Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
51 views
in Algebra by (106k points)
closed by
Find \(|\vec b|\) if \((\vec a + \vec b) \cdot (\vec a - \vec b) = 8\) and \(|\vec a| = 8|\vec b|\) ?
1. \(\frac{2}{3\sqrt 7}\)
2. \(\frac{2}{3}\)
3. \(\frac{3}{2\sqrt 7}\)
4. \(\frac{2\sqrt 2}{3\sqrt 7}\)

1 Answer

0 votes
by (106k points)
selected by
 
Best answer
Correct Answer - Option 4 : \(\frac{2\sqrt 2}{3\sqrt 7}\)

CONCEPT:

  • \(\vec a \cdot \vec a = |\vec a|^2\)
  • \(\vec a \cdot \vec b = \vec b \cdot \vec a\)

CALCULATION:

Given: \((\vec a + \vec b) \cdot (\vec a - \vec b) = 8 \ and \ |\vec a| = 8|\vec b|\)

⇒ \((\vec a + \vec b) \cdot (\vec a - \vec b) = |\vec a|^2 - \vec a \cdot \vec b + \vec b \cdot \vec a - |\vec b|^2 = 8\)

As we know that, \(\vec a \cdot \vec b = \vec b \cdot \vec a\)

⇒ \( |\vec a|^2 - |\vec b|^2 = 8\)

∵ It is given that \(|\vec a| = 8|\vec b|\)

⇒ \(63|\vec b|^2 = 8\)

⇒ \(|\vec b| = \frac{2\sqrt 2}{3\sqrt 7}\)

Hence, option 4 is correct.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...