PART A
The given vectors are,
[tex]v_1 = \: < \: 3 , \: 5 \: > [/tex]
[tex]v_2 = \: < \: - 4 , \: 7 \: > [/tex]
The magnitude of the vector
[tex]v= \: < \: x , \: y \: > \: [/tex]
is given by:
[tex] |v| = \sqrt{ {x}^{2} + {y}^{2} } [/tex]
This implies that,
[tex] |v_1| = \sqrt{ {3}^{2} + {5}^{2} } [/tex]
[tex] |v_1| = \sqrt{9 + 25} [/tex]
[tex]|v_1| = \sqrt{34} [/tex]
[tex]|v_2| = \sqrt{ {( - 4)}^{2} + {7}^{2} } [/tex]
[tex]|v_2| = \sqrt{ 16+ 49} [/tex]
[tex]|v_2| = \sqrt{65} [/tex]
PART B
To find the unit vector in the direction of a given vector, we divide by the magnitude of that vector.
[tex] ^{ - } _{v_1} = \: < \: \frac{3}{ \sqrt{34} } , \: \frac{5}{ \sqrt{34} } \: > [/tex]
Rationalize the denominator.
[tex]^{ - } _{v_1} = \: < \: \frac{3\sqrt{34}}{ 34 } , \: \frac{5\sqrt{34}}{ 34 } \: > [/tex]
Also,
[tex] ^{ - } _{v_2} = \: < \: \frac{ - 4}{ \sqrt{65} } , \: \frac{7}{ \sqrt{65} } \: > [/tex]
[tex]^{ - } _{v_2} = \: < \: \frac{ - 4\sqrt{65}}{ 65 } , \: \frac{7\sqrt{65}}{ 65 } \: > [/tex]
PART C
The sketch of the given vectors as well as their unit vectors are shown in the attachment.
