GSEB Class 12 Maths Solutions Chapter 11 Three Dimensional Geometry Exercise 11.3

Question 1. In each of the following cases, determine the direction cosines of the normal to the plane and the distance of the plane from the origin.
(a) \( z = 2 \)
(b) \( x + y + z = 1 \)
(c) \( 2x + 3y - z = 5 \)
(d) \( 5y + 8 = 0 \)
Answer:
The direction ratios of the normal to a plane \( ax + by + cz = d \) are \( (a, b, c) \). The perpendicular distance from the origin to this plane is given by \( \frac{|d|}{\sqrt{a^{2}+b^{2}+c^{2}}} \).
(a) For the plane \( z = 2 \), the direction cosines are \( (0, 0, 1) \). The distance of the plane from the origin is \( 2 \).
(b) For the plane \( x + y + z = 1 \), the direction ratios of the normal are \( (1, 1, 1) \).
The direction cosines of the normal are \( \frac{1}{\sqrt{1^{2}+1^{2}+1^{2}}}, \frac{1}{\sqrt{1^{2}+1^{2}+1^{2}}}, \frac{1}{\sqrt{1^{2}+1^{2}+1^{2}}} \), which simplifies to \( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \).
The distance from the origin is \( \frac{|1|}{\sqrt{1^{2}+1^{2}+1^{2}}} = \frac{1}{\sqrt{3}} \).
(c) For the plane \( 2x + 3y - z = 5 \), the direction ratios of the normal are \( (2, 3, -1) \).
The magnitude of the normal vector is \( \sqrt{2^{2}+3^{2}+(-1)^{2}} = \sqrt{4+9+1} = \sqrt{14} \).
The direction cosines are \( \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{-1}{\sqrt{14}} \).
The distance of the plane from the origin is \( \frac{|5|}{\sqrt{14}} \).
(d) For the plane \( 5y + 8 = 0 \), which can be written as \( 0x + 5y + 0z = -8 \), the direction ratios of the normal are \( (0, 5, 0) \).
The magnitude is \( \sqrt{0^{2}+5^{2}+0^{2}} = \sqrt{25} = 5 \).
The direction cosines of the normal are \( (\frac{0}{5}, \frac{5}{5}, \frac{0}{5}) \), which means \( (0, 1, 0) \).
The distance of the plane from the origin is \( \frac{|-8|}{5} = \frac{8}{5} \).
In simple words: To find the direction cosines, divide each coefficient by the square root of the sum of their squares. For the distance, divide the constant term by the same square root.

Exam Tip: Remember to use the absolute value of the constant 'd' when calculating the distance from the origin to ensure it's always positive.

Question 2. Find the vector equation of a plane which is at a distance of 7 units from the origin and which is normal to the vector \( 3\hat {i} + 5\hat {j} - 6\hat {k} \) .
Answer:
Let the normal vector be \( \vec{n} = 3\hat {i} + 5\hat {j} - 6\hat {k} \).
The magnitude of this normal vector is \( |\vec{n}| = \sqrt{3^{2} + 5^{2} + (-6)^{2}} \).
\( |\vec{n}| = \sqrt{9 + 25 + 36} = \sqrt{70} \).
The unit normal vector \( \hat{n} \) is \( \frac{\vec{n}}{|\vec{n}|} = \frac{3}{\sqrt{70}}\hat {i} + \frac{5}{\sqrt{70}}\hat {j} - \frac{6}{\sqrt{70}}\hat {k} \).
The given distance from the origin is \( d = 7 \) units.
The required equation of the plane in vector form is \( \vec{r} \cdot \hat{n} = d \).
Substituting the values, we get:
\( \vec{r} \cdot \left( \frac{3}{\sqrt{70}}\hat {i} + \frac{5}{\sqrt{70}}\hat {j} - \frac{6}{\sqrt{70}}\hat {k} \right) = 7 \)
Alternatively, multiplying by \( \sqrt{70} \), the equation can be written as:
\( \vec{r} \cdot (3\hat {i} + 5\hat {j} - 6\hat {k}) = 7\sqrt{70} \).
In simple words: The equation of a plane needs a vector pointing straight out from it (normal vector) and its distance from the origin. You first make the normal vector a "unit" vector (length of 1), then multiply it by the position vector \( \vec{r} \) and set it equal to the distance.

Exam Tip: Always remember that the unit normal vector \( \hat{n} \) is used in the standard form \( \vec{r} \cdot \hat{n} = d \), where 'd' is the perpendicular distance from the origin.

Question 3. Find the cartesian equations of the following planes:
(a) \( \vec{r} \cdot (\hat {i} + \hat {j} + \hat {k}) = 2 \)
(b) \( \vec{r} \cdot (2\hat {i} + 3\hat {j} - 4\hat {k}) = 1 \)
(c) \( \vec{r} \cdot [(s - 2t)\hat {i} + (3 - t)\hat {j} + (2s + t)\hat {k}] = 15 \)
Answer:
(a) Given vector equation: \( \vec{r} \cdot (\hat {i} + \hat {j} - \hat {k}) = 2 \).
Here, \( \vec{r} \) is the position vector of any arbitrary point \( P(x, y, z) \) on the plane, so \( \vec{r} = x\hat {i} + y\hat {j} + z\hat {k} \).
Substitute \( \vec{r} \) into the equation:
\( (x\hat {i} + y\hat {j} + z\hat {k}) \cdot (\hat {i} + \hat {j} - \hat {k}) = 2 \)
This gives: \( (x)(1) + (y)(1) + (z)(-1) = 2 \)
\( x + y - z = 2 \).
This is the required cartesian equation.
(b) Given vector equation: \( \vec{r} \cdot (2\hat {i} + 3\hat {j} - 4\hat {k}) = 1 \).
Substitute \( \vec{r} = x\hat {i} + y\hat {j} + z\hat {k} \) into the equation:
\( (x\hat {i} + y\hat {j} + z\hat {k}) \cdot (2\hat {i} + 3\hat {j} - 4\hat {k}) = 1 \)
This results in: \( (x)(2) + (y)(3) + (z)(-4) = 1 \)
\( 2x + 3y - 4z = 1 \).
This is the required cartesian equation.
(c) Given vector equation: \( \vec{r} \cdot [(s - 2t)\hat {i} + (3 - t)\hat {j} + (2s + t)\hat {k}] = 15 \).
Substitute \( \vec{r} = x\hat {i} + y\hat {j} + z\hat {k} \) into the equation:
\( (x\hat {i} + y\hat {j} + z\hat {k}) \cdot [(s - 2t)\hat {i} + (3 - t)\hat {j} + (2s + t)\hat {k}] = 15 \)
This gives: \( (s - 2t)x + (3 - t)y + (2s + t)z = 15 \).
This is the required cartesian equation.
In simple words: To change a vector equation of a plane into a cartesian equation, replace the position vector \( \vec{r} \) with \( x\hat {i} + y\hat {j} + z\hat {k} \). Then, perform the dot product, which means multiplying the corresponding components of the vectors and summing them up.

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

Question 4. In the following cases, find the coordinates of the foot of perpendicular from the origin:
(a) \( 2x + 3y + 4z - 12 = 0 \)
(b) \( 3y + 4z - 6 = 0 \)
Answer:
(a) Let \( N(x_1, y_1, z_1) \) be the foot of the perpendicular from the origin to the plane \( 2x + 3y + 4z - 12 = 0 \).
The direction ratios of the normal to the plane are \( (2, 3, 4) \).
The direction ratios of the line ON (from origin to \( N \)) are \( (x_1, y_1, z_1) \).
Since ON is normal to the plane, their direction ratios are proportional:
\( \frac{x_1}{2} = \frac{y_1}{3} = \frac{z_1}{4} = k \)
\( \implies x_1 = 2k, y_1 = 3k, z_1 = 4k \). (1)
Since the point \( N(x_1, y_1, z_1) \) lies on the plane \( 2x + 3y + 4z - 12 = 0 \), we substitute these coordinates into the plane's equation:
\( 2(2k) + 3(3k) + 4(4k) - 12 = 0 \)
\( 4k + 9k + 16k - 12 = 0 \)
\( 29k = 12 \)
\( \implies k = \frac{12}{29} \).
Substitute the value of \( k \) back into (1):
\( x_1 = 2 \cdot \frac{12}{29} = \frac{24}{29} \)
\( y_1 = 3 \cdot \frac{12}{29} = \frac{36}{29} \)
\( z_1 = 4 \cdot \frac{12}{29} = \frac{48}{29} \)
Therefore, the foot of the perpendicular from the origin to the given plane is \( (\frac{24}{29}, \frac{36}{29}, \frac{48}{29}) \).
(b) Let \( N(x_1, y_1, z_1) \) be the foot of the perpendicular from the origin to the plane \( 3y + 4z - 6 = 0 \).
The direction ratios of the normal to the plane are \( (0, 3, 4) \).
The direction ratios of ON are \( (x_1, y_1, z_1) \).
\( \frac{x_1}{0} = \frac{y_1}{3} = \frac{z_1}{4} = k \)
\( \implies x_1 = 0, y_1 = 3k, z_1 = 4k \).
Since \( N(x_1, y_1, z_1) \) lies on the plane \( 3y + 4z - 6 = 0 \):
\( 3(3k) + 4(4k) - 6 = 0 \)
\( 9k + 16k - 6 = 0 \)
\( 25k = 6 \)
\( \implies k = \frac{6}{25} \).
Substitute the value of \( k \):
\( x_1 = 0 \)
\( y_1 = 3 \cdot \frac{6}{25} = \frac{18}{25} \)
\( z_1 = 4 \cdot \frac{6}{25} = \frac{24}{25} \)
Thus, the foot of the perpendicular from the origin is \( (0, \frac{18}{25}, \frac{24}{25}) \).
(c) For the plane \( x + y + z = 1 \).
The direction ratios of the normal are \( (1, 1, 1) \).
Let \( N(x_1, y_1, z_1) \) be the foot of the perpendicular.
\( \frac{x_1}{1} = \frac{y_1}{1} = \frac{z_1}{1} = k \)
\( \implies x_1 = k, y_1 = k, z_1 = k \).
Since \( N \) lies on the plane \( x + y + z = 1 \):
\( k + k + k = 1 \)
\( 3k = 1 \)
\( \implies k = \frac{1}{3} \).
Thus, \( x_1 = \frac{1}{3}, y_1 = \frac{1}{3}, z_1 = \frac{1}{3} \).
The coordinates of the foot of the perpendicular are \( (\frac{1}{3}, \frac{1}{3}, \frac{1}{3}) \).
(d) For the plane \( 5y + 8 = 0 \).
The direction ratios of the normal are \( (0, 5, 0) \).
Let \( N(x_1, y_1, z_1) \) be the foot of the perpendicular.
\( \frac{x_1}{0} = \frac{y_1}{5} = \frac{z_1}{0} = k \)
\( \implies x_1 = 0, y_1 = 5k, z_1 = 0 \).
Since \( N \) lies on the plane \( 5y + 8 = 0 \):
\( 5(5k) + 8 = 0 \)
\( 25k = -8 \)
\( \implies k = -\frac{8}{25} \).
Thus, \( x_1 = 0 \)
\( y_1 = 5 \cdot (-\frac{8}{25}) = -\frac{8}{5} \)
\( z_1 = 0 \).
The foot of the perpendicular from the origin is \( (0, -\frac{8}{5}, 0) \).
In simple words: To find where a line from the origin hits a plane at a 90-degree angle, use the plane's normal vector to define the line's direction. Then, find the point on this line that also sits on the plane.

Exam Tip: Remember that the line from the origin to the foot of the perpendicular is parallel to the normal vector of the plane, so their direction ratios will be proportional.

Question 5. Find the vector and cartesian equations of the planes (a) that passes through \( (1, 0, -2) \) and the normal to the plane is \( \hat {i} + \hat {j} - \hat {k} \). (b) that passes through the point \( (1, 4, 6) \) and normal vector to the plane is \( \hat {i} - 2\hat {j} + \hat {k} \).
Answer:
(a) The plane passes through the point \( \vec{a} = \hat {i} - 2\hat {k} \), and its normal vector is \( \vec{n} = \hat {i} + \hat {j} - \hat {k} \).
**Vector form:**
The equation of a plane passing through a point \( \vec{a} \) with normal \( \vec{n} \) is \( (\vec{r} - \vec{a}) \cdot \vec{n} = 0 \).
\( [\vec{r} - (\hat {i} - 2\hat {k})] \cdot (\hat {i} + \hat {j} - \hat {k}) = 0 \)
\( \vec{r} \cdot (\hat {i} + \hat {j} - \hat {k}) - (\hat {i} - 2\hat {k}) \cdot (\hat {i} + \hat {j} - \hat {k}) = 0 \)
\( \vec{r} \cdot (\hat {i} + \hat {j} - \hat {k}) - (1 \cdot 1 + 0 \cdot 1 + (-2) \cdot (-1)) = 0 \)
\( \vec{r} \cdot (\hat {i} + \hat {j} - \hat {k}) - (1 + 0 + 2) = 0 \)
\( \vec{r} \cdot (\hat {i} + \hat {j} - \hat {k}) - 3 = 0 \).
So, the vector equation is \( \vec{r} \cdot (\hat {i} + \hat {j} - \hat {k}) = 3 \).
**Cartesian form:**
Let \( (x_1, y_1, z_1) = (1, 0, -2) \). The direction ratios of the normal are \( (a, b, c) = (1, 1, -1) \).
The cartesian equation is \( a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \).
\( 1(x - 1) + 1(y - 0) - 1(z - (-2)) = 0 \)
\( x - 1 + y - z - 2 = 0 \)
\( x + y - z - 3 = 0 \)
So, the cartesian equation is \( x + y - z = 3 \).
(b) The plane passes through the point \( \vec{a} = \hat {i} + 4\hat {j} + 6\hat {k} \), and its normal vector is \( \vec{n} = \hat {i} - 2\hat {j} + \hat {k} \).
**Vector equation:**
\( (\vec{r} - \vec{a}) \cdot \vec{n} = 0 \)
\( [\vec{r} - (\hat {i} + 4\hat {j} + 6\hat {k})] \cdot (\hat {i} - 2\hat {j} + \hat {k}) = 0 \)
\( \vec{r} \cdot (\hat {i} - 2\hat {j} + \hat {k}) - (\hat {i} + 4\hat {j} + 6\hat {k}) \cdot (\hat {i} - 2\hat {j} + \hat {k}) = 0 \)
\( \vec{r} \cdot (\hat {i} - 2\hat {j} + \hat {k}) - (1 \cdot 1 + 4 \cdot (-2) + 6 \cdot 1) = 0 \)
\( \vec{r} \cdot (\hat {i} - 2\hat {j} + \hat {k}) - (1 - 8 + 6) = 0 \)
\( \vec{r} \cdot (\hat {i} - 2\hat {j} + \hat {k}) - (-1) = 0 \)
So, the vector equation is \( \vec{r} \cdot (\hat {i} - 2\hat {j} + \hat {k}) + 1 = 0 \).
**Cartesian form:**
Let \( (x_1, y_1, z_1) = (1, 4, 6) \). The direction ratios of the normal are \( (a, b, c) = (1, -2, 1) \).
The cartesian equation is \( a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \).
\( 1(x - 1) - 2(y - 4) + 1(z - 6) = 0 \)
\( x - 1 - 2y + 8 + z - 6 = 0 \)
\( x - 2y + z + 1 = 0 \).
In simple words: When you know a point on the plane and a vector that is perpendicular to the plane, you can write its equation. For the vector form, you use the dot product of the position vector \( \vec{r} \) minus the point's vector, with the normal vector. For the Cartesian form, you use the coordinates of the point and the components of the normal vector.

Exam Tip: Make sure to correctly perform the dot product, especially when dealing with negative signs in the components of the vectors.

Question 6. Find the equations of the planes that passes through the following three points:
(a) \( (1, 1, -1), (6, 4, -5), (-4, -2, 3) \)
(b) \( (1, 1, 0), (1, 2, 1), (-2, 2, -1) \)
Answer:
(a) The plane passes through the points \( A(1, 1, -1) \), \( B(6, 4, -5) \), and \( C(-4, -2, 3) \).
Let the equation of the plane passing through \( A(1, 1, -1) \) be \( a(x - 1) + b(y - 1) + c(z - (-1)) = 0 \). (1)
Since \( B(6, 4, -5) \) lies on the plane:
\( a(6 - 1) + b(4 - 1) + c(-5 - (-1)) = 0 \)
\( 5a + 3b - 4c = 0 \). (2)
Since \( C(-4, -2, 3) \) lies on the plane:
\( a(-4 - 1) + b(-2 - 1) + c(3 - (-1)) = 0 \)
\( -5a - 3b + 4c = 0 \)
\( 5a + 3b - 4c = 0 \). (3)
From (2) and (3), we can observe that both equations are identical. This implies that the three points are collinear (lie on the same line).
When three points are collinear, a unique plane cannot be formed through them. In this scenario, infinitely many planes can pass through the line containing these three points.
(b) The plane passes through the points \( R(1, 1, 0) \), \( S(1, 2, 1) \), and \( T(-2, 2, -1) \).
**Method 1: Cartesian form using point-normal equation**
Let the equation of the plane passing through \( R(1, 1, 0) \) be \( a(x - 1) + b(y - 1) + c(z - 0) = 0 \). (1)
Since \( S(1, 2, 1) \) lies on it:
\( a(1 - 1) + b(2 - 1) + c(1) = 0 \)
\( 0a + 1b + 1c = 0 \)
\( \implies b + c = 0 \), so \( b = -c \). (2)
Since \( T(-2, 2, -1) \) lies on it:
\( a(-2 - 1) + b(2 - 1) + c(-1) = 0 \)
\( -3a + b - c = 0 \). (3)
Substitute \( b = -c \) from (2) into (3):
\( -3a + (-c) - c = 0 \)
\( -3a - 2c = 0 \)
\( \implies 3a = -2c \), so \( a = -\frac{2}{3}c \).
Now substitute \( a = -\frac{2}{3}c \) and \( b = -c \) into equation (1):
\( -\frac{2}{3}c(x - 1) + (-c)(y - 1) + c(z) = 0 \)
Divide the entire equation by \( c \) (assuming \( c \neq 0 \)):
\( -\frac{2}{3}(x - 1) - (y - 1) + z = 0 \)
Multiply by -3 to clear the fraction and make the leading term positive:
\( 2(x - 1) + 3(y - 1) - 3z = 0 \)
\( 2x - 2 + 3y - 3 - 3z = 0 \)
\( 2x + 3y - 3z - 5 = 0 \).
So, the equation of the plane is \( 2x + 3y - 3z = 5 \).
**Method 2: Cartesian form using determinant**
The equation of a plane passing through three non-collinear points \( (x_1, y_1, z_1) \), \( (x_2, y_2, z_2) \), and \( (x_3, y_3, z_3) \) is given by the determinant:
\[ \begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{vmatrix} = 0 \]Here, the points are \( R(1, 1, 0) \), \( S(1, 2, 1) \), and \( T(-2, 2, -1) \).
\( x_1 = 1, y_1 = 1, z_1 = 0 \)
\( x_2 = 1, y_2 = 2, z_2 = 1 \)
\( x_3 = -2, y_3 = 2, z_3 = -1 \)
Substitute these values:
\[ \begin{vmatrix} x - 1 & y - 1 & z - 0 \\ 1 - 1 & 2 - 1 & 1 - 0 \\ -2 - 1 & 2 - 1 & -1 - 0 \end{vmatrix} = 0 \]\[ \begin{vmatrix} x - 1 & y - 1 & z \\ 0 & 1 & 1 \\ -3 & 1 & -1 \end{vmatrix} = 0 \]
Expand the determinant along the first row:
\( (x - 1)[(1)(-1) - (1)(1)] - (y - 1)[(0)(-1) - (1)(-3)] + z[(0)(1) - (1)(-3)] = 0 \)
\( (x - 1)(-1 - 1) - (y - 1)(0 + 3) + z(0 + 3) = 0 \)
\( (x - 1)(-2) - (y - 1)(3) + z(3) = 0 \)
\( -2(x - 1) - 3(y - 1) + 3z = 0 \)
\( -2x + 2 - 3y + 3 + 3z = 0 \)
\( -2x - 3y + 3z + 5 = 0 \)
Multiply by -1:
\( 2x + 3y - 3z - 5 = 0 \)
So, \( 2x + 3y - 3z = 5 \).
**Method 3: Vector Equation**
The vector equation of a plane passing through three points \( \vec{a}, \vec{b}, \vec{c} \) is \( (\vec{r} - \vec{a}) \cdot [(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})] = 0 \).
Given points are: \( \vec{a} = (1, 1, 0) \), \( \vec{b} = (1, 2, 1) \), \( \vec{c} = (-2, 2, -1) \).
First, find the vectors \( (\vec{b} - \vec{a}) \) and \( (\vec{c} - \vec{a}) \):
\( \vec{b} - \vec{a} = (1 - 1)\hat {i} + (2 - 1)\hat {j} + (1 - 0)\hat {k} = 0\hat {i} + 1\hat {j} + 1\hat {k} = \hat {j} + \hat {k} \).
\( \vec{c} - \vec{a} = (-2 - 1)\hat {i} + (2 - 1)\hat {j} + (-1 - 0)\hat {k} = -3\hat {i} + 1\hat {j} - 1\hat {k} \).
Next, calculate the cross product \( (\vec{b} - \vec{a}) \times (\vec{c} - \vec{a}) \):
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ -3 & 1 & -1 \end{vmatrix} \]
\( = \hat{i}[(1)(-1) - (1)(1)] - \hat{j}[(0)(-1) - (1)(-3)] + \hat{k}[(0)(1) - (1)(-3)] \)
\( = \hat{i}(-1 - 1) - \hat{j}(0 + 3) + \hat{k}(0 + 3) \)
\( = -2\hat {i} - 3\hat {j} + 3\hat {k} \).
Now, substitute into the plane equation:
\( (\vec{r} - \vec{a}) \cdot (-2\hat {i} - 3\hat {j} + 3\hat {k}) = 0 \)
\( \vec{r} \cdot (-2\hat {i} - 3\hat {j} + 3\hat {k}) - \vec{a} \cdot (-2\hat {i} - 3\hat {j} + 3\hat {k}) = 0 \)
\( \vec{r} \cdot (-2\hat {i} - 3\hat {j} + 3\hat {k}) - (1\hat {i} + 1\hat {j} + 0\hat {k}) \cdot (-2\hat {i} - 3\hat {j} + 3\hat {k}) = 0 \)
\( \vec{r} \cdot (-2\hat {i} - 3\hat {j} + 3\hat {k}) - (1 \cdot (-2) + 1 \cdot (-3) + 0 \cdot 3) = 0 \)
\( \vec{r} \cdot (-2\hat {i} - 3\hat {j} + 3\hat {k}) - (-2 - 3 + 0) = 0 \)
\( \vec{r} \cdot (-2\hat {i} - 3\hat {j} + 3\hat {k}) - (-5) = 0 \)
\( \vec{r} \cdot (-2\hat {i} - 3\hat {j} + 3\hat {k}) + 5 = 0 \).
Multiply by -1 to match the cartesian form:
\( \vec{r} \cdot (2\hat {i} + 3\hat {j} - 3\hat {k}) - 5 = 0 \)
So, the vector equation is \( \vec{r} \cdot (2\hat {i} + 3\hat {j} - 3\hat {k}) = 5 \).
In simple words: To find the equation of a plane through three points, you can use either a direct Cartesian method by assuming the plane equation and substituting the points, or a determinant method for a more structured approach, or a vector method by finding two vectors in the plane, calculating their cross product for the normal, and then using a point-normal form.

Exam Tip: When using the determinant method for three points, ensure correct subtraction of coordinates. For the vector method, double-check your cross product calculation, as errors here will propagate throughout the solution.

Question 7. Find the intercepts cut off by the plane \( 2x + y - z = 5 \) with the coordinate axes.
Answer:
The given equation of the plane is \( 2x + y - z = 5 \).
To find the intercepts, we need to convert the equation into the intercept form: \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \).
Divide the entire equation by 5:
\( \frac{2x}{5} + \frac{y}{5} - \frac{z}{5} = \frac{5}{5} \)
\( \frac{2x}{5} + \frac{y}{5} + \frac{z}{-5} = 1 \)
To match the standard form \( \frac{x}{a} \), we rewrite \( \frac{2x}{5} \) as \( \frac{x}{5/2} \):
\( \frac{x}{5/2} + \frac{y}{5} + \frac{z}{-5} = 1 \).
Comparing this with the intercept form, we get:
\( a = \frac{5}{2} \)
\( b = 5 \)
\( c = -5 \)
Therefore, the intercepts on the X, Y, and Z axes are \( \frac{5}{2} \), \( 5 \), and \( -5 \) respectively.
In simple words: To find where a plane crosses the X, Y, and Z axes, simply rearrange its equation so that one side is equal to 1. The denominators of the \( x, y, z \) terms will then be your intercepts.

Exam Tip: Always make sure the right-hand side of the plane equation is 1 before identifying the intercepts, and be careful with the signs for negative intercepts.

Question 8. Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.
Answer:
A plane parallel to the ZOX plane has an equation of the form \( y = a \), where 'a' is a constant.
The plane has an intercept of 3 on the y-axis, which means it passes through the point \( (0, 3, 0) \).
Substitute this point into the equation \( y = a \):
\( 3 = a \).
Therefore, the equation of the required plane is \( y = 3 \).
In simple words: If a plane is parallel to the ZOX plane, it means it's a flat surface that goes horizontally across the y-axis at a specific point. Since it crosses the y-axis at 3, its equation is simply \( y = 3 \).

Exam Tip: Remember that a plane parallel to a coordinate plane (like ZOX) will have an equation where only the coordinate corresponding to the perpendicular axis (Y-axis for ZOX) is set to a constant value.

Question 9. Find the equation of the plane passing through the intersection of the planes \( 3x - y + 2z - 4 = 0 \) and \( x + y + z - 2 = 0 \) and the point \( (2, 2, 1) \).
Answer:
The equation of a plane passing through the intersection of two planes \( P_1 = 0 \) and \( P_2 = 0 \) is given by \( P_1 + \lambda P_2 = 0 \).
Here, \( P_1 = 3x - y + 2z - 4 \) and \( P_2 = x + y + z - 2 \).
So, the equation of the required plane is:
\( (3x - y + 2z - 4) + \lambda(x + y + z - 2) = 0 \). (1)
The plane passes through the point \( (2, 2, 1) \). Substitute these coordinates into equation (1):
\( (3(2) - 2 + 2(1) - 4) + \lambda(2 + 2 + 1 - 2) = 0 \)
\( (6 - 2 + 2 - 4) + \lambda(3) = 0 \)
\( 2 + 3\lambda = 0 \)
\( \implies 3\lambda = -2 \)
\( \implies \lambda = -\frac{2}{3} \).
Substitute the value of \( \lambda \) back into equation (1):
\( (3x - y + 2z - 4) - \frac{2}{3}(x + y + z - 2) = 0 \)
Multiply the entire equation by 3 to remove the fraction:
\( 3(3x - y + 2z - 4) - 2(x + y + z - 2) = 0 \)
\( 9x - 3y + 6z - 12 - 2x - 2y - 2z + 4 = 0 \)
Combine like terms:
\( (9x - 2x) + (-3y - 2y) + (6z - 2z) + (-12 + 4) = 0 \)
\( 7x - 5y + 4z - 8 = 0 \).
This is the required equation of the plane.
In simple words: To find the equation of a plane that goes through where two other planes meet, plus an extra point, first write a general equation that includes a variable \( \lambda \). Then, use the extra point's coordinates to find the value of \( \lambda \). Finally, put \( \lambda \) back into your general equation to get the plane's specific equation.

Exam Tip: This method ( \( P_1 + \lambda P_2 = 0 \) ) is crucial for finding the equation of a plane passing through the intersection of two given planes. Ensure correct algebraic manipulation when substituting \( \lambda \) back into the equation.

Question 10. Find the vector equation of the plane passing through the line of intersection of the planes \( \vec{r} \cdot (2\hat {i} + 2\hat {j} - 3\hat {k}) = 7 \) and \( \vec{r} \cdot (2\hat {i} + 5\hat {j} + 3\hat {k}) = 9 \) and passing through the point \( (2, 1, 3) \).
Answer:
The vector equation of a plane passing through the line of intersection of two planes \( \vec{r} \cdot \vec{n_1} = d_1 \) and \( \vec{r} \cdot \vec{n_2} = d_2 \) is given by:
\( (\vec{r} \cdot \vec{n_1} - d_1) + \lambda(\vec{r} \cdot \vec{n_2} - d_2) = 0 \).
Given planes are:
Plane 1: \( \vec{r} \cdot (2\hat {i} + 2\hat {j} - 3\hat {k}) = 7 \)
Plane 2: \( \vec{r} \cdot (2\hat {i} + 5\hat {j} + 3\hat {k}) = 9 \)
The equation of the required plane is:
\( [\vec{r} \cdot (2\hat {i} + 2\hat {j} - 3\hat {k}) - 7] + \lambda[\vec{r} \cdot (2\hat {i} + 5\hat {j} + 3\hat {k}) - 9] = 0 \). (1)
Rearrange the terms:
\( \vec{r} \cdot [(2 + 2\lambda)\hat {i} + (2 + 5\lambda)\hat {j} + (-3 + 3\lambda)\hat {k}] - (7 + 9\lambda) = 0 \). (2)
The plane passes through the point \( (2, 1, 3) \), which means \( \vec{r} = 2\hat {i} + 1\hat {j} + 3\hat {k} \).
Substitute this \( \vec{r} \) into equation (2):
\( (2\hat {i} + \hat {j} + 3\hat {k}) \cdot [(2 + 2\lambda)\hat {i} + (2 + 5\lambda)\hat {j} + (-3 + 3\lambda)\hat {k}] - (7 + 9\lambda) = 0 \)
Perform the dot product:
\( 2(2 + 2\lambda) + 1(2 + 5\lambda) + 3(-3 + 3\lambda) - (7 + 9\lambda) = 0 \)
\( 4 + 4\lambda + 2 + 5\lambda - 9 + 9\lambda - 7 - 9\lambda = 0 \)
Combine like terms:
\( (4 + 2 - 9 - 7) + (4\lambda + 5\lambda + 9\lambda - 9\lambda) = 0 \)
\( -10 + 9\lambda = 0 \)
\( \implies 9\lambda = 10 \)
\( \implies \lambda = \frac{10}{9} \).
Substitute the value of \( \lambda \) back into equation (2):
\( \vec{r} \cdot [(2 + 2(\frac{10}{9}))\hat {i} + (2 + 5(\frac{10}{9}))\hat {j} + (-3 + 3(\frac{10}{9}))\hat {k}] - (7 + 9(\frac{10}{9})) = 0 \)
\( \vec{r} \cdot [(2 + \frac{20}{9})\hat {i} + (2 + \frac{50}{9})\hat {j} + (-3 + \frac{30}{9})\hat {k}] - (7 + 10) = 0 \)
\( \vec{r} \cdot [(\frac{18+20}{9})\hat {i} + (\frac{18+50}{9})\hat {j} + (\frac{-27+30}{9})\hat {k}] - 17 = 0 \)
\( \vec{r} \cdot [\frac{38}{9}\hat {i} + \frac{68}{9}\hat {j} + \frac{3}{9}\hat {k}] - 17 = 0 \)
Multiply the entire equation by 9:
\( \vec{r} \cdot (38\hat {i} + 68\hat {j} + 3\hat {k}) - 153 = 0 \)
So, the vector equation of the plane is \( \vec{r} \cdot (38\hat {i} + 68\hat {j} + 3\hat {k}) = 153 \).
In simple words: To find the vector equation of a plane going through the line where two other planes meet, and also through a specific point, you create a combined equation for the two planes using a lambda variable. Then, you plug in the coordinates of the given point to solve for lambda. Finally, substitute lambda back into the combined equation to get your answer.

Exam Tip: Pay close attention to the algebraic simplification of the coefficients when substituting the value of \( \lambda \). A common mistake is arithmetic errors in combining fractions or distributing terms.

Question 11. Find the equation of the plane passing through the line of intersection of the planes \( x + y + z = 1 \) and \( 2x + 3y + 4z = 5 \) and is perpendicular to the plane \( x - y + z = 0 \).
Answer:
The equation of a plane passing through the line of intersection of two planes \( P_1 = 0 \) and \( P_2 = 0 \) is given by \( P_1 + \lambda P_2 = 0 \).
Here, \( P_1 = x + y + z - 1 \) and \( P_2 = 2x + 3y + 4z - 5 \).
So, the equation of the required plane is:
\( (x + y + z - 1) + \lambda(2x + 3y + 4z - 5) = 0 \).
Rearrange the terms by grouping coefficients of \( x, y, z \):
\( (1 + 2\lambda)x + (1 + 3\lambda)y + (1 + 4\lambda)z - (1 + 5\lambda) = 0 \). (1)
The direction ratios of the normal to plane (1) are \( (1 + 2\lambda, 1 + 3\lambda, 1 + 4\lambda) \).
The plane (1) is perpendicular to the plane \( x - y + z = 0 \). (2)
The direction ratios of the normal to plane (2) are \( (1, -1, 1) \).
If two planes are perpendicular, the dot product of their normal vectors is zero ( \( a_1a_2 + b_1b_2 + c_1c_2 = 0 \) ).
\( (1 + 2\lambda)(1) + (1 + 3\lambda)(-1) + (1 + 4\lambda)(1) = 0 \)
\( 1 + 2\lambda - 1 - 3\lambda + 1 + 4\lambda = 0 \)
Combine like terms:
\( (1 - 1 + 1) + (2\lambda - 3\lambda + 4\lambda) = 0 \)
\( 1 + 3\lambda = 0 \)
\( \implies 3\lambda = -1 \)
\( \implies \lambda = -\frac{1}{3} \).
Substitute the value of \( \lambda \) back into equation (1):
\( (1 + 2(-\frac{1}{3}))x + (1 + 3(-\frac{1}{3}))y + (1 + 4(-\frac{1}{3}))z - (1 + 5(-\frac{1}{3})) = 0 \)
\( (1 - \frac{2}{3})x + (1 - 1)y + (1 - \frac{4}{3})z - (1 - \frac{5}{3}) = 0 \)
\( (\frac{3 - 2}{3})x + (0)y + (\frac{3 - 4}{3})z - (\frac{3 - 5}{3}) = 0 \)
\( \frac{1}{3}x + 0y - \frac{1}{3}z - (-\frac{2}{3}) = 0 \)
\( \frac{1}{3}x - \frac{1}{3}z + \frac{2}{3} = 0 \)
Multiply the entire equation by 3:
\( x - z + 2 = 0 \).
This is the required equation of the plane.
In simple words: First, create a general equation for a plane that goes through the intersection of the two given planes, using a variable \( \lambda \). Next, because this new plane is perpendicular to a third plane, their normal vectors must multiply to zero. Use this fact to find the value of \( \lambda \), and then substitute it back into your general equation to get the final plane equation.

Exam Tip: The condition for perpendicular planes (dot product of normals is zero) is key here. Be careful with arithmetic, especially when dealing with fractions for \( \lambda \).

Question 12. Find the angle between the planes whose vector equations are \( \vec{r} \cdot (2\hat {i} + 2\hat {j} - 3\hat {k}) = 5 \) and \( \vec{r} \cdot (3\hat {i} - 3\hat {j} + 5\hat {k}) = 3 \).
Answer:
The angle \( \theta \) between two planes \( \vec{r} \cdot \vec{n_1} = d_1 \) and \( \vec{r} \cdot \vec{n_2} = d_2 \) is given by:
\( \cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|} \).
Here, the normal vectors are:
\( \vec{n_1} = 2\hat {i} + 2\hat {j} - 3\hat {k} \)
\( \vec{n_2} = 3\hat {i} - 3\hat {j} + 5\hat {k} \)
First, calculate the dot product \( \vec{n_1} \cdot \vec{n_2} \):
\( \vec{n_1} \cdot \vec{n_2} = (2)(3) + (2)(-3) + (-3)(5) \)
\( = 6 - 6 - 15 = -15 \).
Next, calculate the magnitudes of \( \vec{n_1} \) and \( \vec{n_2} \):
\( |\vec{n_1}| = \sqrt{2^2 + 2^2 + (-3)^2} = \sqrt{4 + 4 + 9} = \sqrt{17} \).
\( |\vec{n_2}| = \sqrt{3^2 + (-3)^2 + 5^2} = \sqrt{9 + 9 + 25} = \sqrt{43} \).
Now, substitute these values into the formula for \( \cos \theta \):
\( \cos \theta = \frac{|-15|}{(\sqrt{17})(\sqrt{43})} = \frac{15}{\sqrt{17 \cdot 43}} = \frac{15}{\sqrt{731}} \).
Therefore, the angle between the planes is \( \theta = \cos^{-1}\left(\frac{15}{\sqrt{731}}\right) \).
In simple words: To find the angle between two planes, you look at their normal vectors (the vectors pointing straight out from each plane). The cosine of the angle between the planes is found by taking the absolute value of the dot product of these normal vectors, then dividing that by the product of their lengths.

Exam Tip: Remember to use the absolute value of the dot product \( |\vec{n_1} \cdot \vec{n_2}| \) because the angle between planes is usually taken as an acute angle (between 0 and \( \frac{\pi}{2} \)).

Question 13. In the following cases, determine whether the given planes are parallel or perpendicular and in case they are neither, find the angle between them:
(a) \( 7x + 5y + 6z + 30 = 0 \) and \( 3x - y - 10z + 4 = 0 \)
(b) \( 2x + y + 3z - 2 = 0 \) and \( x - 2y + 5 = 0 \)
(c) \( 2x - 2y + 4z + 5 = 0 \) and \( 3x - 3y + 6z - 1 = 0 \)
(d) \( 2x - y + 3z - 1 = 0 \) and \( 2x - y + 3z + 3 = 0 \)
(e) \( 4x + 8y + z - 8 = 0 \) and \( y + z - 4 = 0 \)
Answer:
(a) Given planes: \( P_1: 7x + 5y + 6z + 30 = 0 \) and \( P_2: 3x - y - 10z + 4 = 0 \).
Normal vectors are: \( \vec{n_1} = (7, 5, 6) \) and \( \vec{n_2} = (3, -1, -10) \).
**Parallel check:** Check if direction ratios are proportional: \( \frac{7}{3} \neq \frac{5}{-1} \neq \frac{6}{-10} \). Thus, the planes are not parallel.
**Perpendicular check:** Check if \( \vec{n_1} \cdot \vec{n_2} = 0 \):
\( \vec{n_1} \cdot \vec{n_2} = (7)(3) + (5)(-1) + (6)(-10) = 21 - 5 - 60 = 16 - 60 = -44 \).
Since \( -44 \neq 0 \), the planes are not perpendicular.
**Angle between planes:** Since they are neither, we find the angle \( \theta \) using \( \cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|} \).
\( |\vec{n_1}| = \sqrt{7^2 + 5^2 + 6^2} = \sqrt{49 + 25 + 36} = \sqrt{110} \).
\( |\vec{n_2}| = \sqrt{3^2 + (-1)^2 + (-10)^2} = \sqrt{9 + 1 + 100} = \sqrt{110} \).
\( \cos \theta = \frac{|-44|}{\sqrt{110}\sqrt{110}} = \frac{44}{110} = \frac{4}{10} = \frac{2}{5} \).
\( \implies \theta = \cos^{-1}\left(\frac{2}{5}\right) \).
(b) Given planes: \( P_1: 2x + y + 3z - 2 = 0 \) and \( P_2: x - 2y + 5 = 0 \).
Normal vectors are: \( \vec{n_1} = (2, 1, 3) \) and \( \vec{n_2} = (1, -2, 0) \).
**Parallel check:** \( \frac{2}{1} \neq \frac{1}{-2} \). Not parallel.
**Perpendicular check:** \( \vec{n_1} \cdot \vec{n_2} = (2)(1) + (1)(-2) + (3)(0) = 2 - 2 + 0 = 0 \).
Since \( \vec{n_1} \cdot \vec{n_2} = 0 \), the planes are perpendicular.
(c) Given planes: \( P_1: 2x - 2y + 4z + 5 = 0 \) and \( P_2: 3x - 3y + 6z - 1 = 0 \).
Normal vectors are: \( \vec{n_1} = (2, -2, 4) \) and \( \vec{n_2} = (3, -3, 6) \).
**Parallel check:** Check if direction ratios are proportional: \( \frac{2}{3} = \frac{-2}{-3} = \frac{4}{6} \). Since \( \frac{2}{3} = \frac{2}{3} = \frac{2}{3} \), the direction ratios are proportional.
Thus, the planes are parallel.
(d) Given planes: \( P_1: 2x - y + 3z - 1 = 0 \) and \( P_2: 2x - y + 3z + 3 = 0 \).
Normal vectors are: \( \vec{n_1} = (2, -1, 3) \) and \( \vec{n_2} = (2, -1, 3) \).
**Parallel check:** The normal vectors are identical, which means their direction ratios are equal and therefore proportional. Thus, the planes are parallel.
(e) Given planes: \( P_1: 4x + 8y + z - 8 = 0 \) and \( P_2: y + z - 4 = 0 \) (which is \( 0x + y + z - 4 = 0 \)).
Normal vectors are: \( \vec{n_1} = (4, 8, 1) \) and \( \vec{n_2} = (0, 1, 1) \).
**Parallel check:** \( \frac{4}{0} \) is undefined, so not proportional. Not parallel.
**Perpendicular check:** \( \vec{n_1} \cdot \vec{n_2} = (4)(0) + (8)(1) + (1)(1) = 0 + 8 + 1 = 9 \).
Since \( 9 \neq 0 \), the planes are not perpendicular.
**Angle between planes:**
\( |\vec{n_1}| = \sqrt{4^2 + 8^2 + 1^2} = \sqrt{16 + 64 + 1} = \sqrt{81} = 9 \).
\( |\vec{n_2}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{0 + 1 + 1} = \sqrt{2} \).
\( \cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|} = \frac{|9|}{(9)(\sqrt{2})} = \frac{1}{\sqrt{2}} \).
\( \implies \theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4} \) or \( 45^\circ \).
In simple words: To check if planes are parallel, see if their normal vectors point in the same direction (proportional components). To check if they are perpendicular, see if their normal vectors are at 90 degrees to each other (dot product is zero). If neither, you can use the dot product and magnitudes to find the angle between them.

Exam Tip: Always identify the normal vectors first. For parallel planes, \( \vec{n_1} = k\vec{n_2} \). For perpendicular planes, \( \vec{n_1} \cdot \vec{n_2} = 0 \). If neither condition is met, use the cosine formula for the angle.

Question 14. In the following cases, find the distances of each of the given points from the corresponding given place:
Point Plane
(a) (0,0,0) \( 3x-4y + 12z = 3 \)
(b) (3,2,1) \( 2x-y+2z + 3 = 0 \)
(c) (2,3,-5) \( x + 2y - 2z = 9 \)
(d) (-6,0,0) \( 2x-3y + 6z-2 = 0 \)
Answer:
The distance of a point \( (x_1, y_1, z_1) \) from the plane \( ax + by + cz + d = 0 \) is given by the formula:
\[ \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \]
(a) For the point \( (0, 0, 0) \) and the plane \( 3x - 4y + 12z - 3 = 0 \):
Perpendicular distance \( = \frac{|3.0-4.0 + 12.0-3|}{\sqrt{3^2 + (-4)^2 + (12)^2}} \)
\( = \frac{|-3|}{\sqrt{9+16+144}} \)
\( = \frac{3}{\sqrt{169}} \)
\( = \frac{3}{13} \)
(b) For the point \( (3, 2, 1) \) and the plane \( 2x - y + 2z + 3 = 0 \):
Perpendicular distance \( = \frac{|2.3-(-2)+2.1+3|}{\sqrt{2^2+(-1)^2 + 2^2}} \)
\( = \frac{|6+2+2+3|}{\sqrt{4+1+4}} \)
\( = \frac{13}{\sqrt{9}} \)
\( = \frac{13}{3} \)
(c) For the point \( (2, 3, -5) \) and the plane \( x + 2y - 2z - 9 = 0 \):
Perpendicular distance \( = \frac{|2+2.3-2(-5)-9|}{\sqrt{1^2 +2^2+(-2)^2}} \)
\( = \frac{|2+6+10-9|}{\sqrt{1+4+4}} \)
\( = \frac{9}{\sqrt{9}} \)
\( = 3 \)
(d) For the point \( (-6, 0, 0) \) and the plane \( 2x - 3y + 6z - 2 = 0 \):
Perpendicular distance \( = \frac{|2.(-6)-3.0+6.0-2|}{\sqrt{2^2+(-3)^2 +6^2}} \)
\( = \frac{|-12-2|}{\sqrt{4+9+36}} \)
\( = \frac{|-14|}{\sqrt{49}} \)
\( = \frac{14}{7} \)
\( = 2 \)
In simple words: To find how far a point is from a plane, use the specific formula. Put the point's coordinates and the plane's numbers into the formula, and then calculate the absolute value on top and the square root of the squared coefficients below. Simplify the fraction to get the final distance.

Exam Tip: Remember to correctly substitute the coordinates of the point and the coefficients of the plane into the distance formula. Be careful with signs, especially when calculating the numerator and squaring negative numbers in the denominator.