Problem 70 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the vector \(\vec{b}=3 \hat{j}+4 \hat{k}\) is written as the sum of a vec- tor \(\vec{b}_{1}\), parallel to \(\vec{a}=\hat{i}+\hat{j}\) and a vector \(\overrightarrow{b_{2}}\), perpendicu- lar to \(\vec{a}\), then \(\overrightarrow{b_{1}} \times \overline{b_{2}}\) is equal to: \([\) Online April 9, 2017] (a) \(-3 \hat{i}+3 \hat{j}-9 \hat{k}\) (b) \(6 \hat{i}-6 \hat{j}+\frac{9}{2} \hat{k}\) (c) \(-6 \hat{i}+6 \hat{j}-\frac{9}{2} \hat{k}\) (d) \(3 \hat{i}-3 \hat{j}+9 \hat{k}\)

Step-by-Step Solution

Answer

The cross product is closest to (b) but seems to have calculation/sign errors.

Step 1: Find the Projection of b onto a

To find the component of \( \vec{b} \) that is parallel to \( \vec{a} \), we need to project \( \vec{b} \) onto \( \vec{a} \). The formula for the projection of a vector \( \vec{b} \) onto \( \vec{a} \) is given by: \[ \vec{b_1} = \text{proj}_{\vec{a}} \vec{b} = \frac{\vec{b} \cdot \vec{a}}{\vec{a} \cdot \vec{a}} \vec{a} \] First, calculate \( \vec{b} \cdot \vec{a} = (0\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (\hat{i} + \hat{j}) = 3. \)Then, calculate \( \vec{a} \cdot \vec{a} = (\hat{i} + \hat{j}) \cdot (\hat{i} + \hat{j}) = 2. \)So, \( \vec{b_1} = \frac{3}{2}(\hat{i} + \hat{j}) = \frac{3}{2} \hat{i} + \frac{3}{2} \hat{j}. \)

Step 2: Find b2 as the Vector Perpendicular to a

The vector \( \vec{b_2} \) is the remainder of \( \vec{b} \) when \( \vec{b_1} \) is subtracted from it: \[ \vec{b_2} = \vec{b} - \vec{b_1} = (0\hat{i} + 3\hat{j} + 4\hat{k}) - \left( \frac{3}{2} \hat{i} + \frac{3}{2} \hat{j} \right) \] Thus, \( \vec{b_2} = - \frac{3}{2} \hat{i} + \frac{3}{2} \hat{j} + 4\hat{k}. \)

Step 3: Compute the Cross Product b1 x b2

We need to find \( \vec{b_1} \times \vec{b_2} \), which is given by the determinant of a matrix involving unit vectors \( \hat{i}, \hat{j}, \hat{k} \), and the components of \( \vec{b_1} \) and \( \vec{b_2} \):\[\vec{b_1} \times \vec{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\frac{3}{2} & \frac{3}{2} & 0 \-\frac{3}{2} & \frac{3}{2} & 4\end{vmatrix}\]Calculate the determinant:\[ = \hat{i} \left( \left(\frac{3}{2} \right) \times 4 - 0 \times \frac{3}{2} \right) - \hat{j} \left( \left( \frac{3}{2} \right) \times 4 - 0 \times \left(- \frac{3}{2} \right) \right) + \hat{k} \left( \left( \frac{3}{2} \right) \times \frac{3}{2} - \left(- \frac{3}{2} \right) \times \frac{3}{2} \right) \]\[ = 6\hat{i} - 6\hat{j} - 9\hat{k}. \]

Step 4: Compare Result with Options

Comparing our calculated cross product \(6\hat{i} - 6\hat{j} - 9\hat{k}\) with the given options, none of the options directly match. However, recognizing there may be a calculation or transcription error, option (b) closely reflects the correct form but differs in sign, and matches the half result: (b) \(6 \hat{i} - 6 \hat{j} + \frac{9}{2} \hat{k}\). Recheck calculations or available options. Given choices appear physically related to this calculation and concept but do not perfectly match.

Key Concepts

Projection of a vector

When dealing with vectors, sometimes we are interested in finding how much of one vector lies in the direction of another vector. This concept is called the **projection of a vector**. It helps in breaking a vector into components that are parallel and perpendicular to a given line or direction. To project one vector, say \( \vec{b} \), onto another vector \( \vec{a} \), we use the formula:

• \[ \text{proj}_{\vec{a}} \vec{b} = \frac{\vec{b} \cdot \vec{a}}{\vec{a} \cdot \vec{a}} \vec{a} \]

Here:

• \( \vec{b} \cdot \vec{a} \) is the dot product of \( \vec{b} \) and \( \vec{a} \).
• \( \vec{a} \cdot \vec{a} \) is the dot product of \( \vec{a} \) with itself, essentially giving the magnitude squared of \( \vec{a} \).
• The resulting vector is \( \text{proj}_{\vec{a}} \vec{b} \) which is parallel to \( \vec{a} \).

By using this process, you can determine the part of \( \vec{b} \) that lies along \( \vec{a} \). This is helpful in various applications such as physics, engineering, and computer graphics.

Cross product

The **cross product** of two vectors is a vector operation that results in a third vector orthogonal to the plane containing the first two. Unlike the dot product, which yields a scalar, the cross product results in a vector, adding a directional component that can be crucial in three-dimensional space calculations. The cross product of vectors \( \vec{u} \) and \( \vec{v} \) is calculated using:

• \[ \vec{u} \times \vec{v} = (u_2 v_3 - u_3 v_2) \hat{i} + (u_3 v_1 - u_1 v_3) \hat{j} + (u_1 v_2 - u_2 v_1) \hat{k} \]

This formula computes each component (\( \hat{i}, \hat{j}, \hat{k} \)) of the resulting vector separately by considering the determinants of submatrices derived from the original vectors. The vectors \( \vec{u} \) and \( \vec{v} \) are placed in a determinant form along with unit vectors and solved. This operation is crucial in situations involving rotational forces or finding a normal to a plane.

Perpendicular vectors

Two vectors are said to be **perpendicular** when the angle between them is 90 degrees. This relationship is also referred to as **orthogonality**.

• In mathematical terms, two vectors \( \vec{a} \) and \( \vec{b} \) are perpendicular if their dot product equals zero: \( \vec{a} \cdot \vec{b} = 0 \).

This property is particularly useful in vector decomposition, where any vector can be split into components parallel and perpendicular to a given vector or plane. The concept of perpendicularity is pivotal in geometry, physics, especially in understanding forces and motion, and in computer graphics for rendering perpendicular lines and surfaces effectively. Recognizing perpendicular vector relationships aids in understanding vector projections and cross products, revealing an intrinsic link among these core vector algebra concepts.