a→, b→,c→ are the vectors of lengths 2,3,4, respectvely of a¯,b¯,c¯ are perpendicular to b¯+c¯, c¯+a¯, a¯+b¯ respectively then |a¯+b¯+c¯|`

a¯,b,¯c¯ are perpendicular to b¯+c¯,c¯+a¯,a¯+b¯a¯.(b¯+c¯)+b¯.(c¯+a¯)+c.¯(a¯+b¯)=0⇒2(a¯.b¯+b¯.c¯+c¯.a¯)=0|a¯+b¯+c¯|2=(a¯+b¯+c¯)2=|a¯|2+|b¯|2+|c¯|2+2(a¯.b¯+b¯.c¯+c¯.a¯)=4+9+16+0=29|a¯+b¯+c¯|2=29

a→, b→,c→ are the vectors of lengths 2,3,4, respectvely of a¯,b¯,c¯ are perpendicular to b¯+c¯, c¯+a¯, a¯+b¯ respectively then |a¯+b¯+c¯|`

a¯,b,¯c¯ are perpendicular to b¯+c¯,c¯+a¯,a¯+b¯a¯.(b¯+c¯)+b¯.(c¯+a¯)+c.¯(a¯+b¯)=0⇒2(a¯.b¯+b¯.c¯+c¯.a¯)=0|a¯+b¯+c¯|2=(a¯+b¯+c¯)2=|a¯|2+|b¯|2+|c¯|2+2(a¯.b¯+b¯.c¯+c¯.a¯)=4+9+16+0=29|a¯+b¯+c¯|2=29

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$\vec{a}, \vec{b}, \vec{c} a r e t h e v e c t o r s o f l e n g t h s 2, 3, 4, r e s p e c t v e l y$ of $\bar{a}, \bar{b}, \bar{c}$ are perpendicular to $\bar{b} + \bar{c}, \bar{c} + \bar{a}, \bar{a} + \bar{b}$ respectively then $| \bar{a} + \bar{b} + \bar{c} |$ `

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

-29

$- \sqrt{29}$

$\sqrt{29}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\bar{a}, \bar{b,} \bar{c}$ are perpendicular to $\bar{b} + \bar{c}, \bar{c} + \bar{a}, \bar{a} + \bar{b}$

$\begin{array}{l} \bar{a} . (\bar{b} + \bar{c}) + \bar{b} . (\bar{c} + \bar{a}) + \bar{c .} (\bar{a} + \bar{b}) = 0 \\ \Rightarrow 2 (\bar{a} . \bar{b} + \bar{b} . \bar{c} + \bar{c} . \bar{a}) = 0 \\ {| \bar{a} + \bar{b} + \bar{c} |}^{2} = {(\bar{a} + \bar{b} + \bar{c})}^{2} = {| \bar{a} |}^{2} + {| \bar{b} |}^{2} + {| \bar{c} |}^{2} + 2 (\bar{a} . \bar{b} + \bar{b} . \bar{c} + \bar{c} . \bar{a}) \\ = 4 + 9 + 16 + 0 \\ = 29 \\ {| \bar{a} + \bar{b} + \bar{c} |}^{2} = \sqrt{29} \end{array}$