} - 5\hat{k}) = 0 \)
\( \implies \vec{r} \cdot (3\hat{i} + 4\hat{j} - 5\hat{k}) - (2(3) + 2(4) + 1(-5)) = 0 \)
\( \implies \vec{r} \cdot (3\hat{i} + 4\hat{j} - 5\hat{k}) - (6 + 8 - 5) = 0 \)
\( \implies \vec{r} \cdot (3\hat{i} + 4\hat{j} - 5\hat{k}) - 9 = 0 \)
Thus, the non-parametric vector equation is \( \vec{r} \cdot (3\hat{i} + 4\hat{j} - 5\hat{k}) = 9 \).
To find the Cartesian equation, let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
\( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (3\hat{i} + 4\hat{j} - 5\hat{k}) = 9 \)
\( \implies 3x + 4y - 5z = 9 \)
Therefore, the Cartesian equation of the plane is \( 3x + 4y - 5z - 9 = 0 \). This process helps determine a plane's orientation and position in space, which is critical in various engineering and physics problems.
In simple words: First, find a vector between the two points in the plane. Then, get the normal vector of the plane it's perpendicular to. Cross product these two vectors to find the normal vector of the plane we want. Use this normal and one point to write the vector equation, then convert it to the Cartesian form.

🎯 Exam Tip: When a plane passes through two points and is perpendicular to another plane, the normal vector of the desired plane is parallel to the cross product of the vector connecting the two points and the normal vector of the perpendicular plane.

Question 3. Find parametric form of vector equation and Cartesian equations of the plane passing through the points (2, 2, 1), (1, -2, 3) and parallel to the straight line passing through the points (2, 1, -3) and (-1, 5, -8)
Answer: Solution not provided in the source material.

🎯 Exam Tip: To solve this type of problem, first find a vector \( \vec{u} \) by subtracting the coordinates of the two points on the plane. Then, find another vector \( \vec{v} \) by subtracting the coordinates of the two points on the parallel line. The parametric vector equation will be \( \vec{r} = \vec{a} + s\vec{u} + t\vec{v} \), where \( \vec{a} \) is one of the points on the plane. The Cartesian equation can be found by taking the scalar triple product \( (\vec{r} - \vec{a}) \cdot (\vec{u} \times \vec{v}) = 0 \).

Question 4. Find the non-parametric form of vector equation and Cartesian equation of the plane passing through the point (1, -2, 4) and perpendicular to the plane \( x + 2y - 3z = 11 \) and parallel to the line \( \frac {x+7 }{ 3 } = \frac { y+3 }{ -1 } = \frac { z }{ 1 } \)
Answer:
Let the given point be \( \vec{a} = \hat{i} - 2\hat{j} + 4\hat{k} \).
The plane is perpendicular to the plane \( x + 2y - 3z = 11 \). The normal vector to this plane is \( \vec{b} = \hat{i} + 2\hat{j} - 3\hat{k} \).
The plane is parallel to the line \( \frac {x+7 }{ 3 } = \frac { y+3 }{ -1 } = \frac { z }{ 1 } \). The direction vector of this line is \( \vec{c} = 3\hat{i} - \hat{j} + \hat{k} \).
Since the required plane is perpendicular to the plane with normal \( \vec{b} \) and parallel to the line with direction \( \vec{c} \), its normal vector, \( \vec{n} \), must be perpendicular to both \( \vec{b} \) and \( \vec{c} \).
Thus, we find \( \vec{n} = \vec{b} \times \vec{c} \).
\( \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 3 & -1 & 1 \end{vmatrix} \)
\( \implies \hat{i}((2)(1) - (-3)(-1)) - \hat{j}((1)(1) - (-3)(3)) + \hat{k}((1)(-1) - (2)(3)) \)
\( \implies \hat{i}(2 - 3) - \hat{j}(1 + 9) + \hat{k}(-1 - 6) \)
\( \implies -\hat{i} - 10\hat{j} - 7\hat{k} \)
The non-parametric vector equation of the plane passing through \( \vec{a} \) with normal \( \vec{n} \) is \( (\vec{r} - \vec{a}) \cdot \vec{n} = 0 \).
\( [\vec{r} - (\hat{i} - 2\hat{j} + 4\hat{k})] \cdot (-\hat{i} - 10\hat{j} - 7\hat{k}) = 0 \)
\( \implies \vec{r} \cdot (-\hat{i} - 10\hat{j} - 7\hat{k}) - (\hat{i} - 2\hat{j} + 4\hat{k}) \cdot (-\hat{i} - 10\hat{j} - 7\hat{k}) = 0 \)
\( \implies \vec{r} \cdot (-\hat{i} - 10\hat{j} - 7\hat{k}) - ((1)(-1) + (-2)(-10) + (4)(-7)) = 0 \)
\( \implies \vec{r} \cdot (-\hat{i} - 10\hat{j} - 7\hat{k}) - (-1 + 20 - 28) = 0 \)
\( \implies \vec{r} \cdot (-\hat{i} - 10\hat{j} - 7\hat{k}) - (-9) = 0 \)
\( \implies \vec{r} \cdot (-\hat{i} - 10\hat{j} - 7\hat{k}) = -9 \)
Multiplying by -1, we get the non-parametric vector equation: \( \vec{r} \cdot (\hat{i} + 10\hat{j} + 7\hat{k}) = 9 \).
To find the Cartesian equation, let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).
\( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} + 10\hat{j} + 7\hat{k}) = 9 \)
\( \implies x(1) + y(10) + z(7) = 9 \)
Therefore, the Cartesian equation of the plane is \( x + 10y + 7z = 9 \). This approach is powerful for defining planes with specific orientations relative to other geometric objects, such as other planes and lines.
In simple words: First, find the starting point. The plane's normal vector comes from the cross product of the normal of the perpendicular plane and the direction of the parallel line. Use this normal vector with the point to get the non-parametric vector equation. Then, change this into the Cartesian equation by replacing \( \vec{r} \) with \( x\hat{i} + y\hat{j} + z\hat{k} \).

🎯 Exam Tip: Carefully identify the point the plane passes through and the vectors defining its orientation. If the plane is perpendicular to a given plane, its normal vector will be perpendicular to the normal of the given plane. If it's parallel to a line, its normal will be perpendicular to the direction vector of that line.

Question 5. Find the parametric form of vector equation and Cartesian equations of the plane containing the line \( \overline { r } = (\hat { i } – \hat { j } + 3\hat { k }) + t(2\hat { i } – \hat {j} + 4\hat {k} ) \) and perpendicular to plane \( \overline { r } \cdot (\hat { i } + 2\hat { j } + \hat { k }) = 8 \)
Answer: Solution not provided in the source material.

🎯 Exam Tip: When a plane contains a line, the initial point of the line can be used as a point on the plane. The direction vector of the line is one vector in the plane. If the plane is also perpendicular to another plane, the normal vector of the second plane provides another direction for determining the normal of the desired plane. Use these two direction vectors to find the normal vector of the plane and then form its equation.

Question 6. Find the parametric vector non-parametric vector and Cartesian form of the equations of the plane passing through the three non- collinear points (3, 6, -2), (-1, -2, 6) and (6, 4, -2).
Answer: Solution not provided in the source material.

🎯 Exam Tip: For a plane passing through three non-collinear points \( A, B, C \), first find two vectors in the plane, for example, \( \vec{AB} \) and \( \vec{AC} \). The normal vector to the plane will be \( \vec{n} = \vec{AB} \times \vec{AC} \). Then use one of the points (e.g., A) and the normal vector \( \vec{n} \) to find the non-parametric form: \( (\vec{r} - \vec{a}) \cdot \vec{n} = 0 \). To get the parametric form, use \( \vec{r} = \vec{a} + s(\vec{b} - \vec{a}) + t(\vec{c} - \vec{a}) \).

Question 7. Find the non-parametric form of vector equation and Cartesian equations of the plane \( \overline { r } = (6\hat { i } – \hat { j } + \hat { k }) + s(\hat { -i } + 2\hat { j } + \hat { k }) + t(\hat { -5i } – 4\hat { j } – 5\hat { k }) \)
Answer: Solution not provided in the source material.

🎯 Exam Tip: When given the parametric vector equation of a plane in the form \( \vec{r} = \vec{a} + s\vec{b} + t\vec{c} \), the vector \( \vec{a} \) is a point on the plane, and \( \vec{b} \) and \( \vec{c} \) are two vectors lying in the plane. To find the non-parametric vector form, compute the normal vector \( \vec{n} = \vec{b} \times \vec{c} \), then use the formula \( (\vec{r} - \vec{a}) \cdot \vec{n} = 0 \). From this, the Cartesian equation can be easily derived by substituting \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).