My answer don't match with the given answer.

[Win_odd Dhamnekar]

A vessel A is a sailing with a velocity of 11 knots per hour in the direction south-east and a second vessel B is sailing with a velocity of 13 knots per hour in a direction 30° of north. Find the velocity of A relative to B.

Answer: $\displaystyle v_A=\left[\frac{11}{2}, -\frac{11}{2}\right], v_B =\left[ -\frac{13}{2}, \frac{13\sqrt{3}}{2}\right], v_{A-B}=[14.28,-19.04 ]=23.80$ knots/hr.

Author gave answer 21.29 knots/hr in the direction 74° 47' south of east. How is that computed?

 

[Dr.Peterson]

A vessel A is a sailing with a velocity of 11 knots per hour in the direction south-east and a second vessel B is sailing with a velocity of 13 knots per hour in a direction 30° of north. Find the velocity of A relative to B.

Answer: $\displaystyle v_A=\left[\frac{11}{2}, -\frac{11}{2}\right], v_B =\left[ -\frac{13}{2}, \frac{13\sqrt{3}}{2}\right], v_{A-B}=[14.28,-19.04 ]=23.80$ knots/hr.

Author gave answer 21.29 knots/hr in the direction 74° 47' south of east. How is that computed?

First, though this doesn't affect the work, a "knot per hour" would be an acceleration, not a velocity: "knot" means "nautical mile per hour", not a distance. (But see here.)

Also, when the question says "30° of north", I would take it to mean "30° east of north" (if it is not just an error), not as you apparently did, "30° west of north". But perhaps you omitted the word "west" when you copied the problem?

But your main error is that the magnitude of your v_A is not 11. Check that!

 

[Win_odd Dhamnekar]

First, though this doesn't affect the work, a "knot per hour" would be an acceleration, not a velocity: "knot" means "nautical mile per hour", not a distance. (But see here.)

Also, when the question says "30° of north", I would take it to mean "30° east of north" (if it is not just an error), not as you apparently did, "30° west of north". But perhaps you omitted the word "west" when you copied the problem?

But your main error is that the magnitude of your v_A is not 11. Check that!

$\displaystyle v_A$ is a typographical error. It should be $\displaystyle v_A= \left[\frac{11}{\sqrt{2}},-\frac{11}{\sqrt{2}}\right]$ If we take 30° east of north, then $\displaystyle v_B=\left[\frac{13}{2}, -\frac{13\sqrt{3}}{2}\right]$

Now $\displaystyle v_{A-B}= [1.28, 3.48], ABS(v_{A-B})=3.707$ knots/hr.

That means even after adjusting $\displaystyle v_B$ I don't get author's answer. How is that? is answer given wrong?

 

[Dr.Peterson]

$\displaystyle v_A$ is a typographical error. It should be $\displaystyle v_A= \left[\frac{11}{\sqrt{2}},-\frac{11}{\sqrt{2}}\right]$ If we take 30° east of north, then $\displaystyle v_B=\left[\frac{13}{2}, -\frac{13\sqrt{3}}{2}\right]$

Now $\displaystyle v_{A-B}= [1.28, 3.48], ABS(v_{A-B})=3.707$ knots/hr.

That means even after adjusting $\displaystyle v_B$ I don't get author's answer. How is that? is answer given wrong?

You still have a wrong sign in B.

But I haven't found a way to get their answer, either. Can you confirm the exact wording of the problem (particularly the direction of B), perhaps as an image?

 

[Win_odd Dhamnekar]

You still have a wrong sign in B.

But I haven't found a way to get their answer, either. Can you confirm the exact wording of the problem (particularly the direction of B), perhaps as an image?

$\displaystyle v_B$ describes $\displaystyle \frac{13}{2}\hat{i}$ for 30° east of North and $\displaystyle -\frac{13\sqrt{3}}{2}\hat{j}$ describe 60° north of east. Is that incorrect?

 

[Dr.Peterson]

$\displaystyle v_B$ describes $\displaystyle \frac{13}{2}\hat{i}$ for 30° east of North and $\displaystyle -\frac{13\sqrt{3}}{2}\hat{j}$ describe 60° north of east. Is that incorrect?

Why is north negative?

I recommend drawing a fairly accurate figure to check your work and your answers, particularly since the book appears to be wrong.

 

[Win_odd Dhamnekar]

Why is north negative?

I recommend drawing a fairly accurate figure to check your work and your answers, particularly since the book appears to be wrong.

Okay, We take north of east $\displaystyle \frac{13\sqrt{3}}{2}\hat{j}$ positive in $\displaystyle v_B$. In $\displaystyle v_A, \frac{11}{\sqrt{2}}\hat{i}$ is 45° east of south and $\displaystyle -\frac{11}{\sqrt{2}}\hat{j}$ is 45°south of east. Is this correct?

Why is it wrong to take $\displaystyle \frac{11}{\sqrt{2}}\hat{i}$ 45° east of south negative in $\displaystyle v_A?$

Is it just because we took $\displaystyle \frac{13}{2}\hat{i}$ in $\displaystyle v_A$ positive?

 

[Dr.Peterson]

Okay, We take north of east $\displaystyle \frac{13\sqrt{3}}{2}\hat{j}$ positive in $\displaystyle v_B$. In $\displaystyle v_A, \frac{11}{\sqrt{2}}\hat{i}$ is 45° east of south and $\displaystyle -\frac{11}{\sqrt{2}}\hat{j}$ is 45°south of east. Is this correct?

Why is it wrong to take $\displaystyle \frac{11}{\sqrt{2}}\hat{i}$ 45° east of south negative in $\displaystyle v_A?$

Is it just because we took $\displaystyle \frac{13}{2}\hat{i}$ in $\displaystyle v_A$ positive?

Did I say that both were wrong? Only the one. I think I only said,

You still have a wrong sign in B.

Why is north negative?

But, again, I haven't found a way to get their answer by changing any one thing, so I wouldn't worry about not having their answer. Just draw a picture and see if your answer matches that.