QUESTION 13.
The given vector has the initial point at [tex]P=(-6,1)[/tex]; and terminal point at [tex]Q=(4,-4)[/tex]
We want to write vector [tex]v[/tex] in the form: [tex]ai+bj[/tex].
Using position vectors, we have;
[tex]v=^{\rightarrow}_{OQ}-^{\rightarrow}_{OP}[/tex]
This implies that;
[tex]v=\binom{4}{-4}-\binom{-6}{1}[/tex]
[tex]v=\binom{4--6}{-4-1}[/tex]
[tex]v=\binom{10}{-5}[/tex]
We now write this in the required form;
[tex]v=10i-5j[/tex]
QUESTION 14
It was given that;
[tex]P=(-5,1)[/tex] and Q=(x,-14).
[tex]^{\to}_{PQ}=^{\to}_{OQ}-^{\to}_{OP}[/tex]
[tex]^{\to}_{PQ}=\binom{x}{-14}-\binom{-5}{1}[/tex]
[tex]^{\to}_{PQ}=\binom{x--5}{-14-1}[/tex]
[tex]^{\to}_{PQ}=\binom{x+5}{-15}[/tex]
If the length of this vector is 25, then we can solve the following equation for x.
[tex]\sqrt{(x+5)^2+(-15)^2} =25[/tex]
We square both sides to get;
[tex](x+5)^2+(-15)^2=25^2[/tex]
[tex](x+5)^2+225=625[/tex]
[tex](x+5)^2=625-225[/tex]
[tex](x+5)^2=400[/tex]
Take the square root of both sides to get;
[tex]x+5=\pm \sqrt{400}[/tex]
[tex]x+5=\pm20[/tex]
[tex]x=-5\pm20[/tex]
[tex]x=-25,x=15[/tex]
QUESTION 15
The given vector, has magnitude [tex]||v||=13[/tex] and makes an angle of [tex]\alpha=60\degree[/tex] with the positive x-axis.
The components of this vector is given by
[tex]v=\binom{x=||v|| \cos \alpha}{y= ||v|| \sin \alpha}[/tex]
[tex]v=\binom{13 \cos 60\degree}{13\sin 60\degree}[/tex]
[tex]v=\binom{\frac{13}{2}}{\frac{13\sqrt{3}}{2}}[/tex]
Hence the required form is;
[tex]v=\frac{13}{2}i+\frac{13\sqrt{3}}{2}j[/tex]
QUESTION 16.
The given vector, has magnitude [tex]||v||=5[/tex] and makes an angle of [tex]\alpha=45\degree[/tex] with the positive x-axis.
The components of this vector is given by
[tex]v=\binom{x=||v|| \cos \alpha}{y= ||v|| \sin \alpha}[/tex]
[tex]v=\binom{5 \cos 45\degree}{5\sin 45\degree}[/tex]
[tex]v=\binom{\frac{5\sqrt{2}}{2}}{\frac{13\sqrt{3}}{2}}[/tex]
Hence the required form is;
[tex]v=\frac{5\sqrt{2}}{2}i+\frac{5\sqrt{2}}{2}j[/tex]
QUESTION 17
The given vector, has magnitude [tex]||v||=3[/tex] and makes an angle of [tex]\alpha=270\degree[/tex] with the positive x-axis.
The components of this vector is given by
[tex]v=\binom{x=||v|| \cos \alpha}{y= ||v|| \sin \alpha}[/tex]
[tex]v=\binom{3 \cos 270\degree}{13\sin 270\degree}[/tex]
[tex]v=\binom{0}{-1}[/tex]
Hence the required form is;
[tex]v=0i-j[/tex]
QUESTION 18
The given vector is;
[tex]v=-i+\sqrt{3}j[/tex]
This vector is in the second quadrant, because the x-component is negative and the y-component is positive.
The direction angle is given by;
[tex]\tan(\alpha)=\frac{y}{x}[/tex]
[tex]\tan(\alpha)=\frac{\sqrt{3} }{-1}=-\sqrt{3}[/tex]
This implies that;
[tex]\alpha=120\degree[/tex]
The direction angle of the vector is 120 degrees.
QUESTION 19
The given vector is;
[tex]v=-i+3j[/tex]
This vector is in the second quadrant, because the x-component is negative and the y-component is positive.
The direction angle is given by;
[tex]\tan(\apha)=\frac{y}{x}[/tex]
[tex]\tan(\alpha)=\frac{3}{-1}=-3[/tex]
This implies that;
[tex]\alpha=108.4\degree[/tex]
The direction angle of the vector is 1O8.4 degrees to one decimal place.
QUESTION 20
We want to find the angle between the vectors,
[tex]v=8i+6j[/tex]
and
[tex]w=4i+9j[/tex]
The angle between the two vectors is given by;
[tex]\cos \theta=\frac{v\bullet w}{||v||\:||w||}[/tex]
[tex]\cos \theta=\frac{(8i+6j)\bullet (4i+9j)}{(\sqrt{8^2+6^2} )(\sqrt{4^2+9^2})}[/tex]
[tex]\cos \theta=\frac{(8\times4+6\times9)}{(\sqrt{64+36} )(\sqrt{16+81})}[/tex]
[tex]\cos \theta=\frac{32+54}{(\sqrt{64+36} )(16+81)}[/tex]
[tex]\cos \theta=\frac{86}{(\sqrt{100} )(\sqrt{97})}[/tex]
[tex]\cos \theta=0.8732[/tex]
[tex]\theta=\cos^{-1}(0.8732)[/tex]
[tex]\theta=29.2\degree[/tex]
The angle between the two vectors is 29.2 degrees to one decimal place.
