Question

Let the distance between vectors are given as follows :
$\left(i\right)4i+3j-6k,-2i+j-k$ be $\sqrt{k}$
$\left(ii\right)-2i+3j+5k,7i-k$ be $m\sqrt{n}$
Find $k-(m\ast n)$ ?

• A. 20
• B. 21
• C. 22
• D. 23

Answer 1

The correct option is D 23

(i) Distance between $4i+3j-6k$ and $-2i+j-k$ is given by $\sqrt{(4+2{)}^2+(3-1{)}^2+(-6+1{)}^2}=\sqrt{36+4+25}=\sqrt{65}$

$\Rightarrow k=65$

(ii) Distance between $-2i+3j+5k$ and $7i-k$ would be $\sqrt{(-2-7{)}^2+(3{)}^2+(5+1{)}^2}=\sqrt{81+9+36}=\sqrt{126}=3\sqrt{14}$

$\Rightarrow m=3,n=14$

$\therefore k-(m\ast n)=65-(3\times 14)=65-42=23$