Discussion Forum



1)

If $A= 3\hat{i}-2\hat{j}+\hat{k}, B= \hat{i}-3\hat{J}+5\hat{K} $and $C= 2\hat{i}+\hat{j}-4\hat{k}$ forn a right angled triangle, then out of the following which one is satisfied ?


A) A=B+C and $A^{2}=B^{2}+C^{2}$

B) A=B+C and $B^{2}=A^{2}+C^{2}$

C) B=A+C and $B^{2}=A^{2}+C^{2}$

D) B= A+C and $A^{2}=B^{2}+C^{2}$

Answer:

Option B

Explanation:

Given,

 $A= 3\hat{i}-2\hat{j}+\hat{k}, $

$B= \hat{i}-3\hat{J}+5\hat{K} $

$C= 2\hat{i}+\hat{j}-4\hat{k}$

 Here,  $|A|=\sqrt{(3)^{2}+(-2)^{2}+(1)^{2}}$

or  $|A|=\sqrt{9+4+1}=\sqrt{14}$  .....(i)

$|B|=\sqrt{(1)^{2}+(-3)^{2}+(5)^{2}}$

$|B|=\sqrt{1+9+25}=\sqrt{35}$   ......(ii)

and   

  $|C|=\sqrt{(2)^{2}+(1)^{2}+(-4)^{2}}$

$|C|=\sqrt{4+1+16}=\sqrt{21}$  .........(iii)

 From Eqs.(i) , (ii) and (iii) , we get

 B2= A2+C2

$(\sqrt{35})^{2}=(\sqrt{14})^{2}+(21)^{2}$

Now, A.C= ($3\hat{i}-2\hat{j}+\hat{k}).(2\hat{i}+\hat{j}-4\hat{k}$)

A.C= 6-2-4=0

 Hence, A and C  are perpendicular to each other 

$\therefore$ Resultant of A and C  is B

1562021164_P3.PNG  B= A+C

 [according to triangle law]