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Chapter 25: Problem 16

If \(\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}, \vec{b}=2 \hat{i}+3 \hat{j}-\hat{k}\) and \(\vec{c}=r \hat{i}+\hat{j}+(2 r-1) \hat{k}\) are three vectors such that \(\vec{c}\) is parallel to the plane of \(\vec{a}\) and \(\vec{b}\), then \(r\) is equal to [Online May 19, 2012] (a) 1 (b) \(-1\) (c) 0 (d) 2

Short Answer

Expert verified
The value of \(r\) is 0.

Step by step solution

01

Understand Parallelism in Planes

To determine if vector \(\vec{c}\) is parallel to the plane formed by vectors \(\vec{a}\) and \(\vec{b}\), recognize that any vector parallel to the plane will not have any component along the normal to the plane. The normal vector to the plane can be found by taking the cross-product of \(\vec{a}\) and \(\vec{b}\).
02

Calculate the Cross Product \(\vec{a} \times \vec{b}\)

The cross product \(\vec{a} \times \vec{b}\) is obtained as follows:\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 3 \ 2 & 3 & -1 \end{vmatrix} = \hat{i}((-2)(-1) - (3)(3)) - \hat{j}((1)(-1) - (3)(2)) + \hat{k}((1)(3) - (-2)(2)) \]\[ = \hat{i}(2 - 9) - \hat{j}(-1 - 6) + \hat{k}(3 + 4) \]\[ = -7\hat{i} + 7\hat{j} + 7\hat{k} \].
03

Ensure Vector \(\vec{c}\) is Orthogonal to Normal

\(\vec{c}\) is parallel to the plane if it is orthogonal to the normal vector \(\vec{a} \times \vec{b}\). Therefore, the dot product \(\vec{c} \cdot (\vec{a} \times \vec{b}) = 0\). Calculate this dot product: \[ (r\hat{i} + \hat{j} + (2r-1)\hat{k}) \cdot (-7\hat{i} + 7\hat{j} + 7\hat{k}) = 0 \].
04

Solve for \(r\)

Calculating the dot product and setting it to zero gives: \[ -7r + 7 + 7(2r-1) = 0 \]. Simplify this equation: \[ -7r + 7 + 14r - 7 = 0 \]. Combine terms: \[ 7r = 0 \]. Solve for \(r\): \[ r = 0 \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cross Product of Vectors
The cross product is a fundamental operation in vector calculus, important for various applications such as determining a vector perpendicular to two given vectors. For vectors \(\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}\) and \(\vec{b} = 2\hat{i} + 3\hat{j} - \hat{k}\), the cross product \(\vec{a} \times \vec{b}\) is computed using the determinant method.
  • First, arrange the unit vectors and the components of the two vectors in a 3x3 matrix.
  • The cross product involves expanding this determinant, which yields a new vector perpendicular to both \(\vec{a}\) and \(\vec{b}\).
  • This vector is often considered the normal to the plane defined by \(\vec{a}\) and \(\vec{b}\).
Calculating the cross product results in the vector: \(-7\hat{i} + 7\hat{j} + 7\hat{k}\). This normal vector will be essential in understanding vector parallelism, especially concerning a new vector \(\vec{c}\).
Plane Geometry
Understanding plane geometry is crucial when dealing with vectors in mathematics, particularly in three dimensions. A plane can be defined using two non-parallel vectors, with the plane being all the linear combinations of these vectors.
  • The normal vector, obtained from the cross product, is perpendicular to every vector in the plane.
  • In our context, \(\vec{a} \times \vec{b}\) gives us the normal vector to the plane containing \(\vec{a}\) and \(\vec{b}\).
  • A third vector, like \(\vec{c}\), is said to be parallel to this plane if its direction is perpendicular to this normal vector.
This geometric relationship is used to explore if a vector lies in, or is parallel to, the plane made by other vectors.
Dot Product
The dot product, or scalar product, of two vectors \(\vec{u}\) and \(\vec{v}\) is a measure of their directional alignment. It provides a scalar value obtained via multiplying corresponding components of the vectors and summing these products.
  • Mathematically, this is represented as \(\vec{u} \cdot \vec{v} = u_i v_i + u_j v_j + u_k v_k\).
  • When the dot product is zero, this indicates the vectors are orthogonal or perpendicular.
  • For vector \(\vec{c} = r\hat{i} + \hat{j} + (2r - 1)\hat{k}\), determining if it is orthogonal to \(\vec{a} \times \vec{b}\) requires that its dot product with \(-7\hat{i} + 7\hat{j} + 7\hat{k}\) equals zero.
Applying the dot product here verifies the condition of parallelism for vector \(\vec{c}\). This calculation eventually yields the solution \(r = 0\).
JEE Main Mathematics
JEE Main is a highly competitive engineering entrance examination in India, requiring a deep understanding of mathematical concepts. The use of vectors and operations like the cross and dot product are intrinsic to the type of problems encountered.
  • Such exercises develop abstract thinking and spatial visualization skills.
  • JEE Main questions often require applying multiple mathematical concepts to solve a single problem, such as using both the cross and dot products here.
  • Success in such problems relies on a strong grasp of vector operations, equations of planes, and properties of determinants.
Preparing for JEE Main Mathematics involves practicing problem-solving strategies and understanding the underlying principles, as seen in this vector and plane geometry exercise.

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Most popular questions from this chapter

A vector \(\vec{a}=\alpha \hat{i}+2 \hat{j}+\beta \hat{k}(\alpha, \beta \in \boldsymbol{R})\) lies in the plane of the vectors, \(\vec{b}=\hat{i}+\hat{j}\) and \(\vec{c}=\hat{i}-\hat{j}+4 \hat{k}\). If \(\vec{a}\) bisects the angle between \(\vec{b}\) and \(\vec{c}\), then: [Jan. 7, 2020 (I)] (a) \(\vec{a} \cdot \hat{i}+3=0\) (b) \(\vec{a} \cdot \hat{i}+1=0\) (c) \(\vec{a} \cdot \hat{k}+2=0\) (d) \(\vec{a} \cdot \hat{k}+4=0\)

Let the vectors \(\vec{a}, \vec{b}, \vec{c}\) be such that \(|\vec{a}|=2,|\vec{b}|=4\) and \(|\vec{c}|=4\). If the projection of \(\vec{b}\) on \(\vec{a}\) is equal to the projection of \(\vec{c}\) on \(\vec{a}\) and \(\vec{b}\) is perpendicular to \(\vec{c}\), then the value of \(|\vec{a}+\vec{b}-\vec{c}|\) is

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 \(\hat{i}+\hat{j}+\hat{k}\) and \(\hat{i}+2 \hat{j}+3 \hat{k}\), is : \([\) April 08,2019 (I)] (a) \(\frac{\sqrt{3}}{2}\) (b) \(\sqrt{6}\) (c) \(3 \sqrt{6}\) (d) \(\sqrt{\frac{3}{2}}\)

If \(\vec{a}, \vec{b}, \vec{c}\) are vectors such that \([\vec{a} \vec{b} \vec{c}]=4\) then \([\vec{a} \times \vec{b} \vec{b} \times \vec{c} \vec{c} \times \vec{a}]=\) (a) 16 (b) 64 (c) 4 (d) 8

Let \(\mathrm{ABC}\) be a triangle whose circumcentre is at \(\mathrm{P}\). If the position vectors \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) and \(\mathrm{P}\) are \(\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}\) and \(\frac{\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}}{4}\) respectively, then the position vector of the orthocentre of this triangle, is: \(\quad\) Online April 10, 2016] (a) \(-\left(\frac{\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}}{2}\right)\) (b) \(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}\) (c) \(\left(\frac{\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}}{2}\right)\) (d) \(\overrightarrow{0}\)
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