Mastering 3 Dimensional Geometry Miscellaneous Exercise (NCERT) - Class 12 Maths

The Essence of Miscellaneous Exercise in 3D Geometry

The Miscellaneous Exercise in 3D Geometry is strategically placed at the end of the chapter to serve as a comprehensive review and application ground for all concepts covered. It presents problems that often require integrating knowledge from different sub-topics, such as simultaneously dealing with properties of lines and planes, finding distances under complex conditions, or determining angles between various geometric entities. These problems are designed to test not just your memory of formulas, but your conceptual understanding and analytical skills. For instance, you might be asked to find the equation of a plane that passes through a given point and also contains a specific line, or to calculate the shortest distance between two lines presented in a non-standard form. Success in this exercise means you've truly grasped the interplay between vector and Cartesian forms, understood the geometric interpretations of equations, and can confidently apply theorems to solve real mathematical challenges. Mastering this section is vital for securing top marks in your board exams, as it hones the critical thinking skills necessary for advanced mathematics.

Recap of Essential 3D Geometry Formulas

Equation of a Line (Vector Form)

Passing through a point \(\vec{a}\) and parallel to vector \(\vec{b}\): \(\vec{r} = \vec{a} + \lambda\vec{b}\). Passing through two points \(\vec{a}\) and \(\vec{b}\): \(\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})\).

Equation of a Line (Cartesian Form)

Passing through \((x_1, y_1, z_1)\) and having direction ratios \((a, b, c)\): \(\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}\).

Equation of a Plane (Normal Form)

Distance 'd' from origin, normal vector \(\hat{n}\): \(\vec{r} \cdot \hat{n} = d\). Cartesian: \(lx + my + nz = d\).

Equation of a Plane (Passing through a point)

Passing through \(\vec{a}\) and normal to \(\vec{n}\): \((\vec{r} - \vec{a}) \cdot \vec{n} = 0\). Cartesian: \(A(x-x_1) + B(y-y_1) + C(z-z_1) = 0\).

Shortest Distance between Skew Lines

Lines \(\vec{r}_1 = \vec{a}_1 + \lambda\vec{b}_1\) and \(\vec{r}_2 = \vec{a}_2 + \mu\vec{b}_2\): \(d = \left|\frac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|}\right|\).

Angle between two Lines

Lines parallel to \(\vec{b}_1\) and \(\vec{b}_2\): \(\cos\theta = \left|\frac{\vec{b}_1 \cdot \vec{b}_2}{|\vec{b}_1||\vec{b}_2|}\right|\).

Angle between two Planes

Planes with normals \(\vec{n}_1\) and \(\vec{n}_2\): \(\cos\theta = \left|\frac{\vec{n}_1 \cdot \vec{n}_2}{|\vec{n}_1||\vec{n}_2|}\right|\).

Angle between a Line and a Plane

Line parallel to \(\vec{b}\), plane normal to \(\vec{n}\): \(\sin\theta = \left|\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}\right|\).

Worked Examples from Miscellaneous Exercise

• Example 1: Finding Shortest Distance between Skew Lines Find the shortest distance between the lines \(\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(2\hat{i} + \hat{j} + 2\hat{k})\) and \(\vec{r} = (- \hat{i} - \hat{j} + \hat{k}) + \mu(\hat{i} - \hat{j} + \hat{k})\). Step 1: Identify \(\vec{a}_1, \vec{b}_1, \vec{a}_2, \vec{b}_2\). From the first line: \(\vec{a}_1 = \hat{i} + 2\hat{j} + \hat{k}\), \(\vec{b}_1 = 2\hat{i} + \hat{j} + 2\hat{k}\). From the second line: \(\vec{a}_2 = - \hat{i} - \hat{j} + \hat{k}\), \(\vec{b}_2 = \hat{i} - \hat{j} + \hat{k}\). Step 2: Calculate \(\vec{a}_2 - \vec{a}_1\). \(\vec{a}_2 - \vec{a}_1 = (- \hat{i} - \hat{j} + \hat{k}) - (\hat{i} + 2\hat{j} + \hat{k}) = -2\hat{i} - 3\hat{j} + 0\hat{k}\). Step 3: Calculate \(\vec{b}_1 \times \vec{b}_2\). \(\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 2 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(1 - (-2)) - \hat{j}(2 - 2) + \hat{k}(-2 - 1) = 3\hat{i} - 0\hat{j} - 3\hat{k}\). Step 4: Calculate \(|\vec{b}_1 \times \vec{b}_2|\). \(|\vec{b}_1 \times \vec{b}_2| = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}\). Step 5: Calculate \((\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)\). \((3\hat{i} - 3\hat{k}) \cdot (-2\hat{i} - 3\hat{j}) = (3)(-2) + (0)(-3) + (-3)(0) = -6\). Step 6: Apply the shortest distance formula. \(d = \left|\frac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|}\right| = \left|\frac{-6}{3\sqrt{2}}\right| = \left|\frac{-2}{\sqrt{2}}\right| = \left|-\sqrt{2}\right| = \sqrt{2}\). Final Answer: The shortest distance between the lines is \(\sqrt{2}\) units. ---
• Example 2: Finding the Equation of a Plane Find the vector and Cartesian equations of the plane passing through the points \((2, 5, -3)\), \((-2, -3, 5)\) and \((5, 3, -3)\). Step 1: Define position vectors for the given points. Let \(A = (2, 5, -3)\), \(B = (-2, -3, 5)\), \(C = (5, 3, -3)\). \(\vec{a} = 2\hat{i} + 5\hat{j} - 3\hat{k}\) \(\vec{b} = -2\hat{i} - 3\hat{j} + 5\hat{k}\) \(\vec{c} = 5\hat{i} + 3\hat{j} - 3\hat{k}\) Step 2: Find two vectors lying in the plane. \(\vec{AB} = \vec{b} - \vec{a} = (-2-2)\hat{i} + (-3-5)\hat{j} + (5-(-3))\hat{k} = -4\hat{i} - 8\hat{j} + 8\hat{k}\) \(\vec{AC} = \vec{c} - \vec{a} = (5-2)\hat{i} + (3-5)\hat{j} + (-3-(-3))\hat{k} = 3\hat{i} - 2\hat{j} + 0\hat{k}\) Step 3: Find the normal vector to the plane using the cross product. \(\vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & -8 & 8 \\ 3 & -2 & 0 \end{vmatrix}\) \(\vec{n} = \hat{i}((-8)(0) - (8)(-2)) - \hat{j}((-4)(0) - (8)(3)) + \hat{k}((-4)(-2) - (-8)(3))\) \(\vec{n} = \hat{i}(0 + 16) - \hat{j}(0 - 24) + \hat{k}(8 + 24)\) \(\vec{n} = 16\hat{i} + 24\hat{j} + 32\hat{k}\) We can simplify the normal vector by taking a common factor of 8: \(\vec{n}' = 2\hat{i} + 3\hat{j} + 4\hat{k}\). Step 4: Write the vector equation of the plane. Using point A \((\vec{a})\) and normal vector \(\vec{n}'\): \((\vec{r} - \vec{a}) \cdot \vec{n}' = 0\) \((\vec{r} - (2\hat{i} + 5\hat{j} - 3\hat{k})) \cdot (2\hat{i} + 3\hat{j} + 4\hat{k}) = 0\) Step 5: Write the Cartesian equation of the plane. Let \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\). \(( (x-2)\hat{i} + (y-5)\hat{j} + (z+3)\hat{k} ) \cdot (2\hat{i} + 3\hat{j} + 4\hat{k}) = 0\) \(2(x-2) + 3(y-5) + 4(z+3) = 0\) \(2x - 4 + 3y - 15 + 4z + 12 = 0\) \(2x + 3y + 4z - 7 = 0\) Final Answer: Vector Equation: \((\vec{r} - (2\hat{i} + 5\hat{j} - 3\hat{k})) \cdot (2\hat{i} + 3\hat{j} + 4\hat{k}) = 0\) Cartesian Equation: \(2x + 3y + 4z - 7 = 0\) ---
• Example 3: Distance of a Point from a Plane Find the distance of the point \((3, 4, 5)\) from the plane \(\vec{r} \cdot (2\hat{i} - 3\hat{j} + 6\hat{k}) + 11 = 0\). Step 1: Identify the point and the plane equation. Point \((x_1, y_1, z_1) = (3, 4, 5)\). Plane equation in vector form: \(\vec{r} \cdot (2\hat{i} - 3\hat{j} + 6\hat{k}) + 11 = 0\). Step 2: Convert the plane equation to Cartesian form if necessary, and identify A, B, C, D. The plane equation is \(\vec{r} \cdot \vec{n} = D\). Our given equation is \(\vec{r} \cdot (2\hat{i} - 3\hat{j} + 6\hat{k}) = -11\). Comparing with \(Ax + By + Cz = D\), we have \(A=2, B=-3, C=6\) and \(D=-11\). The formula for distance of a point \((x_1, y_1, z_1)\) from the plane \(Ax + By + Cz + D' = 0\) is \(d = \frac{|Ax_1 + By_1 + Cz_1 + D'|}{\sqrt{A^2 + B^2 + C^2}}\). So, we rewrite the plane equation as \(2x - 3y + 6z + 11 = 0\), where \(D' = 11\). Step 3: Calculate the denominator \(\sqrt{A^2 + B^2 + C^2}\). \(\sqrt{A^2 + B^2 + C^2} = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7\). Step 4: Substitute the point coordinates into the numerator. \(|Ax_1 + By_1 + Cz_1 + D'| = |(2)(3) + (-3)(4) + (6)(5) + 11|\) \(= |6 - 12 + 30 + 11| = |-6 + 30 + 11| = |24 + 11| = |35| = 35\). Step 5: Calculate the distance. \(d = \frac{35}{7} = 5\). Final Answer: The distance of the point \((3, 4, 5)\) from the given plane is 5 units.

Exam Strategies and Common Pitfalls

Navigating the miscellaneous exercise and your board exams requires smart strategies and awareness of common errors. Here's how to maximize your score:

1. Read Carefully, Visualize Clearly: Many errors stem from misinterpreting the question. Is it a line or a plane? Is it shortest distance between skew lines or parallel lines? Draw a rough diagram if it helps you visualize the scenario.
2. Master Vector-Cartesian Conversions: Be adept at switching between vector and Cartesian forms. Some problems are easier to solve in one form than the other, or may require starting in one and converting to another for the final answer. For instance, sometimes finding the normal vector in Cartesian form is quicker with coordinates, then converting it back to vector form.
3. Check for Perpendicularity/Parallelism: These conditions simplify many problems. Remember that for parallel lines, direction vectors are proportional (\(\vec{b}_1 = k\vec{b}_2\)); for perpendicular lines, their dot product is zero (\(\vec{b}_1 \cdot \vec{b}_2 = 0\)). For planes, similar rules apply to their normal vectors.
4. Scalar Triple Product for Coplanarity: If three vectors are coplanar, their scalar triple product is zero. This is a powerful tool for checking if points are coplanar or if lines intersect.
5. Algebraic Accuracy: 3D Geometry involves a lot of vector algebra (dot products, cross products) and arithmetic. A single sign error or calculation mistake can lead to a completely wrong answer. Double-check your calculations, especially cross products.
6. Distance vs. Directed Distance: When asked for 'distance', remember it's always a positive value. Use the absolute value in distance formulas. If a question implies 'signed distance' or a particular side, be careful with the sign.
7. Practice Diverse Problems: The miscellaneous exercise covers a wide range of problem types. Don't just stick to one type you're comfortable with. Practice finding angles, distances, equations of lines/planes under various conditions.

Practice Questions with Solutions

• Q: Find the distance of the point \((2, 3, -5))\) from the plane \(\vec{r} \cdot (\hat{i} + 2\hat{j} - 2\hat{k}) = 9\). A: Step 1: Write the Cartesian equation of the plane. Given vector equation is \(\vec{r} \cdot (\hat{i} + 2\hat{j} - 2\hat{k}) = 9\). Let \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\). So, \((x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} + 2\hat{j} - 2\hat{k}) = 9\). This gives the Cartesian equation: \(x + 2y - 2z = 9\), or \(x + 2y - 2z - 9 = 0\). Step 2: Identify the coordinates of the point \((x_1, y_1, z_1))\) and coefficients of the plane \((A, B, C, D'))\). Point \((x_1, y_1, z_1) = (2, 3, -5)\). Plane coefficients: \(A=1, B=2, C=-2, D'=-9\). Step 3: Use the distance formula for a point from a plane. The distance \(d = \frac{|Ax_1 + By_1 + Cz_1 + D'|}{\sqrt{A^2 + B^2 + C^2}}\) . Step 4: Substitute the values and calculate. Numerator: \(|(1)(2) + (2)(3) + (-2)(-5) - 9| = |2 + 6 + 10 - 9| = |18 - 9| = |9| = 9\). Denominator: \(\sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3\). Distance \(d = \frac{9}{3} = 3\) units. Final answer: The distance of the point from the plane is 3 units. ---
• Q: 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 \((2, 2, 1))\). A: Step 1: Use the equation of a plane passing through the intersection of two planes. The equation of a plane passing through the intersection of planes \(P_1 = 0\) and \(P_2 = 0\) is given by \(P_1 + \lambda P_2 = 0\). Here, \((3x - y + 2z - 4) + \lambda(x + y + z - 2) = 0\). Step 2: Use the given point to find the value of \(\lambda\). The plane passes through \((2, 2, 1))\). Substitute these coordinates into the equation: \((3(2) - 2 + 2(1) - 4) + \lambda(2 + 2 + 1 - 2) = 0\) \((6 - 2 + 2 - 4) + \lambda(3) = 0\) \(2 + 3\lambda = 0\) \(3\lambda = -2 \implies \lambda = -\frac{2}{3}\). Step 3: Substitute the value of \(\lambda\) back into the plane equation. \((3x - y + 2z - 4) - \frac{2}{3}(x + y + z - 2) = 0\) Multiply by 3 to clear the fraction: \(3(3x - y + 2z - 4) - 2(x + y + z - 2) = 0\) \(9x - 3y + 6z - 12 - 2x - 2y - 2z + 4 = 0\) Step 4: Simplify to get the final equation. \(7x - 5y + 4z - 8 = 0\). Final answer: The equation of the plane is \(7x - 5y + 4z - 8 = 0\). ---
• Q: Find the angle between the line \(\frac{x+1}{2} = \frac{y}{3} = \frac{z-3}{6}\) and the plane \(10x + 2y - 11z = 3\). A: Step 1: Identify the direction ratios of the line and the normal vector of the plane. For the line \(\frac{x+1}{2} = \frac{y}{3} = \frac{z-3}{6}\), the direction vector is \(\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}\). For the plane \(10x + 2y - 11z = 3\), the normal vector is \(\vec{n} = 10\hat{i} + 2\hat{j} - 11\hat{k}\). Step 2: Calculate the magnitudes of \(\vec{b}\) and \(\vec{n}\). \(|\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7\). \(|\vec{n}| = \sqrt{10^2 + 2^2 + (-11)^2} = \sqrt{100 + 4 + 121} = \sqrt{225} = 15\). Step 3: Use the formula for the angle between a line and a plane. The angle \(\theta\) between a line with direction vector \(\vec{b}\) and a plane with normal vector \(\vec{n}\) is given by \(\sin\theta = \left|\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}\right|\). Step 4: Calculate the dot product \(\vec{b} \cdot \vec{n}\). \(\vec{b} \cdot \vec{n} = (2)(10) + (3)(2) + (6)(-11) = 20 + 6 - 66 = 26 - 66 = -40\). Step 5: Substitute values into the formula. \(\sin\theta = \left|\frac{-40}{(7)(15)}\right| = \left|\frac{-40}{105}\right| = \frac{40}{105}\). Simplify the fraction: \(\frac{40}{105} = \frac{8 \times 5}{21 \times 5} = \frac{8}{21}\). Step 6: Express \(\theta\). \(\theta = \arcsin\left(\frac{8}{21}\right)\). Final answer: The angle between the line and the plane is \(\arcsin\left(\frac{8}{21}\right)\). ---
• Q: Find the coordinates of the foot of the perpendicular drawn from the point \((1, 0, 3))\) to the line \(\frac{x+2}{3} = \frac{y-3}{-1} = \frac{z+1}{5}\). A: Step 1: Represent a general point on the line. Let the line be \(\frac{x+2}{3} = \frac{y-3}{-1} = \frac{z+1}{5} = \lambda\). Any point R on this line can be written as \((3\lambda - 2, -\lambda + 3, 5\lambda - 1))\). Step 2: Form the vector PR, where P is the given point. Let \(P = (1, 0, 3)\). The vector \(\vec{PR}\) will be \((3\lambda - 2 - 1)\hat{i} + (-\lambda + 3 - 0)\hat{j} + (5\lambda - 1 - 3)\hat{k}\). \(\vec{PR} = (3\lambda - 3)\hat{i} + (-\lambda + 3)\hat{j} + (5\lambda - 4)\hat{k}\). Step 3: Use the condition that PR is perpendicular to the line. The direction vector of the line is \(\vec{b} = 3\hat{i} - \hat{j} + 5\hat{k}\). Since \(\vec{PR}\) is perpendicular to the line, their dot product must be zero: \(\vec{PR} \cdot \vec{b} = 0\). \((3\lambda - 3)(3) + (-\lambda + 3)(-1) + (5\lambda - 4)(5) = 0\) \(9\lambda - 9 + \lambda - 3 + 25\lambda - 20 = 0\) \(35\lambda - 32 = 0\) \(35\lambda = 32 \implies \lambda = \frac{32}{35}\). Step 4: Substitute \(\lambda\) back into the coordinates of R to find the foot of the perpendicular. \(x = 3\left(\frac{32}{35}\right) - 2 = \frac{96}{35} - \frac{70}{35} = \frac{26}{35}\) \(y = -\frac{32}{35} + 3 = -\frac{32}{35} + \frac{105}{35} = \frac{73}{35}\) \(z = 5\left(\frac{32}{35}\right) - 1 = \frac{160}{35} - \frac{35}{35} = \frac{125}{35} = \frac{25}{7}\) Final answer: The coordinates of the foot of the perpendicular are \(\left(\frac{26}{35}, \frac{73}{35}, \frac{25}{7}\right)\). ---

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