If `A=3hat(i)-2hat(j)+hat(k),B=hat(i)-3hat(j)+5hat(k) and C=2hat(i)+hat(j)-4hat(k)` form a right angled triangle, then out of the following which one is satisfed ?

Options

A. `A=B+CandA^(2)=B^(2)+C^(2)`
B. `A=B+CandB^(2)=A^(2)+C^(2)`
C. `B=A+CandB^(2)=A^(2)+C^(2)`
D. `B=A+CandA^(2)=B^(2)+C^(2)`

Answer

B

Explanation

Given,
`A=3hat(i)-2hat(j)+hat(k)`
`B=hat(i)-3hat(j)+5hat(k)`
`C=2hat(i)+hat(j)-4hat(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
`B^(2)=A^(2)+C^(2)`
`(sqrt(35))^(2)=(sqrt(14))^(2)+(21)^(2)`
Now, `A*C=(3hat(i)-2hat(j)+hat(k))*(2hat(i)+hat(j)-4hat(k))`
Hence, A and C are perpendicular to each other `:.` Resultant of A and C is B

`B=A+C`
[according to triangle law]