Problem 4 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k}\) and \(\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}\) be two vectors. If a vector perpendicular to both the vectors \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\) has the magnitude 12 then one such vector is : [April 12, 2019 (I)] (a) \(4(2 \hat{i}+2 \hat{j}+2 \hat{k})\) (b) \(4(2 \hat{i}-2 \hat{j}-\hat{k})\) (c) \(4(2 \hat{i}+2 \hat{j}-\hat{k})\) (d) \(4(-2 \hat{i}-2 \hat{j}+\hat{k})\)

Step-by-Step Solution

Answer

The vector is option (b): \(4(2 \hat{i} - 2 \hat{j} - \hat{k})\).

Step 1: Calculate Sum Vector

First, compute \( \vec{a} + \vec{b} \). Given \( \vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} \) and \( \vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k} \), calculate \( \vec{a} + \vec{b} = (3 + 1)\hat{i} + (2 + 2)\hat{j} + (2 - 2)\hat{k} = 4\hat{i} + 4\hat{j} + 0\hat{k} \).

Step 2: Calculate Difference Vector

Next, compute \( \vec{a} - \vec{b} \). Using \( \vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} - 2\hat{k} \), calculate \( \vec{a} - \vec{b} = (3 - 1)\hat{i} + (2 - 2)\hat{j} + (2 + 2)\hat{k} = 2\hat{i} + 0\hat{j} + 4\hat{k} \).

Step 3: Cross Product of Vector Sum and Difference

Find the cross product \( (\vec{a}+\vec{b}) \times (\vec{a}-\vec{b}) \). This is computed as the determinant:\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & 4 & 0 \ 2 & 0 & 4 \end{vmatrix} \]Calculate: \( \hat{i}(4 \times 4 - 0 \times 0) - \hat{j}(4 \times 4 - 0 \times 2) + \hat{k}(4 \times 0 - 4 \times 2) \), which simplifies to \( 16\hat{i} - 16\hat{j} - 8\hat{k} \).

Step 4: Normalize to Desired Magnitude

The vector \( 16\hat{i} - 16\hat{j} - 8\hat{k} \) needs to be scaled to have a magnitude of 12. First, find the original magnitude:\[ \sqrt{16^2 + (-16)^2 + (-8)^2} = \sqrt{512} \]Simplifying gives \( \sqrt{256 + 256 + 64} = \sqrt{576} = 24 \).Scale the vector by multiplying by \( \frac{12}{24} = \frac{1}{2} \), resulting in the vector \( 8\hat{i} - 8\hat{j} - 4\hat{k} \). Finally, multiply by the given constant 4: \( 4(2\hat{i} - 2\hat{j} - \hat{k}) \).

Step 5: Determine the Correct Option

Compare the resulting vector \( 4(2\hat{i} - 2\hat{j} - \hat{k}) \) with the given options. This matches option (b).

Key Concepts

Cross Product

The cross product is a mathematical operation used in vector calculus, particularly to find a vector that is perpendicular to two given vectors. It is calculated using the formula:\[\vec{c} = \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}\]Here, \(\vec{c}\) is the resulting vector, and \(a_1, a_2, a_3\) and \(b_1, b_2, b_3\) are the components of the vectors \(\vec{a}\) and \(\vec{b}\) respectively. The cross product is not only useful for determining perpendicularity but also plays a significant role in physics, especially in electromagnetism and moment calculations.

• It is important to note that the cross product is only defined in three-dimensional space.
• The magnitude of the cross product gives the area of the parallelogram formed by the two vectors.
• Unlike the dot product, the cross product is not commutative, meaning that \(\vec{a} \times \vec{b} eq \vec{b} \times \vec{a}\).

Understanding the cross product helps in finding a vector perpendicular to two original vectors, which can be crucial in many geometric and physical contexts.

Magnitude of Vector

The magnitude of a vector is a measure of its length and is calculated using the formula:\[|\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}\]where \(v_1, v_2, v_3\) are the components of the vector \(\vec{v}\). The magnitude represents how long the vector is in a multi-dimensional space and is an essential concept when comparing vectors or finding unit vectors.

• Magnitude provides a scalar measure that transforms a vector into a plain number, representing its length.
• Vectors can be normalized to turn them into unit vectors, which have a magnitude of 1, by dividing by their magnitude.
• Understanding magnitude helps in scaling vectors to desired lengths, as seen when we altered a vector’s magnitude to meet a specific criterion.

Calculating the magnitude is simple yet vital, as it often serves as an initial step in many vector-related computations, such as finding forces or velocities in physics.

Vector Addition and Subtraction

Vector addition and subtraction are fundamental operations in vector calculus and are often the first operations students encounter. These operations are performed component-wise:

• Addition: If \(\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}\) and \(\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}\), then \(\vec{a} + \vec{b} = (a_1 + b_1) \hat{i} + (a_2 + b_2) \hat{j} + (a_3 + b_3) \hat{k}\).
• Subtraction: Similarly, for subtraction, \(\vec{a} - \vec{b} = (a_1 - b_1) \hat{i} + (a_2 - b_2) \hat{j} + (a_3 - b_3) \hat{k}\).

These operations allow us to manipulate vectors to find resultant vectors, describe physical systems, or solve geometric problems. Through vector addition, we can combine forces or velocities to determine the total effect or direction of movement. Meanwhile, vector subtraction tells us the difference or direction between two points or states in space.
Understanding vector addition and subtraction is crucial for any field involving vector spaces, such as physics, engineering, and computer graphics.