Book Store

Download books and chapters from book store.
Currently only available for
CBSE Gujarat Board Haryana Board

Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for
Class 10 Class 12

સદિશનું બીજગણિત

નીચેના સદિશનું માન શોધો :
$open parentheses 2 comma space 3 comma space square root of 3 close parentheses$

નીચેના સદિશનું માન શોધો :
$3 space straight i with hat on top space minus space 4 straight k with hat on top$

2 straight i with hat on top space minus space 2 straight j with hat on top space plus space straight k with hat on top

ની દિશામાં એકમ સદિશ શોધો.

20 માનવાળો અને $negative 3 straight i with hat on top space plus space 2 square root of 3 space straight j with hat on top space minus space 2 straight k with hat on top$ ની દિશાની વિરુદ્ધ દિશાનો સદિશ શોધો.

$stack straight x space with bar on top space equals space 3 straight i with hat on top space plus space 4 straight j with hat on top space minus space 5 straight k with hat on top space space bold અન ે bold space space stack straight y space with bar on top space equals space 2 straight i with hat on top space plus space straight j with hat on top comma$ , સદિશો માટે $stack straight x space with bar on top space plus space 2 stack space straight y space with bar on top$ ની દિશામાં એકમ સદિશ મેળવો.

અહીં, $stack straight x space with bar on top space equals space 3 straight i with hat on top space plus space 4 straight j with hat on top space minus space 5 straight k with hat on top space equals space open parentheses 3 comma space 4 comma space minus 5 close parentheses$

તથા $stack straight y space with bar on top space space equals space 2 straight i with hat on top space plus space straight j with hat on top space equals space open parentheses 2 comma space 1 comma space 0 close parentheses$

$therefore space stack straight x space with bar on top space plus space 2 space stack straight y space with bar on top space equals space open parentheses 3 comma space 4 comma space minus 5 close parentheses space plus space 2 space open parentheses 2 comma space 1 comma space 0 close parentheses space space space space space space space space space space space space space space space space space space equals space open parentheses 3 comma space 4 comma space minus 5 close parentheses space plus space open parentheses 4 comma space 2 comma space 0 close parentheses space space space space space space space space space space space space space space space space space space equals space open parentheses 7 comma space 6 comma space minus 5 close parentheses therefore space open vertical bar space stack straight x space with bar on top plus space 2 space stack straight y space with bar on top close vertical bar space equals space square root of 7 squared space plus space 6 squared space plus space open parentheses negative 5 close parentheses squared end root space space space space space space space space space space space space space space space space space space space equals space square root of 49 space plus space 36 plus space 35 end root space equals space square root of 110$

$therefore space stack straight x space with bar on top space plus space 2 space stack straight y space with bar on top$ ની દિશામાં એકમ સદિશ

$equals space fraction numerator stack straight x space with bar on top plus space 2 space stack straight y space with bar on top over denominator open vertical bar space stack straight x space with bar on top space plus space 2 space stack straight y space with bar on top close vertical bar end fraction equals space fraction numerator open parentheses 7 comma space 6 comma space minus 5 close parentheses over denominator square root of 110 end fraction equals space open parentheses fraction numerator 7 over denominator square root of 110 end fraction comma space fraction numerator 6 over denominator square root of 110 end fraction comma space fraction numerator negative 5 over denominator square root of 110 end fraction close parentheses equals space fraction numerator bold 7 over denominator square root of bold 110 end fraction bold space bold i with bold hat on top bold space bold plus bold space fraction numerator bold 6 over denominator square root of bold 110 end fraction bold space bold j with bold hat on top bold space bold minus bold space fraction numerator bold 5 over denominator square root of bold 110 end fraction bold space bold k with bold hat on top$

Switch

Chapter Chosen

Book Chosen

Subject Chosen

Book Store

Previous Year Papers

સદિશનું બીજગણિત

Switch