Projections of a vector. Direction cosines. Cauchy-Schwarz inequality.

Projections of a vector. Direction cosines.

The projection of vector $a$ onto vector $b$ is defined as the number $P{r}_{b}a=|a|\cos \phi ,$ where $\phi =\widehat{(a,b)}$ is the angle between vectors $a$ and $b,$ $0\le \phi \le \pi .$

The coordinates $X,Y,Z$ of vector $a$ in a rectangular basis coincide with the projections of vector $a$ onto the basis vectors $i,j,k$ respectively, and the length of vector $a$ is equal to $|a|=\sqrt{X^2+Y^2+Z^2}.$

Numbers $$\cos \alpha =\cos \widehat{(a,i)}=\frac{X}{\sqrt{X^2+Y^2+Z^2}},$$

$$\cos \beta =\cos \widehat{(a,j)}=\frac{Y}{\sqrt{X^2+Y^2+Z^2}},$$

$$\cos \gamma =\cos \widehat{(a,k)}=\frac{Z}{\sqrt{X^2+Y^2+Z^2}},$$ are called direction cosines of vector $a$.

The direction cosines coincide with the coordinates (projections) of its unit vector $a_0=\frac{a}{|a|}$.

Cauchy-Bunyakovsky Inequality.

The Cauchy-Bunyakovsky Inequality holds for any vectors in Euclidean space. $$|(x,y){|}^2\le (x,x)(y,y).$$

Examples:

1. Given vectors $a_1(-1;2;0),$ $a_2(3;1;1),$ and $a_3(2;0;1),$ and $a=a_1-2a_2+\frac{1}{3}{a}_3.$ Calculate:

a) $|a_1|$ and coordinates of the unit vector $a_{1,0}$ of vector $a_1;$

b) $\cos \widehat{(a,j)};$

c) The $X$ coordinate of vector $a;$

d) $P{r}_{j}a.$

Solution.

a) $|a_1|=\sqrt{(-1{)}^2+2^2}=\sqrt{5};$

$a_{1,0}=\frac{a_1}{|a_1|}=\frac{(-1;2;0)}{\sqrt{5}}=(-\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}},0).$

b) $\cos \widehat{(a_1,j)}=\frac{a_{1y}}{|a_1|}.$

$a_{1y}=2;$

$|a_1|=\sqrt{5}.$

Thus, $\cos \widehat{a_1,j}=\frac{a_{1y}}{|a_1|}=\frac{2}{\sqrt{5}}$

c) $a=2a_1-2a_2+1/3a_3=(-1;2;0)-2(3;1;1)+1/3(2;0;1)=$ $=(-1;2;0)+(-6;-2;-2)+(2/3;0;1/3)=(-19/3;0;-5/3).$

From here $a_x=-19/3.$

d) $P{r}_{j}a=a_y=0.$

Answer: a) $|a_1|=\sqrt{5},$ $a_{1,0}(-\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}},0).$

b) $\cos \widehat{(a_1,j)}=\frac{2}{\sqrt{5}};$

c) $a_x=-19/3;$

d) $P{r}_{j}a=0.$

2. Find the coordinates of the unit vector $a_0,$ given $a(6;7;-6).$

Solution.

$$a_0=\frac{a}{|a|}=\frac{(6;7;-6)}{\sqrt{6^2+7^2+6^2}}=\frac{(6;7;-6)}{\sqrt{121}}=\frac{(6;7;-6)}{11}=(\frac{6}{11},\frac{7}{11},-\frac{6}{11}).$$

Answer: $(\frac{6}{11},\frac{7}{11},-\frac{6}{11}).$

Find the vector $x,$ which forms an angle of $\pi /3$ with the unit vector $j$ and an angle of $2\pi /3$ with the unit vector $k,$ given that $|x|=5\sqrt{2}.$

Solution.

Let $x=(x_1,x_2,x_3).$ Then

$\cos (x,j)=\frac{x_2}{|x|}=\frac{x_2}{5\sqrt{2}}=\frac{1}{2}\Rightarrow x_2=\frac{5\sqrt{2}}{2}=\frac{5}{\sqrt{2}}.$

$\cos (x,k)=\frac{x_3}{|x|}=\frac{x_3}{5\sqrt{2}}=-\frac{1}{2}\Rightarrow x_3=-\frac{5\sqrt{2}}{2}=-\frac{5}{\sqrt{2}}.$

Next, we find $x_1:$

$|x_1|=\sqrt{x_1^2+x_2^2+x_3^2}=\sqrt{x_1^2+(5/\sqrt{2}{)}^2+(5/\sqrt{2}{)}^2}=5\sqrt{2}.$

$x_1^2+\frac{50}{2}=50$

$x_1^2=25$

$x_1=\pm 5$

Answer: $(\pm 5,\frac{5\sqrt{2}}{2},-\frac{5}{\sqrt{2}}).$

Homework.

1. Given vectors $a=2i+3j,b=-3j-2k,c=i+j-k.$ Find

a) the coordinates of the unit vector $a_0;$

b) the coordinates of the vector $a-1/2b+c;$

c)the decomposition of the vector $a+b-2c$ with respect to the basis $B=(i,j,k);$

d) $P{r}_j(a-b).$

Answer: a)$a_0(\frac{2}{\sqrt{13}},\frac{3}{\sqrt{13}},0).$

b) $a-\frac{1}{2}b+c=d(3,11/2,0);$

c) $a+b-2c=-2j;$

d) $P{r}_j(a-b)=6.$

2. Find the length and direction cosines of the vector. $p=3a-5b+c$ если $a=4i+7j+3k;b=i+2j+k;c=2i-3j-k.$

Answer: а)$p=\sqrt{154};$ $\cos \alpha =9/\sqrt{154};$ $\cos \beta =8/\sqrt{154};$ $\cos \gamma =3/\sqrt{154}.$

Tags: Cauchy-Schwarz inequality, vector, direction cosines, projections of a vector