Question

If two sides of a square are along X and Y axes, then what will be the unit vectors along the difference diagonals of square?

Answer 1

hey mate
here's the solution

Answer 2

Let a  square OSPQ with  side a

Let  the  vector  OQ  is  along  x-axis  and  vector OS along y-axis.

Now,

$\vec{OQ}=ai\\\;\\\vec{OS}=aj$

Resultant  of  these  two vector will be  one  of the diagonal vector OP(c),

$\vec{OP}=\vec{OQ}+\vec{OS}\\\;\\\vec{c}=ai-aj\\\;\\|\vec{c}|=\sqrt{a^2+a^2}\\\;\\|\vec{c}|=\sqrt{2}a$

Unit vector of  vector c,

$\bm{\hat{#c}}=\frac{\vec{a}}{|\vec{a}|}$

$\\\;\\\bm{\hat{#c}}=\frac{ai+aj}{\sqrt{2}a}\\\;\\\bm{\hat{#c}}=\frac{1}{\sqrt{2}}(i+j)$

Similarly,

$\vec{QP}=aj\\\;\\\vec{QO}=-ai$

Resultant  of  these  two vector will be  one  of the diagonal vector QS(b),

$\vec{QS}=\vec{QP}+\vec{QO}\\\;\\\vec{b}=aj-ai\\\;\\|\vec{b}|=\sqrt{a^2+a^2}\\\;\\|\vec{b}|=\sqrt{2}a$

Unit vector of  vector b,

$\bm{\hat{#b}}=\frac{\vec{b}}{|\vec{b}|}\\\;\\\bm{\hat{#b}}=\frac{aj-ai}{\sqrt{2}a}\\\;\\\bm{\hat{#b}}=\frac{1}{\sqrt{2}}(j-i)$

Difference  of  vector,:-

$\bm{\hat{#c}}-\bm{\hat{#b}}=\frac{1}{\sqrt{2}}(i+j)-\frac{1}{\sqrt{2}}(j-i)\\\;\\\bm{\hat{#c}}-\bm{\hat{#b}}=\frac{1}{\sqrt{2}}(i+j-j+i)\\\;\\\bm{\hat{#c}}-\bm{\hat{#b}}=\frac{1}{\sqrt{2}}(2i)\\\;\\\bm{\hat{#c}}-\bm{\hat{#b}}=\sqrt{2}i$