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

Literature: Collection of problems in mathematics. Part 1. Edited by A. V. Efimov, B. P. Demidovich.

Projections of a vector. Direction cosines.

The projection of vector $a$ onto vector $b$ is defined as the number $Pr_b a=|a|\cos\varphi,$ where $\varphi=\widehat{(a, b)}$ is the angle between vectors $a$ and $b,$ $0\leq\varphi\leq\pi.$

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. $$|(x, y)|^2\leq(x, x)(y, y).$$

Examples:

2.35. 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) $Pr_j a.$

Solution.

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

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

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

$a_{1y}=2;$

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

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

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

From here $a_x=-19/3.$

г) $Pr_j a=a_y=0.$

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

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

в) $a_x=-19/3;$

г) $Pr_j a=0.$

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

Solution.

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

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

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

Homework.

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

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

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

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

г) $Pr_j (a-b).$

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

б) $a-\frac{1}{2}b+c=d(3, 11/2, 0);$

в) $a+b-2c=-2j;$

г) $Pr_j (a-b)=6.$

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