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Chapter 3: Problem 3

A vector perpendicular to \(\hat{i}+\hat{j}+\hat{k}\) is (1) \(\hat{i}-\hat{j}+\hat{k}\) (2) \(\hat{i}-\hat{j}-\hat{k}\) (3) \(-\hat{i}-\hat{j}-\hat{k}\) (4) \(3 \hat{i}+2 \hat{j}-5 \hat{k}\)

Short Answer

Expert verified
The vector perpendicular to \( \hat{i} + \hat{j} + \hat{k} \) is \( 3 \hat{i} + 2 \hat{j} - 5 \hat{k} \) (option 4).

Step by step solution

01

Understand the Condition for Perpendicularity

Two vectors are perpendicular if their dot product is zero. If we have vectors \( \mathbf{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \) and \( \mathbf{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k} \), then they are perpendicular if \( a_1b_1 + a_2b_2 + a_3b_3 = 0 \). Here, our first vector is \( \hat{i} + \hat{j} + \hat{k} \).
02

Calculate the Dot Product with Each Option

Let's denote \( \mathbf{v} = \hat{i} + \hat{j} + \hat{k} \). We calculate the dot product of \( \mathbf{v} \) with each given option: 1. Dot product with \( \hat{i} - \hat{j} + \hat{k} \): \[ (1 \cdot 1) + (1 \cdot -1) + (1 \cdot 1) = 1 - 1 + 1 = 1 \]2. Dot product with \( \hat{i} - \hat{j} - \hat{k} \): \[ (1 \cdot 1) + (1 \cdot -1) + (1 \cdot -1) = 1 - 1 - 1 = -1 \]3. Dot product with \( -\hat{i} - \hat{j} - \hat{k} \): \[ (1 \cdot -1) + (1 \cdot -1) + (1 \cdot -1) = -1 - 1 - 1 = -3 \]4. Dot product with \( 3\hat{i} + 2\hat{j} - 5\hat{k} \): \[ (1 \cdot 3) + (1 \cdot 2) + (1 \cdot -5) = 3 + 2 - 5 = 0 \]
03

Identify the Perpendicular Vector

The condition for two vectors to be perpendicular is that their dot product equals zero. Comparing the results from Step 2, we see that option 4, \( 3\hat{i} + 2\hat{j} - 5\hat{k} \) yields a dot product of zero when dotted with \( \hat{i} + \hat{j} + \hat{k} \).

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Key Concepts

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

Perpendicular Vectors
In vector mathematics, understanding perpendicular vectors is crucial for various applications including physics and engineering. Two vectors are said to be perpendicular to each other if their dot product equals zero.
This characteristic makes it important in determining directions and solving geometric problems where orthogonal relationships are involved. For any vectors \( \mathbf{a} \) and \( \mathbf{b} \), expressed as \( a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \) and \( b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k} \), they are perpendicular if:
  • \( a_1b_1 + a_2b_2 + a_3b_3 = 0 \)
Finding perpendicular vectors is a fundamental task in vector spaces and helps define angles between lines, moving past the usual perception of right angles.
Dot Product
The dot product, also known as the scalar product, is a vital operation that relates two vectors to a scalar. For vectors \( \mathbf{a} \) and \( \mathbf{b} \), the dot product \( \mathbf{a} \cdot \mathbf{b} \) calculates as:
  • \( a_1b_1 + a_2b_2 + a_3b_3 \)
The result is a scalar value that encapsulates the product of the magnitudes of the two vectors and the cosine of the angle between them.
This makes the dot product extremely useful not only for checking perpendicularity, but also for finding projections, understanding vector components and analyzing performance in physical scenarios such as work done by forces. Here, by ensuring the dot product is zero, you determine orthogonality between vectors.
JEE Advanced Mathematics
The JEE Advanced exam is a challenging engineering entrance test in India that covers complex topics including vector mathematics. Questions related to vectors are commonplace in this exam and often test five key aspects:
  • Computations of dot and cross products
  • Identifying vector magnitudes
  • Understanding directional relationships
  • Solving for perpendicular and parallel vectors
  • Decomposing vectors into components
For students tackling JEE Advanced mathematics, mastering concepts like vector perpendicularity and dot product calculations is a must. These topics not only form a foundation for physics and mathematics problems but also help in developing critical thinking skills. For example, identifying perpendicular vectors feeds into larger problems involving geometrical arrangement and physics-based vector fields.

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Most popular questions from this chapter

Find the component of a \(50 \mathrm{~N}\) force which makes an angle of \(45^{\circ}\) with \(Z\)-axis and whose projection in \(X Y\)-plane makes \(45^{\circ}\) with the \(X\)-axis (1) \(F_{x}=25 \mathrm{~N}\) (2) \(F_{y}=25 \sqrt{2} \mathrm{~N}\) (3) \(F_{y}=25 \mathrm{~N}\) (4) \(F_{z}=25 \sqrt{2} \mathrm{~N}\)

Consider east as positive \(x\)-axis, north as positive \(y\)-axis. A girl walks \(10 \mathrm{~m}\) east first time then \(10 \mathrm{~m}\) in a direction \(30^{\circ}\) west of north for the second time and then third time in unknown direction and magnitude so as to return to her initial position. What is her third displacement in unit vector notation? (1) \(-5 \hat{i}-5 \sqrt{3} \hat{j}\) (2) \(5 \hat{i}-5 \sqrt{3} \hat{j}\) (3) \(-5 \hat{i}+5 \sqrt{3} \hat{j}\) (4) She cannot return

If \(a=3 \hat{i}+4 \hat{j}\) and \(b=4 \hat{i}-3 \hat{j}\), then (1) the magnitude of resultant of \(a\) and \(b\) is \(5 \sqrt{2}\) units (2) the angle between \(a\) and \(b\) is \(90^{\circ}\) (3) the magnitude of resultant of \(a\) and \(b\) is 5 units (4) the angle between \(a\) and \(b\) is zero

Two vectors \(\vec{a}\) and \(\vec{b}\) are at an angle of \(60^{\circ}\) with each other. Their resultant makes an angle of \(45^{\circ}\) with \(\vec{a} .\) If \(|\vec{b}|=2\) units, then \(|\vec{a}|\) is (1) \(\sqrt{3}\) (2) \(\sqrt{3}-1\) (3) \(\sqrt{3}+1\) (4) \(\sqrt{3} / 2\)

The magnetic force on a moving charge is \(\vec{F}=q \vec{v} \times \vec{B}\). Here, \(q=\) electric charge on particle, \(v=\) velocity of the particle, \(B=\) magnetic field. A charged particle of charge \(1 \mathrm{C}\) moving with a velocity \((\hat{i}+\hat{j}-\hat{k}) \mathrm{ms}^{-1}\) in a magnetic field \(B=(2 \hat{i}-\hat{j}+4 \hat{k}) \mathrm{T}\) Choose the correct option(s). (1) The component of magnetic force in the direction of velocity is zero (2) The component of magnetic force in the direction of magnetic field is zero (3) The magnitude of magnetic force is \(\sqrt{54} \mathrm{~N}\) (4) None of the above
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