3 Dimensional Geometry Ex 11.3 for Class 12 Maths NCERT

Understanding the Equation of a Plane

A plane is a flat, two-dimensional surface that extends infinitely in three-dimensional space. Unlike lines, planes are defined by their orientation and position relative to the origin. Exercise 11.3 introduces several ways to represent a plane mathematically, each useful depending on the given information. The Normal Form of the equation of a plane uses the direction cosines of its normal vector and its perpendicular distance from the origin. This form is intuitive when you know how far the plane is from the origin and in what direction its perpendicular points. We also learn about the equation of a plane passing through a given point and perpendicular to a given vector, which is a direct application of the normal vector concept. Furthermore, the Intercept Form is incredibly useful when the plane cuts the coordinate axes at specific points. It simplifies problems where you need to find the equation of a plane given its intercepts. Finally, you'll learn how to find the equation of a plane passing through three non-collinear points, a common scenario in geometry. Mastering these different forms and knowing when to apply each one is key to excelling in this exercise and beyond.

Key Forms and Formulas for Planes

Equation of a Plane in Normal Form

The equation of a plane at a perpendicular distance 'p' from the origin and having $\hat{n}$ as the unit normal vector to the plane is $\vec{r} \cdot \hat{n} = p$. In Cartesian form, if $\hat{n} = l\hat{i} + m\hat{j} + n\hat{k}$, then $lx + my + nz = p$, where $l, m, n$ are the direction cosines of the normal to the plane.

Equation of a Plane Passing Through a Given Point and Normal to a Vector

If a plane passes through a point $A(\vec{a})$ and is normal to vector $\vec{N}$, its vector equation is $(\vec{r} - \vec{a}) \cdot \vec{N} = 0$. In Cartesian form, if $A = (x_1, y_1, z_1)$ and $\vec{N} = A\hat{i} + B\hat{j} + C\hat{k}$, the equation is $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$.

Equation of a Plane Passing Through Three Non-Collinear Points

For three non-collinear points $A(\vec{a})$, $B(\vec{b})$, $C(\vec{c})$, the vector equation of the plane is $(\vec{r} - \vec{a}) \cdot [(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})] = 0$. In Cartesian form, it can be found by solving the determinant: $\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0$.

Intercept Form of the Equation of a Plane

If a plane makes intercepts $a, b, c$ with the X, Y, and Z axes respectively, its equation is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.

Angle Between Two Planes

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 $\cos\theta = \left| \frac{\vec{n_1} \cdot \vec{n_2}}{|\vec{n_1}||\vec{n_2}|} \right|$. In Cartesian form, for planes $A_1x + B_1y + C_1z + D_1 = 0$ and $A_2x + B_2y + C_2z + D_2 = 0$, $\cos\theta = \left| \frac{A_1A_2 + B_1B_2 + C_1C_2}{\sqrt{A_1^2+B_1^2+C_1^2} \sqrt{A_2^2+B_2^2+C_2^2}} \right|$.

Distance of a Point from a Plane

The perpendicular distance 'D' of a point $P(x_1, y_1, z_1)$ from the plane $Ax + By + Cz + D = 0$ is given by $D = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$. For a plane in vector form $\vec{r} \cdot \vec{n} = d$, the distance of a point with position vector $\vec{a}$ is $D = \frac{|\vec{a} \cdot \vec{n} - d|}{|\vec{n}|}$.

Step-by-Step Worked Examples

• Example 1: Find the equation of the plane that passes through the point $(1, -2, 4)$ and has $(2, 1, -1)$ as the direction ratios of its normal. Step 1: Identify the given information. The plane passes through point $P_0(x_0, y_0, z_0) = (1, -2, 4)$. The normal vector $\vec{N}$ has direction ratios $(A, B, C) = (2, 1, -1)$, so $\vec{N} = 2\hat{i} + \hat{j} - \hat{k}$. Step 2: Choose the appropriate formula. We use the Cartesian equation of a plane passing through a point $(x_0, y_0, z_0)$ and normal to a vector $\vec{N} = A\hat{i} + B\hat{j} + C\hat{k}$, which is $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$. Step 3: Substitute the values and simplify. $2(x - 1) + 1(y - (-2)) + (-1)(z - 4) = 0$ $2(x - 1) + (y + 2) - (z - 4) = 0$ $2x - 2 + y + 2 - z + 4 = 0$ $2x + y - z + 4 = 0$ Final Answer: The equation of the plane is $2x + y - z + 4 = 0$.
• Example 2: Find the angle between the planes $2x + y - 2z = 5$ and $3x - 6y - 2z = 7$. Step 1: Identify the normal vectors of the planes. For plane 1: $2x + y - 2z = 5$, the normal vector $\vec{n_1} = 2\hat{i} + \hat{j} - 2\hat{k}$. For plane 2: $3x - 6y - 2z = 7$, the normal vector $\vec{n_2} = 3\hat{i} - 6\hat{j} - 2\hat{k}$. Step 2: Calculate the dot product of the normal vectors. $\vec{n_1} \cdot \vec{n_2} = (2)(3) + (1)(-6) + (-2)(-2) = 6 - 6 + 4 = 4$. Step 3: Calculate the magnitudes of the normal vectors. $|\vec{n_1}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$. $|\vec{n_2}| = \sqrt{3^2 + (-6)^2 + (-2)^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$. Step 4: Use the formula for the angle between two planes. $\cos\theta = \left| \frac{\vec{n_1} \cdot \vec{n_2}}{|\vec{n_1}||\vec{n_2}|} \right| = \left| \frac{4}{3 \cdot 7} \right| = \frac{4}{21}$. * Final Answer: The angle between the planes is $\theta = \cos^{-1}\left(\frac{4}{21}\right)$.
• Example 3: Find the distance of the point $(3, -2, 1)$ from the plane $2x - y + 2z + 3 = 0$. Step 1: Identify the given point and the equation of the plane. Point $(x_1, y_1, z_1) = (3, -2, 1)$. Plane equation $Ax + By + Cz + D = 0$ is $2x - y + 2z + 3 = 0$, so $A=2, B=-1, C=2, D=3$. Step 2: Choose the appropriate formula. The distance 'D' of a point $(x_1, y_1, z_1)$ from the plane $Ax + By + Cz + D = 0$ is $D = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$. Step 3: Substitute the values and calculate. Numerator: $|(2)(3) + (-1)(-2) + (2)(1) + 3| = |6 + 2 + 2 + 3| = |13| = 13$. Denominator: $\sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$. Step 4: Compute the distance. $D = \frac{13}{3}$. * Final Answer: The distance of the point $(3, -2, 1)$ from the plane $2x - y + 2z + 3 = 0$ is $\frac{13}{3}$ units.

Exam Tips for 3D Geometry Ex 11.3

To ace questions from 3D Geometry Ex 11.3, remember these key strategies:

1. Understand the Normal Vector: The normal vector is the backbone of plane equations. Always identify or calculate the normal vector first. Its direction ratios (or direction cosines) are crucial for most formulas.
2. Choose the Right Form: Don't try to fit all problems into one formula. If intercepts are given, use the intercept form. If a point and a normal are given, use the point-normal form. If three points are given, use the determinant form or the vector cross product method. Selecting the most efficient formula saves time and reduces errors.
3. Vector vs. Cartesian: Be comfortable switching between vector and Cartesian forms. Often, one form might be easier for initial setup, while the other is better for calculation. For instance, calculating magnitudes and dot products is often straightforward in vector form, while representing the final plane equation might be clearer in Cartesian.
4. Careful with Signs and Magnitudes: When calculating the distance of a point from a plane or the angle between planes, pay close attention to absolute values and square roots. A common mistake is forgetting the absolute value for distance, which must always be positive.
5. Perpendicular vs. Parallel: Understand that the normal vector of a plane is perpendicular to the plane itself. If two planes are parallel, their normal vectors are parallel. If two planes are perpendicular, their normal vectors are perpendicular (their dot product is zero).

Practice Questions with Solutions

• Q: Find the vector equation of a plane which is at a distance of 7 units from the origin and its normal vector is $3\hat{i} + 5\hat{j} - 6\hat{k}$. A: Step 1: Identify the given distance 'p' and normal vector $\vec{N}$. Given $p = 7$ units. Normal vector $\vec{N} = 3\hat{i} + 5\hat{j} - 6\hat{k}$. Step 2: Find the unit normal vector $\hat{n}$. $|\vec{N}| = \sqrt{3^2 + 5^2 + (-6)^2} = \sqrt{9 + 25 + 36} = \sqrt{70}$. $\hat{n} = \frac{\vec{N}}{|\vec{N}|} = \frac{3\hat{i} + 5\hat{j} - 6\hat{k}}{\sqrt{70}}$. Step 3: Apply the normal form of the plane equation $\vec{r} \cdot \hat{n} = p$. $\vec{r} \cdot \left(\frac{3\hat{i} + 5\hat{j} - 6\hat{k}}{\sqrt{70}}\right) = 7$. Final answer: The vector equation of the plane is $\vec{r} \cdot (3\hat{i} + 5\hat{j} - 6\hat{k}) = 7\sqrt{70}$.
• Q: Determine the intercepts made by the plane $2x + y - 3z = 6$ on the coordinate axes. A: Step 1: Convert the equation to intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Divide the entire equation $2x + y - 3z = 6$ by 6. $\frac{2x}{6} + \frac{y}{6} - \frac{3z}{6} = \frac{6}{6}$. Step 2: Simplify the equation. $\frac{x}{3} + \frac{y}{6} + \frac{z}{-2} = 1$. Step 3: Identify the intercepts. By comparing with the intercept form, $a=3$, $b=6$, $c=-2$. Final answer: The intercepts made by the plane on the X, Y, and Z axes are 3, 6, and -2 respectively.
• Q: Find the equation of the plane passing through the points $(1, 1, 0)$, $(1, 2, 1)$, and $(-2, 2, -1)$. A: Step 1: Define the position vectors of the points. Let $\vec{a} = \hat{i} + \hat{j}$, $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$, $\vec{c} = -2\hat{i} + 2\hat{j} - \hat{k}$. Step 2: Find two vectors in the plane, for example, $(\vec{b} - \vec{a})$ and $(\vec{c} - \vec{a})$. $(\vec{b} - \vec{a}) = (\hat{i} + 2\hat{j} + \hat{k}) - (\hat{i} + \hat{j}) = \hat{j} + \hat{k}$. $(\vec{c} - \vec{a}) = (-2\hat{i} + 2\hat{j} - \hat{k}) - (\hat{i} + \hat{j}) = -3\hat{i} + \hat{j} - \hat{k}$. Step 3: Find the normal vector to the plane by taking the cross product of these two vectors. $(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ -3 & 1 & -1 \end{vmatrix} = \hat{i}(-1-1) - \hat{j}(0-(-3)) + \hat{k}(0-(-3)) = -2\hat{i} - 3\hat{j} + 3\hat{k}$. Step 4: Use the point-normal form $(\vec{r} - \vec{a}) \cdot \vec{N} = 0$. $(\vec{r} - (\hat{i} + \hat{j})) \cdot (-2\hat{i} - 3\hat{j} + 3\hat{k}) = 0$. Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. $((x-1)\hat{i} + (y-1)\hat{j} + z\hat{k}) \cdot (-2\hat{i} - 3\hat{j} + 3\hat{k}) = 0$. $-2(x-1) - 3(y-1) + 3(z) = 0$. $-2x + 2 - 3y + 3 + 3z = 0$. $-2x - 3y + 3z + 5 = 0$ or $2x + 3y - 3z - 5 = 0$. Final answer: The equation of the plane is $2x + 3y - 3z - 5 = 0$.
• Q: Find the perpendicular distance from the origin $(0, 0, 0)$ to the plane $3x - 4y + 12z = 39$. A: Step 1: Identify the plane equation and the point. Plane: $3x - 4y + 12z - 39 = 0$. So $A=3, B=-4, C=12, D=-39$. Point: $(x_1, y_1, z_1) = (0, 0, 0)$. Step 2: Apply the distance formula $D = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$. Numerator: $|(3)(0) + (-4)(0) + (12)(0) - 39| = |-39| = 39$. Denominator: $\sqrt{3^2 + (-4)^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$. Step 3: Calculate the distance. $D = \frac{39}{13} = 3$. Final answer: The perpendicular distance from the origin to the plane is 3 units.

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