Question

(a) Find the value of m if the angle between a(3,m) and b (2,-1)is 360 degrees

/ 3marks

(b) Find the longest side of right triangle whose other sides is:(√2+1)cm and (√2-1)cm long

Answer 1

Solution:

(a)

Given:  a(3,m) and b (2,-1)

Using dot product formula of vectors:

$a.b=|a||b|\cos \theta$

$\\\Rightarrow 3(2)+m(-1)=\sqrt{3^2+m^2}\times\sqrt{2^2+(-1)^2}\times \cos(360\degree) \\\Rightarrow 6-m=\sqrt{(9+m^2)5}\times(1) \\\Rightarrow (6-m)^2=5(9+m^2) \\\Rightarrow 36+m^2-12m=45+5m^2 \\\Rightarrow 4m^2+12m+9=0 \\\Rightarrow 4m^2+6m+6m+9=0 \\\Rightarrow (4m+3)(2m+3)=0 \\\Rightarrow m=-\dfrac34,-\dfrac32$

(b)

Two sides are :p=(√2+1)cm and b=(√2-1) cm

We need to find the Hypotenuse (h).

Using Pythagoras theorem,

$h^2=p^2+b^2$

$\Rightarrow h^2=(\sqrt2+1)^2+(\sqrt2-1)^2 \\\Rightarrow h^2=2+1+2\sqrt2+2+1-2\sqrt2 \\\Rightarrow h^2=6 \\\Rightarrow h=\sqrt6\ cm$