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
19.6k views
in Vectors by (31.5k points)
closed by

If a = 1/√10 ( 3i + k) and b = 1/7 ( 2i + 3j - 6k), then the value of (2a - b).[(a x b ) x ( a + 2b)] is 

(a) -3

(b) 5

(c) 3

(d) -5 

1 Answer

+1 vote
by (45.0k points)
selected by
 
Best answer

(a) -3

\((2\vec{a}-\vec{b}).\{(\vec{a}\times \vec{b})\times(\vec{a}+2\vec{b})\}\)

\((\vec{a}\times \vec{b})\times(\vec{a}+2\vec{b})\)

\(=(\vec{a}\times \vec{b})\times \vec{a}+2\,(\vec{a}\times\vec{b})\times\vec{b}\)

\(=(\vec{a}.\vec{a})\vec{b}-(\vec{b}.\vec{a})\vec{a}+2(\vec{a}.\vec{b})\vec{b}-(\vec{b}.\vec{b})\vec{a}\)

\(=|\vec{a}|^ 2\,\vec{b}-(\vec{a}.\vec{b})\vec{a}+2(\vec{a}.\vec{b})\vec{b}-|\vec{b}|^ 2\,\vec{a}\)

\(=\vec{b}-(\vec{a}.\vec{b})\vec{a}+2(\vec{a}.\vec{b})\vec{b}-\vec{a}\)

\((2\vec{a}-\vec{b}).\{\vec{b}-(\vec{a}.\vec{b})\vec{a}+2(\vec{a}.\vec{b})\vec{b}-\vec{a}\}\)

\(2\,\vec{a}.\vec{b}-2(\vec{a}.\vec{b})\,\vec{a}.\vec{a}+4(\vec{a}.\vec{b})(\vec{a}.\vec{b})-\vec{a}.(2\vec{a})\)

\(-\{\vec{b}.\vec{b}-(\vec{a}.\vec{b})(\vec{a}.\vec{b})+2(\vec{a}.\vec{b})(b.\vec{b})-\vec{a}.\vec{b}\}\)

Let 

a - b = α

= 2α - 2α + 4α2 - 2 - (1 - α2 + 2α - α)

= 4α2 - 2 - (1 - α2 + α)

= 4α2 - 2 - 1 + α2 - x

= (5α2 - α - 3) = -3

\(\vec{a}.\vec{b}=\frac{2}{7}.\frac{3}{\sqrt{10}}-\frac{6}{7}\times \frac{1}{\sqrt{10}}\)

\(=\frac{6}{7\sqrt{10}}-\frac{6}{7\sqrt{10}}=0\)

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.

Categories

...