Problem 63 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\vec{a}=3 \hat{i}+2 \hat{j}+x \hat{k}\) and \(\vec{b}=\hat{i}-\hat{j}+\hat{k}\), for some real \(x\). Then \(|\vec{a} \times \vec{b}|=\mathrm{r}\) is possible if : \(\quad\) [April08, 2019 (II)] (a) \(\sqrt{\frac{3}{2}}

Step-by-Step Solution

Answer

(a) \( \sqrt{\frac{3}{2}}

Step 1: Compute the Cross Product

To find the cross product \( \vec{a} \times \vec{b} \), use the determinant formula:\[\vec{a} \times \vec{b} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \3 & 2 & x \1 & -1 & 1\end{vmatrix}\]The result of this determinant is:\[\vec{a} \times \vec{b} = \hat{i}(2 \times 1 - (-1) \times x) - \hat{j}(3 \times 1 - x \times 1) + \hat{k}(3 \times (-1) - 2 \times 1)\]Simplify the expression:\[\vec{a} \times \vec{b} = (2 + x)\hat{i} - (3 - x)\hat{j} - 5\hat{k}\]

Step 2: Compute the Magnitude of the Cross Product

The magnitude of the cross product \(|\vec{a} \times \vec{b}|\) is given by:\[|\vec{a} \times \vec{b}| = \sqrt{(2 + x)^2 + (3 - x)^2 + (-5)^2}\]This simplifies to:\[|\vec{a} \times \vec{b}| = \sqrt{(2 + x)^2 + (3 - x)^2 + 25}\]Compute the expressions under the square root:\[|\vec{a} \times \vec{b}| = \sqrt{4 + 4x + x^2 + 9 - 6x + x^2 + 25}\]Combine like terms:\[|\vec{a} \times \vec{b}| = \sqrt{2x^2 - 2x + 38}\]

Step 3: Analyze the Expression for Possible Values

Consider the range given in the options for \( r \) and use the expression to find which range is possible.The expression for \( |\vec{a} \times \vec{b}| \) is \( \sqrt{2x^2 - 2x + 38} \). This number needs to be analyzed across the options:1. For small possible values near 0: Check if \( r \leq \sqrt{\frac{3}{2}} \).2. For larger values: Compute when \( r \geq 5\sqrt{\frac{3}{2}} \).3. Middle ranges.Plugging in extreme values like \( x = 0 \) and \( x = 5 \), calculate the lower bound \( |\vec{a} \times \vec{b}| \) and check which interval is feasible.

Key Concepts

Magnitude of Cross Product

The magnitude of the cross product of two vectors, \(\vec{a}\) and \(\vec{b}\), provides a scalar value that showcases the area of the parallelogram that the vectors span in three-dimensional space. To find this magnitude, we first need to compute the cross product, \(\vec{a} \times \vec{b}\), and then calculate its magnitude. The mathematical formula to find the magnitude of the cross product is:

• \(|\vec{a} \times \vec{b}| = \sqrt{c_1^2 + c_2^2 + c_3^2}\)

Here, \(c_1\), \(c_2\), and \(c_3\) are the components of \(\vec{a} \times \vec{b}\). For the vectors given in the exercise, \(\vec{a} = 3 \hat{i} + 2 \hat{j} + x \hat{k}\) and \(\vec{b} = \hat{i} - \hat{j} + \hat{k}\), the cross product results in \((2 + x)\hat{i} - (3 - x)\hat{j} - 5\hat{k}\). Thus, the magnitude is \(|\vec{a} \times \vec{b}| = \sqrt{(2+x)^2 + (3-x)^2 + 25}\). This formula accounts for the perpendicularity between the vectors, as the cross product measures the extent to which vector \(\vec{b}\) is perpendicular to vector \(\vec{a}\).

Determinant and Cross Product

The use of determinants is a crucial step in calculating the cross product of two vectors. The determinant of a 3x3 matrix can help visualize and solve the cross product operation. With the vector components of \(\vec{a}\) and \(\vec{b}\), the cross product can be expressed as:

• \(\vec{a} \times \vec{b} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \ 3 & 2 & x \ 1 & -1 & 1\end{vmatrix}\)

Calculating this determinant involves expanding along the top row, \(\hat{i}, \hat{j}, \hat{k}\), and then simplifying. Each entry involves finding a minor and calculating the respective cofactor. Thus, \(\vec{a} \times \vec{b} = \hat{i}(2 \times 1 - (-1) \times x) - \hat{j}(3 \times 1 - x \times 1) + \hat{k}(3 \times (-1) - 2 \times 1)\), leading to the components \((2+x), (3-x), -5\). This method not only simplifies finding cross products but enables us to accurately compute the orientation of these vectors in space.

Vector Algebra in 3D

Vector algebra in three dimensions is a powerful mathematical tool that helps describe spatial relationships and physical quantities such as force, velocity, and more. Vectors in 3D are expressed with three components along the \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) unit vectors, corresponding to the x, y, and z axes respectively. When dealing with vector operations like cross products, determining the resulting vector involves using these components to find their algebraic relationships. The cross product itself is a key operation that results in a new vector perpendicular to the plane defined by \(\vec{a}\) and \(\vec{b}\). This concept reveals how magnitude and direction are intertwined, allowing vectors to represent not just static quantities but also changes and interactions.Understanding the algebraic rules and geometric insights of vectors in 3D allows us to solve complex physics problems and analyze geometric systems. This is particularly useful for determining forces in engineering and rotational dynamics. By using cross products, we gain deeper insight into the spatial dynamics and can calculate quantities like torque and angular momentum effectively.

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Problem 62

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Problem 64

Other exercises in this chapter

Problem 61

Let \(\alpha=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}\) and \(\vec{\beta}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}\). If \(\vec{\beta}=\vec{\beta}_{1}-

View solution

Problem 62

The magnitude of the projection of the vector \(2 \hat{i}+3 \hat{j}+\hat{k}\) on the vector perpendicular to the plane containing the vectors \(\hat{i}+\hat{j}+

View solution

Problem 64

Let \(\vec{a}, \vec{b}\) and \(\vec{c}\) be three unit vectors, out of which vectors \(\vec{b}\) and \(\vec{c}\) are non-parallel. If \(\alpha\) and \(\beta\) a

View solution

Problem 65

Let \(\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}, \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\) and \(\ove

View solution

Key Concepts

Magnitude of Cross Product

The magnitude of the cross product of two vectors, \(\vec{a}\) and \(\vec{b}\), provides a scalar value that showcases the area of the parallelogram that the vectors span in three-dimensional space. To find this magnitude, we first need to compute the cross product, \(\vec{a} \times \vec{b}\), and then calculate its magnitude. The mathematical formula to find the magnitude of the cross product is:

• \(|\vec{a} \times \vec{b}| = \sqrt{c_1^2 + c_2^2 + c_3^2}\)

Here, \(c_1\), \(c_2\), and \(c_3\) are the components of \(\vec{a} \times \vec{b}\). For the vectors given in the exercise, \(\vec{a} = 3 \hat{i} + 2 \hat{j} + x \hat{k}\) and \(\vec{b} = \hat{i} - \hat{j} + \hat{k}\), the cross product results in \((2 + x)\hat{i} - (3 - x)\hat{j} - 5\hat{k}\). Thus, the magnitude is \(|\vec{a} \times \vec{b}| = \sqrt{(2+x)^2 + (3-x)^2 + 25}\). This formula accounts for the perpendicularity between the vectors, as the cross product measures the extent to which vector \(\vec{b}\) is perpendicular to vector \(\vec{a}\).

Determinant and Cross Product

The use of determinants is a crucial step in calculating the cross product of two vectors. The determinant of a 3x3 matrix can help visualize and solve the cross product operation. With the vector components of \(\vec{a}\) and \(\vec{b}\), the cross product can be expressed as:

• \(\vec{a} \times \vec{b} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \ 3 & 2 & x \ 1 & -1 & 1\end{vmatrix}\)

Calculating this determinant involves expanding along the top row, \(\hat{i}, \hat{j}, \hat{k}\), and then simplifying. Each entry involves finding a minor and calculating the respective cofactor. Thus, \(\vec{a} \times \vec{b} = \hat{i}(2 \times 1 - (-1) \times x) - \hat{j}(3 \times 1 - x \times 1) + \hat{k}(3 \times (-1) - 2 \times 1)\), leading to the components \((2+x), (3-x), -5\). This method not only simplifies finding cross products but enables us to accurately compute the orientation of these vectors in space.

Vector Algebra in 3D

Vector algebra in three dimensions is a powerful mathematical tool that helps describe spatial relationships and physical quantities such as force, velocity, and more. Vectors in 3D are expressed with three components along the \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) unit vectors, corresponding to the x, y, and z axes respectively. When dealing with vector operations like cross products, determining the resulting vector involves using these components to find their algebraic relationships. The cross product itself is a key operation that results in a new vector perpendicular to the plane defined by \(\vec{a}\) and \(\vec{b}\). This concept reveals how magnitude and direction are intertwined, allowing vectors to represent not just static quantities but also changes and interactions.Understanding the algebraic rules and geometric insights of vectors in 3D allows us to solve complex physics problems and analyze geometric systems. This is particularly useful for determining forces in engineering and rotational dynamics. By using cross products, we gain deeper insight into the spatial dynamics and can calculate quantities like torque and angular momentum effectively.