ML Aggarwal Class 12 Maths Solutions Section B Chapter 03 Planes

Planes

Introduction

We have an intuitive understanding of what a plane is and know many of its key features. A specific plane can be determined in various ways:

• Exactly one plane passes through three non-collinear points, so three non-collinear points uniquely determine a plane.
• Exactly one plane contains two lines that intersect, so two intersecting lines uniquely determine a plane.
• Exactly one plane contains two lines that are parallel, so two parallel lines uniquely determine a plane.
• Exactly one plane is perpendicular to a given direction and sits at a specified distance from the origin, so a normal direction to a plane and its distance from the origin uniquely determine a plane.
• Exactly one plane passes through a given point and is perpendicular to a given direction, so a point on the plane and its normal direction uniquely determine a plane.
• Exactly one plane passes through a given point and is parallel to two given lines, so a point on the plane and two lines parallel to the plane uniquely determine a plane.

Although there are many other ways to specify a plane, methods (iv) and (v) listed above prove most useful because they give rise to a particularly simple form of the vector equation of a plane.

Additionally, it is important to note: if a line is perpendicular to a plane, then it is perpendicular to every line contained in the plane.

Formally, a plane is a flat surface with the following property - whenever any two distinct points lie on it, the entire line through those points also lies within it (that is, every point on the line is in the plane).

3.1 Equation of a Plane in Different Forms

In this section, we find vector and cartesian equations of a plane in various forms.

3.1.1 Plane perpendicular to a given direction and at a given distance from the origin

Vector equation

Consider a plane p1 that is at distance p from the origin O, and let \( \hat{n} \) be a unit vector perpendicular to the plane (with \( \hat{n} \) directed away from O), as shown in the figure.

Let N be the foot of the perpendicular dropped from O onto the plane p1. Then \( \overrightarrow{ON} = p\hat{n} \).

Let P be any point with position vector \( \vec{r} \). The point P lies in the plane if and only if \( \overrightarrow{NP} \perp \hat{n} \) (because \( \hat{n} \) is perpendicular to the plane p1, it must be perpendicular to every line within the plane).

This gives us: \( \overrightarrow{NP} \cdot \hat{n} = 0 \)

Which means: \( (\vec{r} - p\hat{n}) \cdot \hat{n} = 0 \)

Simplifying: \( \vec{r} \cdot \hat{n} - p\hat{n} \cdot \hat{n} = 0 \)

Since \( \hat{n} \cdot \hat{n} = 1 \), we get: \( \vec{r} \cdot \hat{n} = p \) ... (1)

This is the required vector equation of plane p1.

Cartesian equation

Let \( \langle l, m, n \rangle \) denote the direction cosines of \( \overrightarrow{ON} \), and let P be the point \( (x, y, z) \). Then:

\( \hat{n} = l\hat{i} + m\hat{j} + n\hat{k} \) and \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \)

Substituting into equation (1):

\( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (l\hat{i} + m\hat{j} + n\hat{k}) = p \)

Therefore: \( lx + my + nz = p \) ... (2)

This is known as the normal form.

Corollary 1. Standard form of the vector equation

If \( \vec{n} \) is a vector perpendicular to plane p1, then \( \hat{n} = \frac{\vec{n}}{|\vec{n}|} \).

The vector equation (1) of plane p1 can be rewritten as:

\( \vec{r} \cdot \frac{\vec{n}}{|\vec{n}|} = p \)

or \( \vec{r} \cdot \vec{n} = d \), where \( d = |\vec{n}| p \)

The equation \( \vec{r} \cdot \vec{n} = d \) is called the standard form of the vector equation of a plane.

Corollary 2. General form of the cartesian equation

If \( \langle A, B, C \rangle \) are the direction numbers of vector \( \vec{n} \), then:

\( \vec{n} = A\hat{i} + B\hat{j} + C\hat{k} \)

The equation \( \vec{r} \cdot \vec{n} = d \) becomes:

\( Ax + By + Cz = d \)

Which can be written as: \( Ax + By + Cz + D = 0 \) where \( D = -d \)

Conversely, the equation \( Ax + By + Cz + D = 0 \) can be written as:

\( \vec{r} \cdot (A\hat{i} + B\hat{j} + C\hat{k}) = d \)

where \( d = -D \), representing a plane with \( \langle A, B, C \rangle \) as direction numbers of a normal.

It follows that every first-degree equation in x, y, z (that is, \( Ax + By + Cz + D = 0 \) where A, B, C are not all zero) represents a plane, and \( \langle A, B, C \rangle \) are the direction numbers of a normal to this plane.

The equation \( Ax + By + Cz + D = 0 \) is called the general form of the cartesian equation.

Direction numbers of any normal to a plane are called attitude numbers of the plane.

3.1.2 Plane passing through a given point and perpendicular to a given direction

Vector form

Let plane p1 pass through point A with position vector \( \vec{a} \) and be perpendicular to a direction along vector \( \vec{n} \).

Let P be any point with position vector \( \vec{r} \). Then P lies in the plane if and only if \( \overrightarrow{AP} \perp \vec{n} \).

This gives us: \( \overrightarrow{AP} \cdot \vec{n} = 0 \)

Which means: \( (\vec{r} - \vec{a}) \cdot \vec{n} = 0 \) ... (1)

Cartesian form

Let the given point A be \( (x_1, y_1, z_1) \) and P be \( (x, y, z) \), with \( \langle A, B, C \rangle \) being the direction numbers of \( \vec{n} \). Then:

\( \vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k} \), \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \)

\( \vec{r} - \vec{a} = (x - x_1)\hat{i} + (y - y_1)\hat{j} + (z - z_1)\hat{k} \)

\( \vec{n} = A\hat{i} + B\hat{j} + C\hat{k} \)

Substituting into equation (1):

\( [(x - x_1)\hat{i} + (y - y_1)\hat{j} + (z - z_1)\hat{k}] \cdot (A\hat{i} + B\hat{j} + C\hat{k}) = 0 \)

Therefore: \( A(x - x_1) + B(y - y_1) + C(z - z_1) = 0 \) ... (2)

This is known as the one point form.

3.1.3 Position of a line with respect to a plane

We now find the condition(s) for a line to (i) lie in a plane, (ii) be parallel to a plane, or (iii) meet a plane at a single point.

Vector form

Let the given line be \( \vec{r} = \vec{a} + \lambda\vec{b} \) ... (1)

Let the given plane be \( \vec{r} \cdot \vec{n} = d \) ... (2)

Any point on line (1) is \( \vec{a} + \lambda\vec{b} \). This point lies in plane (2) if and only if:

\( (\vec{a} + \lambda\vec{b}) \cdot \vec{n} = d \)

Which gives: \( (\vec{a} \cdot \vec{n} - d) + \lambda(\vec{b} \cdot \vec{n}) = 0 \) ... (3)

(i) The line (1) lies in plane (2) if every point on line (1) lies in plane (2), which means equation (3) must be satisfied for all real values of \( \lambda \). This happens when (3) is an identity, requiring:

\( \vec{a} \cdot \vec{n} = d \) and \( \vec{b} \cdot \vec{n} = 0 \)

(ii) The line (1) is parallel to plane (2) if no point of line (1) lies in plane (2), which means equation (3) is not satisfied by any real value of \( \lambda \). This requires:

\( \vec{b} \cdot \vec{n} = 0 \) and \( \vec{a} \cdot \vec{n} \neq d \)

(iii) The line (1) meets plane (2) at a single point if equation (3) has exactly one value of \( \lambda \), which happens when:

\( \vec{b} \cdot \vec{n} \neq 0 \)

Cartesian form

Let the given line be \( \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \) ... (1)

Let the given plane be \( Ax + By + Cz + D = 0 \) ... (2)

Any point on line (1) is \( (x_1 + a\lambda, y_1 + b\lambda, z_1 + c\lambda) \).

This point lies in plane (2) if:

\( A(x_1 + a\lambda) + B(y_1 + b\lambda) + C(z_1 + c\lambda) + D = 0 \)

Which gives: \( (Ax_1 + By_1 + Cz_1 + D) + (Aa + Bb + Cc)\lambda = 0 \) ... (3)

(i) The line (1) lies in plane (2) if equation (3) is satisfied for all real values of \( \lambda \), requiring:

\( Ax_1 + By_1 + Cz_1 + D = 0 \) and \( Aa + Bb + Cc = 0 \)

(ii) The line (1) is parallel to plane (2) if equation (3) is satisfied by no real value of \( \lambda \), requiring:

\( Aa + Bb + Cc = 0 \) and \( Ax_1 + By_1 + Cz_1 + D \neq 0 \)

(iii) The line (1) meets plane (2) at a single point if equation (3) has exactly one solution for \( \lambda \), which happens when:

\( Aa + Bb + Cc \neq 0 \)

Illustrative Examples

Example 1. 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 \( 3\hat{i} + 5\hat{j} - 6\hat{k} \).
Solution: Here \( p = 7 \) and \( \vec{n} = 3\hat{i} + 5\hat{j} - 6\hat{k} \).

\( \hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{3\hat{i} + 5\hat{j} - 6\hat{k}}{\sqrt{3^2 + 5^2 + (-6)^2}} = \frac{3\hat{i} + 5\hat{j} - 6\hat{k}}{\sqrt{70}} \)

The vector equation of the plane is \( \vec{r} \cdot \hat{n} = p \)

\( \vec{r} \cdot \frac{3\hat{i} + 5\hat{j} - 6\hat{k}}{\sqrt{70}} = 7 \) or \( \vec{r} \cdot (3\hat{i} + 5\hat{j} - 6\hat{k}) = 7\sqrt{70} \)

Example 2. Find the direction cosines of the perpendicular from the origin to the plane \( \vec{r} \cdot (6\hat{i} - 3\hat{j} - 2\hat{k}) + 1 = 0 \).
Solution: Rewriting the plane equation:

\( \vec{r} \cdot (6\hat{i} - 3\hat{j} - 2\hat{k}) = -1 \) or \( \vec{r} \cdot (-6\hat{i} + 3\hat{j} + 2\hat{k}) = 1 \)

This is in the form \( \vec{r} \cdot \vec{n} = d \), where \( \vec{n} \) is the normal vector from the origin to the plane.

Here \( \vec{n} = -6\hat{i} + 3\hat{j} + 2\hat{k} \), so the direction numbers of the normal from origin to the plane are \( \langle -6, 3, 2 \rangle \).

Dividing by \( \sqrt{(-6)^2 + 3^2 + 2^2} = 7 \), the direction cosines of the perpendicular from origin to the given plane are \( \left\langle -\frac{6}{7}, \frac{3}{7}, \frac{2}{7} \right\rangle \)

Example 3. Find the length of perpendicular from the origin to the plane \( \vec{r} \cdot (3\hat{i} - 4\hat{j} - 12\hat{k}) + 39 = 0 \). Also write the unit vector normal to this plane directed from the origin to the plane.
Solution: Rewriting:

\( \vec{r} \cdot (3\hat{i} - 4\hat{j} - 12\hat{k}) = -39 \) or \( \vec{r} \cdot (-3\hat{i} + 4\hat{j} + 12\hat{k}) = 39 \)

or \( \vec{r} \cdot \vec{n} = 39 \), where \( \vec{n} = -3\hat{i} + 4\hat{j} + 12\hat{k} \)

or \( \vec{r} \cdot \frac{\vec{n}}{|\vec{n}|} = \frac{39}{|\vec{n}|} \), where \( |\vec{n}| = \sqrt{(-3)^2 + 4^2 + 12^2} = 13 \)

or \( \vec{r} \cdot \hat{n} = p \), where \( p = \frac{39}{|\vec{n}|} = \frac{39}{13} = 3 \)

Hence, the perpendicular distance from the origin to the plane is 3 units, and the unit normal vector directed from origin to the plane is:

\( \hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{1}{13}(-3\hat{i} + 4\hat{j} + 12\hat{k}) \)

Example 4. Find the cartesian equation of the plane whose vector equation is \( \vec{r} \cdot (5\hat{i} + 7\hat{j} - 11\hat{k}) = 14 \).
Solution: Since \( \vec{r} \) is the position vector of any point P(x, y, z) on the given plane:

\( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \)

Substituting into the plane equation:

\( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (5\hat{i} + 7\hat{j} - 11\hat{k}) = 14 \)

Therefore: \( 5x + 7y - 11z = 14 \), which is the cartesian equation of the plane.

Example 5. Find the vector equation of the plane whose cartesian equation is \( 3x - 7y + 9z + 12 = 0 \).
Solution: If \( \vec{r} \) is the position vector of any point P(x, y, z) on the plane, the equation can be rewritten as:

\( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (3\hat{i} - 7\hat{j} + 9\hat{k}) + 12 = 0 \)

Therefore: \( \vec{r} \cdot (3\hat{i} - 7\hat{j} + 9\hat{k}) + 12 = 0 \), which is the vector equation of the given plane.

Example 6. In each of the following problems, determine the direction cosines of the normal to the plane and its distance from the origin: (i) \( 2x + 3y - z - 5 = 0 \) (ii) \( 5y + 8 = 0 \).
Solution.

(i) Rewriting the plane equation: \( 2x + 3y - z = 5 \) ... (1)

If \( \vec{r} \) is the position vector of point P(x, y, z) on the plane, equation (1) becomes:

\( \vec{r} \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = 5 \) or \( \vec{r} \cdot \vec{n} = 5 \), where \( \vec{n} = 2\hat{i} + 3\hat{j} - \hat{k} \)

or \( \vec{r} \cdot \frac{\vec{n}}{|\vec{n}|} = \frac{5}{|\vec{n}|} \), where \( |\vec{n}| = \sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{14} \)

or \( \vec{r} \cdot \hat{n} = p \), where \( p = \frac{5}{\sqrt{14}} \) and \( \hat{n} = \frac{2\hat{i} + 3\hat{j} - \hat{k}}{\sqrt{14}} = \frac{2}{\sqrt{14}}\hat{i} + \frac{3}{\sqrt{14}}\hat{j} - \frac{1}{\sqrt{14}}\hat{k} \)

Hence, the direction cosines of the normal to the plane are \( \left\langle \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, -\frac{1}{\sqrt{14}} \right\rangle \) and the perpendicular distance from the origin to the plane is \( \frac{5}{\sqrt{14}} \) units.

(ii) Rewriting: \( 0x - 5y + 0z = 8 \) ... (1)

If \( \vec{r} \) is the position vector of point P(x, y, z) on the plane, equation (1) becomes:

\( \vec{r} \cdot (0\hat{i} - 5\hat{j} + 0\hat{k}) = 8 \) or \( \vec{r} \cdot \vec{n} = 8 \), where \( \vec{n} = 0\hat{i} - 5\hat{j} + 0\hat{k} \)

or \( \vec{r} \cdot \frac{\vec{n}}{|\vec{n}|} = \frac{8}{|\vec{n}|} \), where \( |\vec{n}| = \sqrt{0^2 + (-5)^2 + 0^2} = 5 \)

or \( \vec{r} \cdot \hat{n} = p \), where \( p = \frac{8}{5} \) and \( \hat{n} = \frac{0\hat{i} - 5\hat{j} + 0\hat{k}}{5} = 0\hat{i} - \hat{j} + 0\hat{k} \)

Hence, the direction cosines of the normal to the plane are \( \langle 0, -1, 0 \rangle \) and the perpendicular distance from the origin to the plane is \( \frac{8}{5} \) units.

Example 7. Find the vector as well as cartesian equation of the plane which is at a distance of 6 units from the origin and the direction numbers of a normal to this plane are \( \langle 2, 2, -1 \rangle \).
Solution: Direction numbers of a normal to the plane are \( \langle 2, 2, -1 \rangle \).

Dividing by \( \sqrt{2^2 + 2^2 + (-1)^2} = 3 \), direction cosines of a normal to the plane are \( \left\langle \frac{2}{3}, \frac{2}{3}, -\frac{1}{3} \right\rangle \).

The unit normal vector is \( \frac{2}{3}\hat{i} + \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k} \).

With distance from origin = 6 units (given), the vector equation of the plane is \( \vec{r} \cdot \hat{n} = p \)

\( \vec{r} \cdot \left(\frac{2}{3}\hat{i} + \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k}\right) = 6 \) or \( \vec{r} \cdot (2\hat{i} + 2\hat{j} - \hat{k}) = 18 \)

Its cartesian equation is \( 2x + 2y - z = 18 \)

Example 8. Find the equation of the plane such that the length of perpendicular from the origin to the plane is 5 units and this perpendicular makes angles of 60° and 45° with the x-axis and y-axis respectively.
Solution: Let \( \langle l, m, n \rangle \) be the direction cosines of the perpendicular drawn from origin to the plane. Then:

\( l = \cos 60° = \frac{1}{2} \) and \( m = \cos 45° = \frac{1}{\sqrt{2}} \)

We know that \( l^2 + m^2 + n^2 = 1 \)

\( \Rightarrow \left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + n^2 = 1 \Rightarrow n^2 = 1 - \frac{1}{4} - \frac{1}{2} = \frac{1}{4} \Rightarrow n = \pm\frac{1}{2} \)

Also, perpendicular distance from origin to the plane = \( p = 5 \) (given).

Since n has two values, there are two planes satisfying the given conditions.

The equations of the required planes are:

\( \frac{1}{2}x + \frac{1}{\sqrt{2}}y \pm \frac{1}{2}z = 5 \)

or \( x + \sqrt{2}y + z = 10 \), \( x + \sqrt{2}y - z = 10 \)

Example 9. Find the vector and the cartesian equations of the plane passing through the point (5, 2, -4) and perpendicular to the line with direction ratios \( \langle 2, 3, -1 \rangle \).
Solution: The plane passes through the point with position vector \( \vec{a} = 5\hat{i} + 2\hat{j} - 4\hat{k} \) and has a normal vector \( \vec{n} = 2\hat{i} + 3\hat{j} - \hat{k} \).

The equation of the plane is \( (\vec{r} - \vec{a}) \cdot \vec{n} = 0 \)

\( ((\vec{r} - (5\hat{i} + 2\hat{j} - 4\hat{k})) \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = 0 \)

or \( \vec{r} \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = (5\hat{i} + 2\hat{j} - 4\hat{k}) \cdot (2\hat{i} + 3\hat{j} - \hat{k}) \)

or \( \vec{r} \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = 5 \cdot 2 + 2 \cdot 3 + (-4)(-1) = 20 \)

or \( \vec{r} \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = 20 \)

Its cartesian equation is \( 2x + 3y - z = 20 \)

Example 10. The position vectors of two points A and B are \( 3\hat{i} + \hat{j} + 2\hat{k} \) and \( \hat{i} - 2\hat{j} - 4\hat{k} \) respectively. Find the vector equation of the plane passing through B and perpendicular to the vector \( \overrightarrow{AB} \).
Solution: \( \overrightarrow{AB} = \text{P.V. of B} - \text{P.V. of A} = (\hat{i} - 2\hat{j} - 4\hat{k}) - (3\hat{i} + \hat{j} + 2\hat{k}) = -2\hat{i} - 3\hat{j} - 6\hat{k} \)

As the plane passes through B and is perpendicular to vector \( \overrightarrow{AB} \), its equation is:

\( (\vec{r} - (\hat{i} - 2\hat{j} - 4\hat{k})) \cdot (-2\hat{i} - 3\hat{j} - 6\hat{k}) = 0 \)

or \( (\vec{r} - (\hat{i} - 2\hat{j} - 4\hat{k})) \cdot (2\hat{i} + 3\hat{j} + 6\hat{k}) = 0 \) (dividing by -1)

or \( \vec{r} \cdot (2\hat{i} + 3\hat{j} + 6\hat{k}) = (\hat{i} - 2\hat{j} - 4\hat{k}) \cdot (2\hat{i} + 3\hat{j} + 6\hat{k}) \)

or \( \vec{r} \cdot (2\hat{i} + 3\hat{j} + 6\hat{k}) = 1 \cdot 2 + (-2) \cdot 3 + (-4) \cdot 6 = 2 - 6 - 24 = -28 \)

or \( \vec{r} \cdot (2\hat{i} + 3\hat{j} + 6\hat{k}) + 28 = 0 \)

Example 11. If the line drawn from the point (-10, 5, 4) meets a plane at right angles at the point (4, -1, 2), find the equation of the plane.
Solution: Let the given points (4, -1, 2) and (-10, 5, 4) be A and B respectively.

Direction ratios of line AB are: \( \langle -10 - 4, 5 + 1, 4 - 2 \rangle = \langle -14, 6, 2 \rangle = \langle -7, 3, 1 \rangle \) (after simplification)

The required plane passes through A and is perpendicular to line AB, which means line AB (with direction ratios \( \langle -7, 3, 1 \rangle \)) acts as the normal. Therefore, the equation of the plane is:

\( -7(x - 4) + 3(y - (-1)) + 1(z - 2) = 0 \)

or \( -7(x - 4) + 3(y + 1) + (z - 2) = 0 \)

or \( -7x + 28 + 3y + 3 + z - 2 = 0 \)

or \( -7x + 3y + z + 29 = 0 \)

or \( 7x - 3y - z - 29 = 0 \)

Example 12. Find the vector equation of the straight line passing through (1, 2, 3) and perpendicular to the plane \( \vec{r} \cdot (\hat{i} + 2\hat{j} - 5\hat{k}) + 9 = 0 \).
Solution: The plane equation is \( \vec{r} \cdot (\hat{i} + 2\hat{j} - 5\hat{k}) + 9 = 0 \), which gives:

\( \vec{r} \cdot (-\hat{i} - 2\hat{j} + 5\hat{k}) = 9 \), of the form \( \vec{r} \cdot \vec{n} = d \)

The vector \( \vec{n} = -\hat{i} - 2\hat{j} + 5\hat{k} \) points along a normal to the given plane. The required line is therefore oriented along the direction of vector \( -\hat{i} - 2\hat{j} + 5\hat{k} \).

The required line passes through the point with position vector \( \hat{i} + 2\hat{j} + 3\hat{k} \), so the vector equation of the line is:

\( \vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} + \lambda(-\hat{i} - 2\hat{j} + 5\hat{k}) \), where \( \lambda \) is a parameter.

Example 13. Show that the line \( \vec{r} = 4\hat{i} - 7\hat{k} + \lambda(4\hat{i} - 2\hat{j} + 3\hat{k}) \) is parallel to the plane \( \vec{r} \cdot (5\hat{i} + 4\hat{j} - 4\hat{k}) = 7 \).
Solution: Comparing the line \( \vec{r} = 4\hat{i} - 7\hat{k} + \lambda(4\hat{i} - 2\hat{j} + 3\hat{k}) \) with \( \vec{r} = \vec{a} + \lambda\vec{b} \):

\( \vec{a} = 4\hat{i} - 7\hat{k} \) and \( \vec{b} = 4\hat{i} - 2\hat{j} + 3\hat{k} \)

Comparing the plane \( \vec{r} \cdot (5\hat{i} + 4\hat{j} - 4\hat{k}) = 7 \) with \( \vec{r} \cdot \vec{n} = d \):

\( \vec{n} = 5\hat{i} + 4\hat{j} - 4\hat{k} \) and \( d = 7 \)

The line is parallel to the plane if \( \vec{b} \cdot \vec{n} = 0 \) and \( \vec{a} \cdot \vec{n} \neq d \)

\( (4\hat{i} - 2\hat{j} + 3\hat{k}) \cdot (5\hat{i} + 4\hat{j} - 4\hat{k}) = 4 \cdot 5 + (-2) \cdot 4 + 3 \cdot (-4) = 20 - 8 - 12 = 0 \) ✓

\( (4\hat{i} - 7\hat{k}) \cdot (5\hat{i} + 4\hat{j} - 4\hat{k}) = 4 \cdot 5 + 0 \cdot 4 + (-7) \cdot (-4) = 20 + 0 + 28 = 48 \neq 7 \) ✓

Both conditions are satisfied, hence the given line is parallel to the given plane.

Example 14. Show that the line \( \vec{r} = 2\hat{i} + 3\hat{j} + \lambda(7\hat{i} - 5\hat{k}) \) lies in the plane \( \vec{r} \cdot (5\hat{i} - 3\hat{j} + 7\hat{k}) = 1 \).
Solution: Comparing the line \( \vec{r} = 2\hat{i} + 3\hat{j} + \lambda(7\hat{i} - 5\hat{k}) \) with \( \vec{r} = \vec{a} + \lambda\vec{b} \):

\( \vec{a} = 2\hat{i} + 3\hat{j} \) and \( \vec{b} = 7\hat{i} - 5\hat{k} \)

Comparing the plane \( \vec{r} \cdot (5\hat{i} - 3\hat{j} + 7\hat{k}) = 1 \) with \( \vec{r} \cdot \vec{n} = d \):

\( \vec{n} = 5\hat{i} - 3\hat{j} + 7\hat{k} \) and \( d = 1 \)

The line lies in the plane if \( \vec{b} \cdot \vec{n} = 0 \) and \( \vec{a} \cdot \vec{n} = d \)

\( (7\hat{i} - 5\hat{k}) \cdot (5\hat{i} - 3\hat{j} + 7\hat{k}) = 7 \cdot 5 + 0 \cdot (-3) + (-5) \cdot 7 = 35 + 0 - 35 = 0 \) ✓

\( (2\hat{i} + 3\hat{j}) \cdot (5\hat{i} - 3\hat{j} + 7\hat{k}) = 2 \cdot 5 + 3 \cdot (-3) + 0 \cdot 7 = 10 - 9 + 0 = 1 \) ✓

Both conditions are satisfied, hence the given line lies in the given plane.

Example 15. Find the coordinates of the point where the line \( \frac{x + 1}{2} = \frac{y + 2}{3} = \frac{z + 3}{4} \) meets the plane \( x + y + 4z = 6 \).
Solution: Any point on the line is \( P(-1 + 2\lambda, -2 + 3\lambda, -3 + 4\lambda) \), where \( \lambda \) is an arbitrary real number.

For this point to lie on the plane \( x + y + 4z = 6 \):

\( (-1 + 2\lambda) + (-2 + 3\lambda) + 4(-3 + 4\lambda) = 6 \)

\( -1 + 2\lambda - 2 + 3\lambda - 12 + 16\lambda = 6 \)

\( 21\lambda = 21 \Rightarrow \lambda = 1 \)

Substituting \( \lambda = 1 \): \( P(-1 + 2 \times 1, -2 + 3 \times 1, -3 + 4 \times 1) = (1, 1, 1) \)

Example 16. Find the coordinates of the point where the line joining the points (1, -2, 3) and (2, -1, 5) cuts the plane \( x - 2y + 3z = 19 \). Hence, find the distance of this point from the point (5, 4, 1). (I.S.C. 2008)
Solution: The equation of the line passing through (1, -2, 3) and (2, -1, 5) is:

\( \frac{x - 1}{2 - 1} = \frac{y - (-2)}{-1 - (-2)} = \frac{z - 3}{5 - 3} \), i.e., \( \frac{x - 1}{1} = \frac{y + 2}{1} = \frac{z - 3}{2} \)

Any point on this line is \( P(1 + t, -2 + t, 3 + 2t) \).

For this point to lie on the plane \( x - 2y + 3z = 19 \):

\( (1 + t) - 2(-2 + t) + 3(3 + 2t) = 19 \)

\( 1 + t + 4 - 2t + 9 + 6t = 19 \)

\( 5t = 5 \Rightarrow t = 1 \)

Substituting \( t = 1 \): the point is \( P(2, -1, 5) \)

The line joining the given points intersects the given plane at (2, -1, 5).

Required distance = distance between (2, -1, 5) and (5, 4, 1)

\( = \sqrt{(5 - 2)^2 + (4 - (-1))^2 + (1 - 5)^2} = \sqrt{3^2 + 5^2 + 4^2} = \sqrt{9 + 25 + 16} = \sqrt{50} = 5\sqrt{2} \text{ units} \)

Example 17. Find the distance of the point (1, -2, 3) from the plane \( x - y + z = 5 \) measured along a line parallel to \( \frac{x}{2} = \frac{y}{3} = \frac{z}{-6} \).
Solution: Let A be (1, -2, 3). The equations of the line through A and parallel to the given line are:

\( \frac{x - 1}{2} = \frac{y + 2}{3} = \frac{z - 3}{-6} \)

Any point on this line is \( P(1 + 2\lambda, -2 + 3\lambda, 3 - 6\lambda) \), where \( \lambda \) is an arbitrary real number.

For this point to lie on the plane \( x - y + z = 5 \):

\( (1 + 2\lambda) - (-2 + 3\lambda) + (3 - 6\lambda) = 5 \)

\( 1 + 2\lambda + 2 - 3\lambda + 3 - 6\lambda = 5 \)

\( 6 - 7\lambda = 5 \Rightarrow \lambda = \frac{1}{7} \)

Substituting this value: the point of intersection \( P\left(\frac{9}{7}, -\frac{11}{7}, \frac{15}{7}\right) \)

Required distance \( = AP = \sqrt{\left(\frac{9}{7} - 1\right)^2 + \left(-\frac{11}{7} - (-2)\right)^2 + \left(\frac{15}{7} - 3\right)^2} \)

\( = \sqrt{\left(\frac{2}{7}\right)^2 + \left(\frac{3}{7}\right)^2 + \left(-\frac{6}{7}\right)^2} = \sqrt{\frac{4 + 9 + 36}{49}} = \sqrt{\frac{49}{49}} = 1 \text{ unit} \)

Example 18. Find the coordinates of the foot of perpendicular drawn from the origin to the plane \( 2x - 3y + 4z - 6 = 0 \).
Solution: Direction ratios of a normal to the plane are \( \langle 2, -3, 4 \rangle \).

The equation of the line through origin O(0, 0, 0) and perpendicular to the plane is:

\( \frac{x - 0}{2} = \frac{y - 0}{-3} = \frac{z - 0}{4} \)

Any point on this line is \( N(2\lambda, -3\lambda, 4\lambda) \), where \( \lambda \) is an arbitrary real number.

If this point lies on the plane \( 2x - 3y + 4z - 6 = 0 \):

\( 2(2\lambda) - 3(-3\lambda) + 4(4\lambda) - 6 = 0 \)

\( 4\lambda + 9\lambda + 16\lambda - 6 = 0 \)

\( 29\lambda = 6 \Rightarrow \lambda = \frac{6}{29} \)

The foot of perpendicular from origin to the given plane is:

\( N\left(2 \cdot \frac{6}{29}, -3 \cdot \frac{6}{29}, 4 \cdot \frac{6}{29}\right) = \left(\frac{12}{29}, -\frac{18}{29}, \frac{24}{29}\right) \)

Example 19. Find the length and the foot of perpendicular from the point (1, 1, 2) to the plane \( 2x - 2y + 4z + 5 = 0 \).
Solution: Direction numbers of a normal to the plane are \( \langle 2, -2, 4 \rangle \) or equivalently \( \langle 1, -1, 2 \rangle \).

The equations of the line through point P(1, 1, 2) and perpendicular to the plane are:

\( \frac{x - 1}{1} = \frac{y - 1}{-1} = \frac{z - 2}{2} \)

Any point on this line is \( N(1 + \lambda, 1 - \lambda, 2 + 2\lambda) \).

If it lies on the plane \( 2x - 2y + 4z + 5 = 0 \):

\( 2(1 + \lambda) - 2(1 - \lambda) + 4(2 + 2\lambda) + 5 = 0 \)

\( 2 + 2\lambda - 2 + 2\lambda + 8 + 8\lambda + 5 = 0 \)

\( 12\lambda + 13 = 0 \Rightarrow \lambda = -\frac{13}{12} \)

The foot of perpendicular from P to the plane is:

\( N\left(1 - \frac{13}{12}, 1 + \frac{13}{12}, 2 - \frac{26}{12}\right) = \left(-\frac{1}{12}, \frac{25}{12}, -\frac{1}{6}\right) \)

The length of perpendicular from P to the plane is:

\( NP = \sqrt{\left(-\frac{1}{12} - 1\right)^2 + \left(\frac{25}{12} - 1\right)^2 + \left(-\frac{1}{6} - 2\right)^2} \)

\( = \sqrt{\left(-\frac{13}{12}\right)^2 + \left(\frac{13}{12}\right)^2 + \left(-\frac{13}{6}\right)^2} = \sqrt{\frac{169 + 169 + 676}{144}} = \sqrt{\frac{1014}{144}} = \frac{13\sqrt{6}}{12} \text{ units} \)

Example 20. Find the image of the point (1, 2, 3) in the plane \( x + 2y + 4z = 38 \).
Solution: First, we verify that the point P(1, 2, 3) does not lie in the plane \( x + 2y + 4z = 38 \).

Let M be the foot of perpendicular from P to the plane. Direction ratios of a normal to the plane are \( \langle 1, 2, 4 \rangle \).

The equations of line MP are:

\( \frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{4} \)

Any point on this line is \( Q(1 + \lambda, 2 + 2\lambda, 3 + 4\lambda) \), where \( \lambda \) is an arbitrary real number.

For Q to become the image of P in the plane, M must be the midpoint of segment PQ. The coordinates of M are:

\( M\left(\frac{1 + 1 + \lambda}{2}, \frac{2 + 2 + 2\lambda}{2}, \frac{3 + 3 + 4\lambda}{2}\right) = M\left(\frac{2 + \lambda}{2}, 2 + \lambda, 3 + 2\lambda\right) \)

Since M lies on the plane:

\( \frac{2 + \lambda}{2} + 2(2 + \lambda) + 4(3 + 2\lambda) = 38 \)

\( \frac{2 + \lambda}{2} + 4 + 2\lambda + 12 + 8\lambda = 38 \)

\( \frac{2 + \lambda}{2} + 2\lambda + 8\lambda + 16 = 38 \)

\( \frac{2 + \lambda + 4\lambda + 16\lambda}{2} = 22 \)

\( 2 + 21\lambda = 44 \Rightarrow 21\lambda = 42 \Rightarrow \lambda = 2 \)

Substituting \( \lambda = 2 \), the image point Q is (3, 6, 11).

Exercise 3.1

1. Find the direction cosines of the normal to the plane \( 2x + 3y - z - 7 = 0 \).

2. Find a unit normal vector to the plane \( 2x - 2y - z - 5 = 0 \).

3. Find the cartesian equation of the following planes:

(i) \( \vec{r} \cdot (\hat{i} + \hat{j} - \hat{k}) = 2 \)
(ii) \( \vec{r} \cdot (2\hat{i} + 3\hat{j} - 4\hat{k}) = 1 \)
(iii) \( \vec{r} \cdot ((s - 2t)\hat{i} + (3 - t)\hat{j} + (2s + t)\hat{k}) = 15 \)

4. Find the vector equations of the following planes:

(i) \( 3x + 7y - 6z = 5 \)
(ii) \( 9x - 2y - 5z + 3 = 0 \)

5. If O be the origin and the coordinates of P be (1, 2, -3), then find the equation of the plane through P and perpendicular to OP.

6. Find the equation of the plane passing through the point (2, -3, 1) and perpendicular to the line whose direction ratios are \( \langle 3, -1, 5 \rangle \).

7. Find the equation of the plane passing through the point (1, 2, 3) and perpendicular to the line \( \frac{x}{-2} = \frac{y}{4} = \frac{z}{3} \). (I.S.C. 2002)

8. Show that the line \( \frac{x - 2}{1} = \frac{y - 1}{-1} = \frac{z + 4}{4} \) is parallel to the plane \( x + 5y + z = 7 \).

9. If the line \( \frac{x + 1}{3} = \frac{y - 2}{4} = \frac{z + 6}{5} \) is parallel to the plane \( 2x - 3y + kz = 0 \), then find the value of k.

10. If (3, 6, 11) is the image of the point (1, 2, 3) in the plane \( x + 2y + 4z + k = 0 \), then find the value of k.

11. If (3, k, 6) is the image of the point (1, 3, 4) in the plane \( x - y + z = 5 \), then find the value of k.

12. Write the image of the point (-2, 3, 5) in the plane XOY.

13. Find the vector equation of a plane which is at a distance of 5 units from the origin and which is normal to the vector \( 3\hat{i} + 2\hat{j} - 6\hat{k} \).

14. If the vector equation of a plane is \( \vec{r} \cdot (2\hat{i} + 2\hat{j} - \hat{k}) = 21 \), find the length of perpendicular from the origin to the plane.

15. Find the vector equation of the plane which is at a distance of \( \frac{6}{\sqrt{29}} \) units from the origin and its normal vector from the origin is \( 2\hat{i} - 3\hat{j} + 4\hat{k} \). Also find its cartesian equation.

16. In each of the following problems, determine the direction cosines of the normal to the plane and its distance from the origin:

(i) \( 2x - 3y + 4z - 6 = 0 \)
(ii) \( x + y + z = 1 \)
(iii) \( z = 2 \)
(iv) \( 3x + 5 = 0 \)

17. Find the vector equation of a plane which is at a distance of 5 units from the origin and has \( \langle 2, -1, 2 \rangle \) as the direction numbers of a normal to the plane.

18. Find the vector equation of a plane which is at a distance of \( 2\sqrt{3} \) units from the origin and a normal to the plane makes equal acute angles with coordinate axes.

Hint: Since a normal makes equal acute angle with coordinate axes, \( l = m = n \), where l, m, n are direction cosines of the normal. But \( l^2 + m^2 + n^2 = 1 \Rightarrow 3l^2 = 1 \Rightarrow l = \frac{1}{\sqrt{3}} \). Use \( \vec{r} \cdot \hat{n} = p \).

19. Find the vector and the cartesian equations of the plane

(i) that passes through the point (1, 0, -2) and normal vector to the plane is \( \hat{i} + \hat{j} - \hat{k} \).
(ii) that passes through the point (1, 4, 6) and normal vector to the plane is \( \hat{i} - 2\hat{j} + \hat{k} \).

20. Find the vector and cartesian equation of the plane passing through the point (1, 2, 3) and perpendicular to the line with direction ratios \( \langle 2, 3, -4 \rangle \).

21. (i) Find the vector and the cartesian equation of the plane through the point with position vector \( 2\hat{i} - \hat{j} + \hat{k} \) and perpendicular to the vector \( 4\hat{i} + 2\hat{j} - 3\hat{k} \).
(ii) If the foot of perpendicular drawn from the origin to a plane is (12, -4, -3), find the equation of the plane in vector as well as cartesian form.