/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 If vector \(\mathbf{a}=2 \hat{\m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If vector \(\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}\) and vector \(\mathbf{b}=-2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}\) then \(\frac{\text { Projection of vector a on vector } b}{\text { Projection of vector } \mathbf{b} \text { on vector a }}\) is equal to (a) \(\frac{3}{7}\) (b) \(\frac{7}{3}\) (c) 3 (d) 7

Short Answer

Expert verified
The ratio is \(\frac{7}{3}\), so the answer is (b).

Step by step solution

01

Find the Dot Product of Vectors a and b

First, we must find the dot product of vectors \(\mathbf{a}\) and \(\mathbf{b}\). The dot product is given by \[ \mathbf{a} \cdot \mathbf{b} = (2)(-2) + (-3)(2) + (6)(-1).\]Calculating this, we get \[ \mathbf{a} \cdot \mathbf{b} = -4 - 6 - 6 = -16. \]
02

Calculate the Magnitude of Vector b

The magnitude of vector \(\mathbf{b}\) can be found using the formula:\[ |\mathbf{b}| = \sqrt{(-2)^2 + 2^2 + (-1)^2}. \]This simplifies to \[ |\mathbf{b}| = \sqrt{4 + 4 + 1} = \sqrt{9} = 3. \]
03

Projection of a on b

The projection of \(\mathbf{a}\) on \(\mathbf{b}\) is given by:\[ \text{Proj}_{\mathbf{b}}\mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2} \cdot \mathbf{b}.\]Using the result from Step 1 and Step 2, this becomes:\[ \text{Proj}_{\mathbf{b}}\mathbf{a} = \frac{-16}{3^2} \cdot |\mathbf{b}| = \frac{-16}{9}. \]
04

Calculate the Magnitude of Vector a

The magnitude of vector \(\mathbf{a}\) is given by:\[ |\mathbf{a}| = \sqrt{(2)^2 + (-3)^2 + (6)^2}. \]Simplifying this, we find\[ |\mathbf{a}| = \sqrt{4 + 9 + 36} = \sqrt{49} = 7. \]
05

Projection of b on a

The projection of \(\mathbf{b}\) on \(\mathbf{a}\) is given by:\[ \text{Proj}_{\mathbf{a}}\mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|^2} \cdot \mathbf{a}.\]Using the previously calculated dot product and magnitude of \(\mathbf{a}\), this is:\[ \text{Proj}_{\mathbf{a}}\mathbf{b} = \frac{-16}{7^2} \cdot |\mathbf{a}| = \frac{-16}{49}. \]
06

Find the Ratio of Projections

Finally, to find the desired ratio, we divide the projection of \(\mathbf{a}\) on \(\mathbf{b}\) by the projection of \(\mathbf{b}\) on \(\mathbf{a}\):\[ \frac{\text{Projection of } \mathbf{a} \text{ on } \mathbf{b}}{\text{Projection of } \mathbf{b} \text{ on } \mathbf{a}} = \frac{\frac{-16}{9}}{\frac{-16}{49}}= \frac{49}{9}.\]This simplifies to:\[ \frac{49}{9} = \frac{7}{3}. \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Dot Product
The dot product is an essential operation in vector mathematics, especially when dealing with vector projection problems. To calculate the dot product for two vectors, simply multiply corresponding components of the vectors and sum those products.
For two vectors \( \mathbf{a} = a_1 \hat{\mathbf{i}} + a_2 \hat{\mathbf{j}} + a_3 \hat{\mathbf{k}} \) and \( \mathbf{b} = b_1 \hat{\mathbf{i}} + b_2 \hat{\mathbf{j}} + b_3 \hat{\mathbf{k}} \), the dot product is:
\[ \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3. \]
This operation results in a scalar quantity, which is a single number. It's worth noting that the dot product measures the degree to which two vectors are aligned. If the dot product is zero, vectors are perpendicular to each other.
Magnitude of a Vector
The magnitude of a vector is like finding the "length" of that vector and is crucial for performing operations such as normalization and projection. For a vector \( \mathbf{v} = v_1 \hat{\mathbf{i}} + v_2 \hat{\mathbf{j}} + v_3 \hat{\mathbf{k}} \), its magnitude \|\mathbf{v}\| is calculated using:
\[ |\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}. \]
Magnitudes are always non-negative and give you an idea of how "long" or "strong" a vector is. In vector projection problems, knowing the magnitude of both vectors involved is critical in dividing or scaling to get projection results.
  • Magnitude helps in normalizing vectors, converting them into unit vectors.
  • It's foundational in calculating projections and angles between vectors.
Ratio of Projections
The concept of ratio of projections is used in vector calculus to compare how one vector projects onto another, relative to the opposite scenario. To find this, we calculate each projection separately:
- Projection of vector \( \mathbf{a} \) on \( \mathbf{b} \) is:
\[ \text{Proj}_{\mathbf{b}}\mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2} \cdot \mathbf{b}. \]
- Projection of vector \( \mathbf{b} \) on \( \mathbf{a} \) is:
\[ \text{Proj}_{\mathbf{a}}\mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|^2} \cdot \mathbf{a}. \]
The ratio of these projections thus becomes:
\[ \frac{\text{Proj}_{\mathbf{b}}\mathbf{a}}{\text{Proj}_{\mathbf{a}}\mathbf{b}}. \]
This operation provides insight into relative orientations and strengths of the vectors' projections.
Mathematics for IIT JEE
Preparing for IIT JEE involves understanding sophisticated mathematical concepts, among which vector algebra is significant. Mastery of vector operations like dot products, magnitudes, and projections is crucial, as they frequently appear in physics and mathematics problems in the exam.
  • Practicing problems on vector projections is essential for strong conceptual clarity.
  • Vectors are utilized across subjects, making them a versatile topic to focus on.
Proactive practice with real IIT JEE problems helps bridge the gap between theoretical understanding and practical application. Here, deeper insight into each element of vector operations and consistent practice can distinguish an average problem-solving skill from an exceptional one.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(\mathbf{a}, \mathbf{b}\) and \(\mathbf{c}\) be three mutually perpendicular vectors of equal magnitude. If the vector \(x\) satisfies the equation. \(a \times\\{(x-b) \times a\\}+b \times\\{(x-c) \times b\\}+c \times\\{(x-a) \times c\\}=0\) then find \(\mathbf{x}\)

If \(\mathbf{a}\) and \(\mathbf{b}\) are two unit vectors perpendicular to each other and \(\mathbf{c}=\lambda_{1} \mathbf{a}+\lambda_{2} \mathbf{b}+\lambda_{3}(\mathbf{a} \times \mathbf{b})\), then the following is (are) true (a) \(\lambda_{1}=\mathbf{a} \cdot \mathbf{c}\) (b) \(\lambda_{2}=|\overrightarrow{\mathrm{b}} \times \overline{\mathrm{a}}|\) (c) \(\lambda_{3}=\mid(\mathrm{a} \times \mathbf{b}) \times \mathbf{c}\) (d) \(\lambda_{1}+\lambda_{2}+\lambda_{3}=(a+b+a \times b) \cdot c\)

Let \(\mathrm{a}\), b, \(\mathbf{c}\) are three vectors along the adjacent edges of a tetrahedron, if \(|\mathbf{a}|=|\mathbf{b}|=|\mathbf{c}|=2\) and \(\mathbf{a} \cdot \mathbf{b}=\mathbf{b} \cdot \mathbf{c}=\mathbf{c} \cdot \mathbf{a}=\mathbf{2}\) then volume of tetrahedron is (a) \(\frac{1}{\sqrt{2}}\). (b) \(\frac{2}{\sqrt{3}}\) (c) \(\frac{\sqrt{3}}{2}\) (d) \(\frac{2 \sqrt{2}}{3}\)

The magnitude of the vector \((\mathbf{p} \cdot \mathbf{s})(\mathbf{q} \times \mathbf{r})+(\mathbf{q} \cdot \mathbf{s})(\mathbf{r} \times \mathbf{p})+(\mathbf{r}-\mathbf{s})(\mathbf{p} \times \mathbf{q})\) is (a) 4 (b) 8 (c) 18 (d) 2

\(\mathbf{A}, \mathbf{B}\) and \(\mathbf{C}\) are three vectors given by \(2 \hat{\mathbf{i}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}\) Then, find \(\mathbf{R}\), which satisfies the relation \(\mathbf{R} \times \mathbf{B}=\mathbf{C} \times \mathbf{B}\) and \(\mathbf{R} \cdot \mathbf{A}=\mathbf{0}\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.