NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra (Ex 10.4) Exercise 10.4

NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 is the solution of Exercise 10.4. This page is dedicated to the PDF solution and information on the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4. These solutions are made by collaborating all the important resources and data researched on the topic of Vector Algebra. Students have been provided with the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 on this website to make their preparation for the Class 12 board examinations an effortless journey.

Students have areas of strength and areas that need improvement. Each one of them has their own performance capacities and learning strategies. Expecting all students to understand the same NCERT books with the same results is not right. Therefore, every learner requires a standard resource that has been carefully chosen to correspond with the NCERT Syllabus. NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 provides a holistic way of learning. Every student can use NCERT Solutions to assist them based on their needs. High-calibre study resources are created by knowledgeable faculty members at Extramarks. Extramarks’ NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 have been made to be clear, precise, and useful. Every study material is appropriate for each class level and therefore, helpful for students’ utilization. The PDF of the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 is accessible on the Extramarks’ website for the examinees.

NCERT Solutions for K-12:

Given below are the descriptions of the NCERT Solutions that are offered by Extramarks for each class. These solutions are crucial and beneficial for children in all grades, from 1 to 12.

NCERT Solutions Class 12

Extramarks has study material for all the subjects from the various streams, including Arts/Humanities, Science, and Commerce. The NCERT Solutions for all the subjects can be easily downloaded from the website. All the study materials, includingthe NCERT Solutions are divided into the appropriate categories according to class, topic, and resource type. A student’s career is much advanced by the time they reach Class 12, therefore studying cannot be neglected or treated casually. Students in Intermediate receive continuous guidance from NCERT Solutions Class 12. At this phase of their careers, it is unfair to ask them to sift through irrelevant information on the internet that is provided by questionable sources. Hence, Extramarks can be used as round-the-clock assistance for students. The NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 is a good way to judge the pattern and way of handling questions by Extramarks. Students and their guardians can download NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 and see the efforts that are put into each of the exercises of each chapter. The PDF of NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 is accessible on the Extramarks’ website for the examinees.

NCERT Solutions Class 11

Class 11 students tend to take their studies lightly. After the hectic board examination of Class 10, students wish to take a break and study casually. Students need to understand the importance of the Class 11 examination and the future consequences it may have on the Class 12 board examination. Class 11 is the path that takes students to Class 12. This is why Class 11 should be taken truly seriously. Having NCERT Solutions Class 11 reduces the burden for students. The NCERT Solutions by Extramarks simply lower the pressure on students by assisting them in overcoming doubts.

NCERT Solutions Class 10

Everyone understands the significance of the Class 10 board examinations and is aware of the impact they have on the child’s career. Not only is the effect after the results exhilarating and exhausting, but so is the preparation phase.The development of the NCERT Solutions for Class 10 involves much research and pedagogical knowledge at Extramarks. Extramarks employs highly skilled and knowledgeable teachers who are aware of the needs of every student. Students need to incorporate other materials in their study plan in addition to the NCERT Solutions, such as past years’ papers, sample papers, revision notes, and a host of other things. All the NCERT Solutions for Class 10 are created with the highest care for the convenience of students, just as the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4

NCERT Solutions Class 9

For the approaching board examinations, class 9 serves as the foundational stage. The Class 9 curriculum is extensive and thorough. The concepts and subjects covered in Class 9 are helpful in the upcoming examinations of Class 10 & Class 12. The CBSE guidelines are the basis of Extramarks’ NCERT Solutions, which provide concept-based learning. These are highly beneficial and should be used as a step-by-step solution for each chapter during the revision and preparation period for the Class 9 examinations.

NCERT Solutions Class 8

Class 8 children need to simultaneously learn their academic syllabus and have fun with extracurricular activities. This is why they need to utilise resources for quick and smart solutions rather than difficult ones that take up all their time. Extramarks’ NCERT Solutions for Class 8 strictly adhere to the criteria for CBSE Examinations.

Extramarks Class 8 NCERT Solutions PDF assists students in developing a strong conceptual foundation that will benefit them in their exams.

NCERT Solutions Class 7

Class 7 students are still relatively young and often do not take their academics very seriously. They are unaware of the long-term effects of disregarding their studies. Hence, while students are still working out their study strategies, using NCERT Solutions Class 7 can aid and direct them with their studies.

NCERT Solutions Class 6

In many nations, receiving a free primary education is a fundamental right. If not properly guided, a primary-aged student may become confused and lose sight of their professional path. The NCERT Solutions Class 6 by Extramarks contain all the data required for a thorough comprehension of each subject’s lessons of Class 6. Instead of just going through the NCERT Solutions all by themselves, students should use a combination of the study materials provided by Extramarks and their own hard work to get the results they aspire for.

NCERT Solutions Class 5

The NCERT Solutions from Extramarks provide children with a fun and simple method of learning. When young students of Class 5 receive the right outside assistance, they continue to grow as a whole without being psychologically worn out. The NCERT Solutions Class 5 help clear up their doubts while studying and solve problems with little effort.

NCERT Solutions Class 4

When students receive the help they require, both their academic and extracurricular performances improve. NCERT Solutions Class 4 lets students learn in their own way and at their own pace. When students have easy access to NCERT Solutions, they are more likely to attempt the task on their own because they know they can always refer back to their study materials.

NCERT Solutions Class 3

Class 3 is harder than it seems. For students, it is a significant transition. Studies become more difficult. For the benefit of kids, Extramarks’ NCERT Solutions Class 3 can be downloaded in PDF format. Young children can relieve study stress by reading interesting and educational information.

NCERT Solutions Class 2

Young children’s minds begin to develop, and with the help of their parents, teachers, and the right guidance, they can grow up to be independent adults. Hence, NCERT Solutions Class 2 contain great learning materials for children in Class 2.

NCERT Solutions Class 1

Smart devices like phones and tablets are now commonly used by young children. Parents need to supervise their children’s use of technology and encourage them to use materials that will be helpful in their growth, like Extramarks’ fun study materials. The NCERT Solutions Class 1 are entertaining and interesting, and students will gain a lot of understanding.

Importance of the NCERT Solutions Class 12 in the preparation for board examination:

Using NCERT Solutions to prepare for the board examinations has a wide range of benefits. The NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 has a distinct page where the exercise solution can be downloaded. The NCERT Solutions are organised on the Extramarks’ website in a way that is simple to access and utilise. For Class 12 students who are expected to juggle numerous courses and subjects all at once, the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 and other NCERT Solutions are designed in a way that makes studying simple. Teachers with the necessary training and experience have created Extramarks’ NCERT Solutions for Class 12 Mathematics. These answers are well-suited to the CBSE guidelines. Using the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 and other Extramarks resources is like being tutored by the best teachers in the country by just using a laptop or mobile phone. For clarification and review, students should use the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4.

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra (Ex 10.4) Exercise 10.4

NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 is in downloadable PDF format, and all the exercises in the chapter, are solved by Extramarks experts. Extramarks’ PDFs are developed in line with the guidelines provided by CBSE and the standard of answers a CBSE examiner would expect. The NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 with all the problems and solutions are helpful in the revision of the complete syllabus of the chapter, which enhances the scoring abilities of the students’ answers.

Access NCERT Solutions for Class 12 Maths Chapter-10 – Vector Algebra

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.4

The NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 are very informative and resourceful. Using the NCERT Solutions for the Ex 10.4 Class 12 Mathematics is one of the best options for CBSE students in Class 12. The chapter on Vector Algebra has multiple exercises. This page particularly deals with the Ex 10.4 Class 12 NCERT Solutions. Students can easily download the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 from the Extramarks website, along with other exercises in the chapter.The problems and questions from the exercise have been meticulously handled by the internal subject matter experts at Extramarks while adhering to all CBSE guidelines. Every student of Class 12 can easily follow the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 and comprehend all the problems and solutions for the same. Students also get to understand the questions that are important for their examination and the pattern that will be followed in it. They get informed about the grade distribution and time needed for solving each type of question.Download the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 from this page and get access to all the essential pointers of the chapter.

Q.1

Find | \vec{a} \times \vec{b} |, if \vec{a} = \hat{i} - 7 \hat{j} + 7 \hat{k} and \vec{b} = 3 \hat{i} - 2 \hat{j} + 2 \hat{k} .

Ans

$\begin{array}{l} Given vectors are: \vec{a} = \hat{i} - 7 \hat{j} + 7 \hat{k} and \vec{b} = 3 \hat{i} - 2 \hat{j} + 2 \hat{k} \\ Then, \\ \vec{a} \times \vec{b} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 7 & 7 \\ 3 & - 2 & 2 \end{matrix} | \\ = (- 14 + 14) \hat{i} + (2 - 21) \hat{j} + (- 2 + 21) \hat{k} \\ = - 19 \hat{j} + 19 \hat{k} \\ ∴ | \vec{a} \times \vec{b} | = | - 19 \hat{j} + 19 \hat{k} | \\ = \sqrt{{(- 19)}^{2} + {(19)}^{2}} \\ = 2 \sqrt{19} \end{array}$

Q.2

$\begin{array}{l} Find a unit vector perpendicular to each of the vector \vec{a} + \vec{b} \\ and \vec{a} - \vec{b}, where \vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} and \vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k} . \end{array}$

Ans

$\begin{array}{l} The given vectors are \vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} and \vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k} . \\ Then, \vec{a} + \vec{b} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} + \hat{i} + 2 \hat{j} - 2 \hat{k} \\ = 4 \hat{i} + 4 \hat{j} \\ and \vec{a} - \vec{b} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} - \hat{i} - 2 \hat{j} + 2 \hat{k} \\ = 2 \hat{i} + 4 \hat{k} \\ (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) \\ = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{matrix} | \\ = (16 - 0) \hat{i} - (16 - 0) \hat{j} + (0 - 8) \hat{k} \\ = 16 \hat{i} - 16 \hat{j} - 8 \hat{k} \\ | (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) | \\ = | 16 \hat{i} - 16 \hat{j} - 8 \hat{k} | \\ = \sqrt{{(16)}^{2} + {(- 16)}^{2} + {(- 8)}^{2}} \\ = 24 \\ Hence, the unit vector perpendicular to each of the vectors \\ (\vec{a} + \vec{b}) and (\vec{a} - \vec{b}) is given by the relation, \\ = \pm \frac{(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})}{| (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) |} \\ = \pm \frac{16 \hat{i} - 16 \hat{j} - 8 \hat{k}}{24} \\ = \pm \frac{2 \hat{i} - 2 \hat{j} - \hat{k}}{3} \end{array}$

Q.3

$\begin{array}{l} If a unit vector \vec{a} makes angles \frac{π}{3} with \hat{i}, \frac{π}{4} with \hat{j} and an \\ acute angle θ with \hat{k}, then find θ and hence, the components of \vec{a} . \end{array}$

Ans

$\begin{array}{l} Let unit vector \vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} and | \vec{a} | = 1 . \\ It is given that \vec{a} makes angles \frac{π}{3} with \hat{i}, \frac{π}{4} with \hat{j} and an acute \\ angle θ with \hat{k} . \\ Then, we have: \\ cos \frac{π}{3} = \frac{a_{1}}{| \vec{a} |} \Rightarrow \frac{1}{2} = \frac{a_{1}}{1} \\ \Rightarrow a_{1} = \frac{1}{2} \\ cos \frac{π}{4} = \frac{a_{2}}{| \vec{a} |} \Rightarrow \frac{1}{\sqrt{2}} = \frac{a_{2}}{1} \\ \Rightarrow a_{2} = \frac{1}{\sqrt{2}} \\ cos θ = \frac{a_{3}}{| \vec{a} |} \Rightarrow cos θ = \frac{a_{3}}{1} \\ \Rightarrow a_{3} = cos θ \\ Since, | \vec{a} | = 1 \\ \Rightarrow \sqrt{{a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2}} = 1 \\ \Rightarrow {a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2} = 1 \\ {(\frac{1}{2})}^{2} + {(\frac{1}{\sqrt{2}})}^{2} + {cos}^{2} θ = 1 \\ {cos}^{2} θ = 1 - \frac{1}{4} - \frac{1}{2} \\ {cos}^{2} θ = \frac{1}{4} \\ \Rightarrow cos θ = \frac{1}{2} = \cos \frac{π}{3} \\ So, θ = \frac{π}{3} \\ Thus, θ = \frac{π}{3} and and the components of \vec{a} are (\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2}) . \end{array}$

Q.4

Show that (\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2 (\vec{a} \times \vec{b})

Ans

$\begin{array}{l} L . H . S . = (\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) \\ = \vec{a} \times \vec{a} + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - \vec{b} \times \vec{b} \\ = 0 + \vec{a} \times \vec{b} + \vec{a} \times \vec{b} - 0 [∵ \vec{b} \times \vec{a} = \vec{a} \times \vec{b}] \\ = 2 (\vec{a} \times \vec{b}) = R . H . S . \end{array}$

Q.5

Find λ and μ if (2 \hat{i} + 6 \hat{j} + 27 \hat{k}) \times (\hat{i} + λ \hat{j} + μ \hat{k}) = 0 .

Ans

$\begin{array}{l} (2 \hat{i} + 6 \hat{j} + 27 \hat{k}) \times (\hat{i} + λ \hat{j} + μ \hat{k}) = 0 \\ \Rightarrow | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & λ & μ \end{matrix} | = 0 \\ (6 μ - 27 λ) \hat{i} - (2 μ - 27) \hat{j} + (2 λ - 6) \hat{k} = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} \\ On comparing the corresponding components, we have : \\ 6 μ - 27 λ = 0, 2 μ - 27 = 0 and 2 λ - 6 = 0 \\ \Rightarrow μ = \frac{27}{2} and λ = 3 \end{array}$

Q.6

$\begin{array}{l} Given that \vec{a} . \vec{b} = 0 and \vec{a} \times \vec{b} = 0 . What can you conclude \\ about the vectors \vec{a} and \vec{b} ? \end{array}$

Ans

$\begin{array}{l} (i) \vec{a} . \vec{b} = 0 \\ Either | \vec{a} | = 0 or | \vec{b} | = 0, \\ or \vec{a} ⊥ \vec{b} (if \vec{a} and \vec{b} are non-zero.) \\ (ii) \vec{a} \times \vec{b} = 0 \\ Either | \vec{a} | = 0 or | \vec{b} | = 0, \\ or \vec{a} ∥ \vec{b} (if \vec{a} and \vec{b} are non-zero.) \\ But, \vec{a} and \vec{b} cannot be perpendicular and parallel simultaneously. \\ Hence, | \vec{a} | = 0 or | \vec{b} | = 0 . \end{array}$

Q.7

$\begin{array}{l} Let the vectors \vec{a}, \vec{b}, \vec{c} be given as a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}, \\ b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}, c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k} . Then show that \\ \vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} . \end{array}$

Ans

$\begin{array}{l} The given vectors are: a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}, b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}, c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k} . \\ L . H . S . = \vec{a} \times (\vec{b} + \vec{c}) \\ = (a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}) \times (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} + c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}) \\ = (a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}) \times {(b_{1} + c_{1}) \hat{i} + (b_{2} + c_{2}) \hat{j} + (b_{3} + c_{3}) \hat{k}} \\ = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ (b_{1} + c_{1}) & (b_{2} + c_{2}) & (b_{3} + c_{3}) \end{matrix} | \\ = {a_{2} (b_{3} + c_{3}) - a_{3} (b_{2} + c_{2})} \hat{i} + {a_{1} (b_{3} + c_{3}) - a_{3} (b_{1} + c_{1})} \hat{j} \\ + {a_{1} (b_{2} + c_{2}) - a_{2} (b_{3} + c_{3})} \hat{k} \\ = {a_{2} b_{3} + a_{2} c_{3} - a_{3} b_{2} - a_{3} c_{2}} \hat{i} + {a_{1} b_{3} + a_{1} c_{3} - a_{3} b_{1} - a_{3} c_{1}} \hat{j} \\ + {a_{1} b_{2} + a_{1} c_{2} - a_{2} b_{3} - a_{2} c_{3}} \hat{k} . .. (i) \\ \vec{a} \times \vec{b} = (a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}) \times (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) \\ = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{matrix} | \\ = (a_{2} b_{3} - a_{3} b_{2}) \hat{i} + (a_{1} b_{3} - a_{3} b_{1}) \hat{j} + (a_{1} b_{2} - a_{2} b_{1}) \hat{k} \\ \vec{a} \times \vec{c} = (a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}) \times (c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}) \\ = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix} | \\ = (a_{2} b_{3} - a_{3} b_{2}) \hat{i} + (a_{1} b_{3} - a_{3} b_{1}) \hat{j} + (a_{1} b_{2} - a_{2} b_{1}) \hat{k} \\ \vec{a} \times \vec{b} + \vec{a} \times \vec{c} \\ = (a_{2} b_{3} - a_{3} b_{2}) \hat{i} + (a_{1} b_{3} - a_{3} b_{1}) \hat{j} + (a_{1} b_{2} - a_{2} b_{1}) \hat{k} + (a_{2} b_{3} - a_{3} b_{2}) \hat{i} \\ + (a_{1} b_{3} - a_{3} b_{1}) \hat{j} + (a_{1} b_{2} - a_{2} b_{1}) \hat{k} \\ = (a_{2} b_{3} - a_{3} b_{2} + a_{2} b_{3} - a_{3} b_{2}) \hat{i} + (a_{1} b_{3} - a_{3} b_{1} + a_{1} b_{3} - a_{3} b_{1}) \hat{j} \\ + (a_{1} b_{2} - a_{2} b_{1} + a_{1} b_{2} - a_{2} b_{1}) \hat{k} \\ = (a_{2} b_{3} + a_{2} b_{3} - a_{3} b_{2} - a_{3} b_{2}) \hat{i} + (a_{1} b_{3} + a_{1} b_{3} - a_{3} b_{1} - a_{3} b_{1}) \hat{j} \\ + (a_{1} b_{2} + a_{1} b_{2} - a_{2} b_{1} - a_{2} b_{1}) \hat{k} . .. (ii) \\ From equation (i) and equation (ii), we have \\ \vec{a} \times (\vec{b} \times \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} . \\ Hence, the given result is proved. \end{array}$

Q.8

$\begin{array}{l} If either vector \vec{a} = 0 or \vec{b} = 0, then \vec{a} . \vec{b} = 0 . But the converse \\ need not be true. Justify your answer with an example. \end{array}$

Ans

$\begin{array}{l} Let \vec{a} = 3 \hat{i} + 4 \hat{j} + 3 \hat{k} and \vec{b} = 2 \hat{i} + 3 \hat{j} - 6 \hat{k} \\ Then, \\ \vec{a} . \vec{b} = (3 \hat{i} + 4 \hat{j} + 3 \hat{k}) (2 \hat{i} + 3 \hat{j} - 6 \hat{k}) \end{array}$

$\begin{array}{l} = 6 + 12 - 18 \\ = 0 \\ And \\ | \vec{a} | = | 3 \hat{i} + 4 \hat{j} + 3 \hat{k} | \\ = \sqrt{3^{2} + 4^{2} + 3^{2}} \\ = \sqrt{9 + 16 + 9} \\ = \sqrt{34} \neq 0 \\ | \vec{b} | = | 2 \hat{i} + 3 \hat{j} - 6 \hat{k} | \\ = \sqrt{2^{2} + 3^{2} + {(- 6)}^{2}} \\ = \sqrt{4 + 9 + 36} \\ = \sqrt{49} \\ = 7 \neq 0 \\ Hence, the converse of the given statement need not be true. \end{array}$

Q.9 Find the area of the triangle with vertices A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5).

Ans

$\begin{array}{l} The vertices of triangle ABC are given as A (1, 1, 2), B (2, 3, 5) \\ and C (1, 5, 5). \\ The adjacent sides \vec{AB} and \vec{BC} of Δ ABC are given as: \\ \vec{AB} = (2 - 1) \hat{i} + (3 - 1) \hat{j} + (5 - 2) \hat{k} \\ = \hat{i} + 2 \hat{j} + 3 \hat{k} \\ \vec{BC} = (1 - 2) \hat{i} + (5 - 3) \hat{j} + (5 - 5) \hat{k} \\ = - \hat{i} + 2 \hat{j} \\ ∴ \vec{AB} \times \vec{BC} \\ = \frac{1}{2} | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ - 1 & 2 & 0 \end{matrix} | \\ = \frac{1}{2} {(0 - 6) \hat{i} - (0 + 3) \hat{j} + (2 + 2) \hat{k}} \\ = \frac{1}{2} (- 6 \hat{i} - 3 \hat{j} + 4 \hat{k}) \\ The area ΔABC \\ = \frac{1}{2} | \vec{AB} \times \vec{BC} | \\ = \frac{1}{2} | - 6 \hat{i} - 3 \hat{j} + 4 \hat{k} | \\ = \frac{1}{2} \sqrt{{(- 6)}^{2} + {(- 3)}^{2} + 4^{2}} \\ = \frac{1}{2} \sqrt{36 + 9 + 16} \\ = \frac{1}{2} \sqrt{61} square units \\ Thus, the area of Δ ABC is \frac{1}{2} \sqrt{61} square units . \end{array}$

Q.10

$\begin{array}{l} Find the area of the parallelogram whose adjacents sides \\ are determined by the vectors \vec{a} = \hat{i} - \hat{j} + 3 \hat{k} and \vec{b} = 2 \hat{i} - 7 \hat{j} + \hat{k} . \end{array}$

Ans

$\begin{array}{l} The area of the parallelogram whose adjacent sides are \vec{a} and \vec{b} \\ is | \vec{a} \times \vec{b} | . \\ Adjacent sides are in vector form, which are as follows: \\ \vec{a} = \hat{i} - \hat{j} + 3 \hat{k} and \\ \vec{b} = 2 \hat{i} - 7 \hat{j} + \hat{k} \\ ∴ \vec{a} \times \vec{b} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & 3 \\ 2 & - 7 & 1 \end{matrix} | \\ = (- 1 + 21) \hat{i} + (1 - 6) \hat{j} + (- 7 + 2) \hat{k} \\ = 20 \hat{i} - 5 \hat{j} - 5 \hat{k} \\ | \vec{a} \times \vec{b} | = | 20 \hat{i} - 5 \hat{j} - 5 \hat{k} | \\ = \sqrt{20^{2} + {(- 5)}^{2} + {(- 5)}^{2}} \\ = \sqrt{400 + 25 + 25} \\ = \sqrt{450} \\ = 15 \sqrt{2} \\ Thus, the area of the given parallelogram is 15 \sqrt{2} square units. \end{array}$

Q.11

$\begin{array}{l} Let the vectors \vec{a} and \vec{b} be such that | \vec{a} | = 3 and | \vec{b} | = \frac{\sqrt{3}}{2}, then \\ \vec{a} \times \vec{b} is a unit vector, if the angle between \vec{a} and \vec{b} is \\ (A) \frac{π}{6} \\ (B) \frac{π}{4} \\ (C) \frac{π}{3} \\ (D) \frac{π}{2} \end{array}$

Ans

$\begin{array}{l} Given : | \vec{a} | = 3 and | \vec{b} | = \frac{\sqrt{2}}{3} \\ Since, \vec{a} \times \vec{b} = | \vec{a} | | \vec{b} | sinθ \hat{n}, where \hat{n} is a unit vector perpendicular \\ to both \vec{a} and \vec{b} and θ is the angle between \vec{a} and \vec{b} . \\ Since, \vec{a} \times \vec{b} is a unit vector, so | \vec{a} \times \vec{b} | = 1 . \\ | \vec{a} \times \vec{b} | = 1 \\ \Rightarrow | | \vec{a} | | \vec{b} | sinθ \hat{n} | = 1 \\ \Rightarrow | \vec{a} | | \vec{b} | | sinθ | = 1 \\ 3 \times \frac{\sqrt{2}}{3} \times sinθ = 1 \\ sinθ = \frac{1}{\sqrt{2}} = \sin \frac{π}{4} \\ \Rightarrow θ = \frac{π}{4} \\ Hence, \vec{a} \times \vec{b} is a unit vector if the angle between \vec{a} and \vec{b} is \frac{π}{4} . \\ The correct answer is B. \end{array}$

Q.12

$\begin{array}{l} Area of a rectangle having vertices A, B, C and D with \\ position vector - \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \hat{i} - \frac{1}{2} \hat{j} + 4 \hat{k} \\ and - \hat{i} - \frac{1}{2} \hat{j} + 4 \hat{k} respectively is \\ (A) \frac{1}{2} \\ (B) 1 \\ (C) 2 \\ (D) 4 \end{array}$

Ans

$\begin{array}{l} The position vectors of vertices A, B, C, and D of rectangle \\ ABCD are given as: \\ \vec{OA} = - \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \vec{OB} = \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \vec{OC} = \hat{i} - \frac{1}{2} \hat{j} + 4 \hat{k} and \\ \vec{OD} = - \hat{i} - \frac{1}{2} \hat{j} + 4 \hat{k} \\ The adjacent sides \vec{AB} and \vec{BC} of the given rectangle are given as: \\ \vec{AB} = (1 + 1) \hat{i} + (\frac{1}{2} - \frac{1}{2}) \hat{j} + (4 - 4) \hat{k} = 2 \hat{i} \\ \vec{BC} = (1 - 1) \hat{i} + (- \frac{1}{2} - \frac{1}{2}) \hat{j} + (4 - 4) \hat{k} = - \hat{j} \\ ∴ \vec{AB} \times \vec{BC} \\ = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 0 \\ 0 & - 1 & 0 \end{matrix} | \\ = (- 2 - 0) \hat{k} \\ = - 2 \hat{k} \\ | \vec{AB} \times \vec{BC} | \\ = | - 2 \hat{k} | \\ = 2 \\ Now, it is known that the area of a parallelogram whose adjacent \\ sides are \vec{a} and \vec{b} is | \vec{a} \times \vec{b} | . \\ Hence, the area of the given rectangle is | \vec{AB} \times \vec{BC} | = 2 square units. \\ The correct answer is C. \end{array}$

Please register to view this section

FAQs (Frequently Asked Questions)

1. How many problems and solutions are there in the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4?

The NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 has 12 questions all of which have solutions provided by Extramarks.

2. In what ways is the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 useful for Class 12 Mathematics preparation?

Students should the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 as a guide while solving the chapter on Vector Algebra. The NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 can be used during the zero-hour preparation of the examination or be used as revision material before the examination. Class 12 Mathematics can be challenging; having the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 by the side can be helpful when students need help and do not have time to waste. This keeps the students’ flow while solving Exercise 10.4 Class 12 intact.

3. What is the best procedure to prepare for CBSE Class 12 Mathematics?

The preparation for CBSE Class 12 Mathematics is as follows:

Analyse the NCERT(CBSE) Syllabus thoroughly.
Create a plan according to one’s study pattern.
Use past years’ papers and study material like the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 to understand the level of preparation.
Gain strength over weak areas by continuous revision.
Constant self-assessment by using the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 and other revision material to enhance preparation.

4. What topics are dealt with in the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4?

The NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 deals with the chapter on Vector Algebra. The NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 has clear step-by-step solutions with detailed explanations for all the questions.

5. Is the NCERT Solutions useful for Class 12 students?

Class 12 serves as a bridge between a student’s high school and college experiences. NCERT Solutions are a very useful resource to perform remarkably. Extramarks’ NCERT Solutions are accurate, easy and reliable. The preparations will benefit evidently from using study tools provided by Extramarks like the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4.

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra (Ex 10.4) Exercise 10.4

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra (Ex 10.4) Exercise 10.4

Access NCERT Solutions for Class 12 Maths Chapter-10 – Vector Algebra

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.4

Please register to view this section

FAQs (Frequently Asked Questions)

1. How many problems and solutions are there in the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4?

2. In what ways is the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 useful for Class 12 Mathematics preparation?

3. What is the best procedure to prepare for CBSE Class 12 Mathematics?

4. What topics are dealt with in the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4?

5. Is the NCERT Solutions useful for Class 12 students?

NCERT Solutions Related Links

NCERT Solutions

Solved Board Papers

CBSE Class

ICSE Class

Sample Paper

Exam Weightage

Board Paper Solution 2020

Test Preparation

Other Services

About Us

Terms & Conditions

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra (Ex 10.4) Exercise 10.4

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra (Ex 10.4) Exercise 10.4

Access NCERT Solutions for Class 12 Maths Chapter-10 – Vector Algebra

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.4

Share

Please register to view this section

FAQs (Frequently Asked Questions)

1. How many problems and solutions are there in the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4?

2. In what ways is the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 useful for Class 12 Mathematics preparation?

3. What is the best procedure to prepare for CBSE Class 12 Mathematics?

4. What topics are dealt with in the NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4?

5. Is the NCERT Solutions useful for Class 12 students?

Share

NCERT Solutions Related Links

NCERT Solutions

Solved Board Papers

CBSE Class

ICSE Class

Sample Paper

Exam Weightage

Board Paper Solution 2020

Test Preparation

Other Services

About Us

Terms & Conditions