Find 2 unit vectors that make an angle of 60 with v={3,4}

[ahorn]

>>Find 2 unit vectors that make an angle of \(\displaystyle 60^\text{o} \text{ with } \vec v=\langle 3,4 \rangle\).

My working:

\(\displaystyle \cos{60^\text{o}}=\frac{1}{2}= \frac{\langle u_1,u_2\rangle\cdot\langle 3,4\rangle}{5\sqrt{u_1^2+u_2^2}}=\frac{3u_1+4u_2}{5\sqrt{u_1^2+u_2^2}}\)

Now, \(\displaystyle u_1^2+u_2^2=1\quad\Rightarrow u_1=\pm\sqrt{1-u_2^2}\)

So, \(\displaystyle \frac{5}{2}=\pm3\sqrt{1-u_2^2}+4u_2\\\text{ }\\
36(1-u_2^2)=25-80u_2+64u_2^2\\\text{ }\\
100u_2^2-80u_2-11=0\\\text{ }\\
u_2=\frac{4\pm3\sqrt{3}}{10}\approx-0.120 \quad \text{or}\quad 0.920\\\text{ }\\
\therefore u_1\approx -0.393 \quad \text{or} \quad 0.993\\
\text{ }\\
\text{i.e. the solutions are}\quad \langle-0.393,0.920\rangle \text{ and } \langle0.993, -0.120\rangle\)


Are my solutions correct?

(Stewart: Calculus and Concepts, Section 9.3 no. 26)

 

[Ishuda]

For a problem like this is is nice to know one form of the definition of a dot product: "Geometrically, it is the product of the Euclidean magnitudes of the two vectors and the cosine of the angle between them."
http://en.wikipedia.org/wiki/Dot_product
which is what you used to solve for the vectors. Now do the dot products and divide by the magnitude, i.e
(-0.393, 0.920) \(\displaystyle \bullet\) (3, 4) = 2.501
\(\displaystyle \sqrt{25 * (.393^2+.920^2)}\) = 5.0021220497
\(\displaystyle \frac{2.01}{5.0021220497} = .4999878002\)

Since the answer was supposed to be 0.5, that is not quite the right answer, (0.392820323, 0.919615242) is a little closer. Given that, the other answer is probably not quite right either, you might check.

 

[ahorn]

For a problem like this is is nice to know one form of the definition of a dot product: "Geometrically, it is the product of the Euclidean magnitudes of the two vectors and the cosine of the angle between them."
http://en.wikipedia.org/wiki/Dot_product
which is what you used to solve for the vectors. Now do the dot products and divide by the magnitude, i.e
(-0.393, 0.920) \(\displaystyle \bullet\) (3, 4) = 2.501
\(\displaystyle \sqrt{25 * (.393^2+.920^2)}\) = 5.0021220497
\(\displaystyle \frac{2.01}{5.0021220497} = .4999878002\)

Since the answer was supposed to be 0.5, that is not quite the right answer, (0.392820323, 0.919615242) is a little closer. Given that, the other answer is probably not quite right either, you might check.

Thank you. But, I was obviously rounding off to 3 decimal places.

 

[HallsofIvy]

A completely different way to do this is to note that the vector <3, 4> makes angle arctan(4/3)= 53.13 degrees with the positive x-axis. So vectors "that make an angle of 60 (degrees)" with it must make angles 53.13+ 60= 113.13 degrees and 53.13- 60= -6.87 degrees with the positive x-axis.

Unit vectors in those directions are <cos(113.13), sin(113.13)>= <-0.393, 0.920> and <cos(-6.87), sin(-6.87)>= <0.993, -0.120>.