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Find the components of the vector v}with given initial point P and terminal point Q. Find |v|. Sketch |v|. Find the unit vector in the direction of v.

P:(1,0,1.2),Q:(0,0,6.2)P:(1,0,1.2), \quad Q:(0,0,6.2)

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It \textbf v is a vector with the initial point P: (x1,y1,z1)P:\ (x_1, y_1, z_1) and the terminal point Q: (x2,y2,z2)Q:\ (x_2, y_2, z_2), then its components are given by

v1=x2x1,v2=y2y1,v3=z2z1,v_1=x_2-x_1,\quad v_2=y_2-y_1,\quad v_3=z_2-z_1,

and we denote this vector by v=[v1,v2,v3]\textbf v=[v_1, v_2,v_3]. So the components of the vector with the given initial and terminal point PP and QQ are

v1=01,v2=00,v3=6.21.2v1=1,v2=0,v3=5 .\begin{align*} &v_1=0-1,\quad v_2=0-0,\quad v_3=6.2-1.2\\ \Longrightarrow\quad &\boxed{v_1=-1,\quad v_2=0,\quad v_3=5}\ . \end{align*}

The length of the vector v=[v1,v2,v3]\textbf v=[v_1,v_2,v_3] is given by

v=v12+v22+v32,|\textbf v|=\sqrt{v_1^2+v_2^2+v_3^2},

so the length of our vector is

v=(1)2+02+52v=26 .\begin{align*} &|\textbf v|=\sqrt{(-1)^2+0^2+5^2}\\ \Longrightarrow\quad &\boxed{|\textbf v|=\sqrt{26}}\ . \end{align*}

We can obtain the unit vector in the direction of \textbf v by dividing the vector \textbf v with its norm (length).

x=vv=[1,0,5]26x=[1/26,0,5/26] .\begin{align*} &\textbf x=\frac{\textbf v}{|\textbf v|}=\frac{[-1,0,5]}{\sqrt{26}}\\ \Longrightarrow\quad & \boxed{\textbf x=\left[-1/\sqrt{26},0,5/\sqrt{26}\right]}\ . \end{align*}

The sketch of the vector \textbf v is given in the figure below.

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