Mastering Vector Algebra Miscellaneous Exercise - Class 12 NCERT

Understanding the Miscellaneous Exercise

The miscellaneous exercise in Vector Algebra is strategically placed at the end of the chapter to serve as a consolidated test of your understanding. Unlike previous exercises that might focus on specific operations like dot products or cross products, this section presents problems that often require integrating knowledge from various sub-topics. You might encounter questions that involve finding angles, areas, volumes, projections, determining collinearity or coplanarity, or even proving vector identities, all within a single problem.

This integrative approach is crucial for developing strong problem-solving skills. It trains you to not just recall formulas but to analyze a problem, identify the underlying vector principles, and formulate a step-by-step solution. It's an excellent way to prepare for the board exams, where questions often test your ability to apply multiple concepts in an unfamiliar context. Through these problems, you'll learn to see vectors not as isolated operations but as a powerful tool for describing and solving geometric problems in three-dimensional space.

Key Vector Concepts You'll Need

Scalar Product (Dot Product)

The dot product of two vectors $\vec{a}$ and $\vec{b}$ is defined as $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta$, where $\theta$ is the angle between them. It is a scalar quantity and is used to find the angle between vectors or the projection of one vector onto another.

Vector Product (Cross Product)

The cross product of two vectors $\vec{a}$ and $\vec{b}$ is defined as $\vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin\theta \hat{n}$, where $\hat{n}$ is a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$. It is a vector quantity and its magnitude represents the area of the parallelogram formed by the vectors.

Scalar Triple Product

For three vectors $\vec{a}$, $\vec{b}$, $\vec{c}$, the scalar triple product is $\vec{a} \cdot (\vec{b} \times \vec{c})$. Geometrically, its absolute value represents the volume of the parallelepiped formed by the three vectors. If the scalar triple product is zero, the three vectors are coplanar.

Projection of a Vector

The projection of vector $\vec{a}$ on vector $\vec{b}$ is given by $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$. This scalar value represents the component of $\vec{a}$ along the direction of $\vec{b}$.

Collinearity and Coplanarity

Vectors are collinear if they are parallel (one is a scalar multiple of the other). Vectors are coplanar if they lie on the same plane, which can be checked using the scalar triple product.

Step-by-Step Approach to Miscellaneous Problems

1. Step 1: Analyze the Problem Statement — Read the problem carefully, identifying all given vectors, magnitudes, directions, and conditions (e.g., perpendicularity, collinearity, area, volume). Visualize the scenario if possible, perhaps by sketching a diagram in 3D space.
2. Step 2: Identify Relevant Concepts and Formulas — Based on the given information and what needs to be found, determine which vector algebra concepts are most applicable. For example, 'angle' suggests dot product, 'area' suggests cross product, 'volume' or 'coplanar' suggests scalar triple product, and 'perpendicular' suggests a zero dot product.
3. Step 3: Formulate a Plan — Break down the problem into smaller, manageable steps. If multiple concepts are involved, decide the order of operations. For example, if you need to find the area of a triangle formed by three points, you first need to form two vectors representing two sides originating from a common vertex, then apply the cross product formula.
4. Step 4: Execute Calculations Systematically — Perform the vector operations and algebraic manipulations carefully. Pay close attention to signs, scalar multipliers, and properties of vector operations (e.g., non-commutativity of cross product). Show all your intermediate steps clearly.
5. Step 5: Verify and Interpret Your Result — Once you have an answer, check if it's reasonable within the context of the problem. Does the magnitude make sense? Is the direction correct (if it's a vector)? Ensure the final answer is in the format requested by the question (scalar, vector, etc.).

Worked Examples

• Example 1: Proving Coplanarity using Vector Sum Problem: Show that the vectors $\vec{a}$, $\vec{b}$, $\vec{c}$ are coplanar if $\vec{a} + \vec{b} + \vec{c} = \vec{0}$. Step 1: Understand the condition for coplanarity. Three vectors are coplanar if their scalar triple product is zero, i.e., $\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$. Step 2: Use the given condition. We are given $\vec{a} + \vec{b} + \vec{c} = \vec{0}$. This implies $\vec{a} = -(\vec{b} + \vec{c})$. Step 3: Substitute into the scalar triple product expression. Consider $\vec{a} \cdot (\vec{b} \times \vec{c})$. Substitute $\vec{a} = -(\vec{b} + \vec{c})$: $= -(\vec{b} + \vec{c}) \cdot (\vec{b} \times \vec{c})$ Step 4: Apply properties of dot and cross products. $= - [\vec{b} \cdot (\vec{b} \times \vec{c}) + \vec{c} \cdot (\vec{b} \times \vec{c})]$ (Distributive property of dot product) Recall that the scalar triple product is zero if any two vectors are identical. So, $\vec{b} \cdot (\vec{b} \times \vec{c}) = 0$ and $\vec{c} \cdot (\vec{b} \times \vec{c}) = 0$ (since $\vec{b} \times \vec{c}$ is perpendicular to both $\vec{b}$ and $\vec{c}$). Therefore, their dot product with $\vec{b}$ or $\vec{c}$ is zero. Step 5: Conclude. $= - [0 + 0] = 0$. Since $\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$, the vectors $\vec{a}$, $\vec{b}$, $\vec{c}$ are coplanar. Final Answer: The given condition $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ leads to a zero scalar triple product, thus proving that the vectors $\vec{a}$, $\vec{b}$, $\vec{c}$ are coplanar.
• Example 2: Area of a Parallelogram Problem: Find the area of the parallelogram whose adjacent sides are given by the vectors $\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$. Step 1: Recall the formula for the area of a parallelogram. The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by $|\vec{a} \times \vec{b}|$. Step 2: Calculate the cross product $\vec{a} \times \vec{b}$. $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{vmatrix}$ $= \hat{i}(1 \cdot 1 - 4 \cdot (-1)) - \hat{j}(3 \cdot 1 - 4 \cdot 1) + \hat{k}(3 \cdot (-1) - 1 \cdot 1)$ $= \hat{i}(1 + 4) - \hat{j}(3 - 4) + \hat{k}(-3 - 1)$ $= 5\hat{i} - (-1)\hat{j} + (-4)\hat{k}$ $= 5\hat{i} + \hat{j} - 4\hat{k}$ Step 3: Calculate the magnitude of the cross product. $|\vec{a} \times \vec{b}| = \sqrt{5^2 + 1^2 + (-4)^2}$ $= \sqrt{25 + 1 + 16}$ $= \sqrt{42}$ Step 4: State the final area. Final Answer: The area of the parallelogram is $\sqrt{42}$ square units.
• Example 3: Perpendicular Vectors Problem: If $\vec{a} = 2\hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{b} = -\hat{i} + 2\hat{j} + \hat{k}$, and $\vec{c} = 3\hat{i} + \hat{j}$, such that $\vec{a} + \lambda\vec{b}$ is perpendicular to $\vec{c}$, find the value of $\lambda$. Step 1: Understand the condition for perpendicularity. Two vectors are perpendicular if their dot product is zero. Given that $(\vec{a} + \lambda\vec{b})$ is perpendicular to $\vec{c}$, we have $(\vec{a} + \lambda\vec{b}) \cdot \vec{c} = 0$. Step 2: Calculate the vector $\vec{a} + \lambda\vec{b}$. $\vec{a} + \lambda\vec{b} = (2\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(-\hat{i} + 2\hat{j} + \hat{k})$ $= (2 - \lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (3 + \lambda)\hat{k}$ Step 3: Perform the dot product with $\vec{c}$. $(\vec{a} + \lambda\vec{b}) \cdot \vec{c} = [(2 - \lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (3 + \lambda)\hat{k}] \cdot (3\hat{i} + \hat{j} + 0\hat{k})$ $= (2 - \lambda)(3) + (2 + 2\lambda)(1) + (3 + \lambda)(0)$ $= 6 - 3\lambda + 2 + 2\lambda + 0$ $= 8 - \lambda$ Step 4: Set the dot product to zero and solve for $\lambda$. Since $(\vec{a} + \lambda\vec{b}) \cdot \vec{c} = 0$, $8 - \lambda = 0$ $\lambda = 8$ Final Answer: The value of $\lambda$ is 8.

Exam Tips for Vector Algebra Miscellaneous Exercise

To excel in the miscellaneous exercise and secure top marks in your board exams, keep these tips in mind:

• Conceptual Clarity is King: Don't just memorize formulas. Understand why and when to use them. For instance, know that a zero dot product implies perpendicularity, and a zero scalar triple product implies coplanarity. This fundamental understanding will help you choose the correct approach for diverse problems.
• Geometric Visualization: Whenever possible, try to visualize the vectors and their operations in 3D space. A quick sketch can often clarify the problem and guide you towards the correct solution. For example, understanding that $\vec{a} \times \vec{b}$ is a vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$ can simplify many proofs.
• Systematic Problem Solving: Break down complex problems into smaller, manageable steps. Don't rush into calculations. First, understand the question, then identify the required concepts, formulate a plan, execute, and finally verify your answer. This structured approach minimizes errors.
• Practice Variety: The miscellaneous exercise is designed to be comprehensive. Ensure you practice a wide range of problems covering dot products, cross products, scalar triple products, projections, and their applications to geometry (area, volume, angles, collinearity, coplanarity). Don't skip problems that seem unconventional.
• Careful with Calculations and Signs: Vector operations, especially cross products and scalar triple products, involve multiple terms and signs. One small error can lead to a completely wrong answer. Double-check your arithmetic and determinant expansions. Showing all intermediate steps helps in identifying and correcting mistakes.

Practice Questions with Solutions

• Q: Find the projection of the vector $\vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ on the vector $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$. A: Step 1: Recall the formula for projection. The projection of vector $\vec{a}$ on vector $\vec{b}$ is given by $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$. Step 2: Calculate the dot product $\vec{a} \cdot \vec{b}$. $\vec{a} \cdot \vec{b} = (2)(1) + (3)(2) + (2)(1) = 2 + 6 + 2 = 10$. Step 3: Calculate the magnitude of $\vec{b}$. $|\vec{b}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$. Step 4: Substitute values into the projection formula. Projection $= \frac{10}{\sqrt{6}}$. Final answer: The projection of $\vec{a}$ on $\vec{b}$ is $\frac{10}{\sqrt{6}}$ or $\frac{10\sqrt{6}}{6} = \frac{5\sqrt{6}}{3}$.
• Q: Find the area of the triangle having the points A(1, 1, 1), B(1, 2, 3), and C(2, 3, 1) as its vertices. A: Step 1: Form two vectors representing two sides of the triangle from a common vertex. Let's choose vertex A. The vectors are $\vec{AB}$ and $\vec{AC}$. $\vec{AB} = (1-1)\hat{i} + (2-1)\hat{j} + (3-1)\hat{k} = 0\hat{i} + 1\hat{j} + 2\hat{k}$. $\vec{AC} = (2-1)\hat{i} + (3-1)\hat{j} + (1-1)\hat{k} = 1\hat{i} + 2\hat{j} + 0\hat{k}$. Step 2: Calculate the cross product of these two vectors. $\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix}$ $= \hat{i}(1 \cdot 0 - 2 \cdot 2) - \hat{j}(0 \cdot 0 - 2 \cdot 1) + \hat{k}(0 \cdot 2 - 1 \cdot 1)$ $= \hat{i}(0 - 4) - \hat{j}(0 - 2) + \hat{k}(0 - 1)$ $= -4\hat{i} + 2\hat{j} - \hat{k}$. Step 3: Calculate the magnitude of the cross product. $|\vec{AB} \times \vec{AC}| = \sqrt{(-4)^2 + 2^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21}$. Step 4: Apply the formula for the area of a triangle. Area $= \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} \sqrt{21}$. Final answer: The area of the triangle is $\frac{\sqrt{21}}{2}$ square units.
• Q: If $\vec{a}$, $\vec{b}$, $\vec{c}$ are three vectors such that $|\vec{a}|=3$, $|\vec{b}|=4$, $|\vec{c}|=5$ and each one of them is perpendicular to the sum of the other two, find $|\vec{a}+\vec{b}+\vec{c}|$. A: Step 1: Interpret the perpendicularity condition. Given that each vector is perpendicular to the sum of the other two, we have: $\vec{a} \cdot (\vec{b} + \vec{c}) = 0 \Rightarrow \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 0$ (Eq 1) $\vec{b} \cdot (\vec{a} + \vec{c}) = 0 \Rightarrow \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{c} = 0$ (Eq 2) $\vec{c} \cdot (\vec{a} + \vec{b}) = 0 \Rightarrow \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} = 0$ (Eq 3) Step 2: Sum the three equations. $(\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}) + (\vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{c}) + (\vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b}) = 0$ Using commutativity of dot product ($\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$): $2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$ This implies $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = 0$ (Eq 4) Step 3: Use the formula for the magnitude of the sum of vectors. $|\vec{a} + \vec{b} + \vec{c}|^2 = (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c})$ $= |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$ Step 4: Substitute known values. We are given $|\vec{a}|=3$, $|\vec{b}|=4$, $|\vec{c}|=5$. And from Eq 4, $2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$. $|\vec{a} + \vec{b} + \vec{c}|^2 = 3^2 + 4^2 + 5^2 + 0$ $= 9 + 16 + 25 = 50$. Step 5: Find the magnitude. $|\vec{a} + \vec{b} + \vec{c}| = \sqrt{50} = 5\sqrt{2}$. Final answer: The magnitude $|\vec{a}+\vec{b}+\vec{c}|$ is $5\sqrt{2}$.
• Q: Show that $(\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) = 2(\vec{a} \times \vec{b})$. A: Step 1: Expand the left-hand side (LHS) using the distributive property of the cross product. LHS $= (\vec{a}-\vec{b}) \times (\vec{a}+\vec{b})$ $= \vec{a} \times (\vec{a}+\vec{b}) - \vec{b} \times (\vec{a}+\vec{b})$ $= (\vec{a} \times \vec{a}) + (\vec{a} \times \vec{b}) - (\vec{b} \times \vec{a}) - (\vec{b} \times \vec{b})$ Step 2: Apply properties of the cross product. We know that the cross product of a vector with itself is the zero vector: $\vec{a} \times \vec{a} = \vec{0}$ and $\vec{b} \times \vec{b} = \vec{0}$. Also, the cross product is anti-commutative: $\vec{b} \times \vec{a} = - (\vec{a} \times \vec{b})$. Step 3: Substitute these properties into the expanded expression. LHS $= \vec{0} + (\vec{a} \times \vec{b}) - (- (\vec{a} \times \vec{b})) - \vec{0}$ $= (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{b})$ $= 2(\vec{a} \times \vec{b})$. Step 4: Compare with the right-hand side (RHS). The LHS is equal to $2(\vec{a} \times \vec{b})$, which is the RHS. Final answer: Thus, it is shown that $(\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) = 2(\vec{a} \times \vec{b})$.

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