Question

Given that A = 2i + 3j - k, B = i - j +2k and C = 3i + 4j + k

a)find the unit vector along the direction of nA, for n<0

b) find the length of the projection of vector A on vector C

c)find the length of projection of vector C and vector A

d) find the sine of the angle between vector A and vector C

Answer 1

Given that $\overrightarrow{A} = 2\overrightarrow{i} + 3\overrightarrow{j} - \overrightarrow{k}$ , $\overrightarrow{B} = \overrightarrow{i} -\overrightarrow{j} +2 \overrightarrow{k}$ and $\overrightarrow{C} = 3\overrightarrow{i} + 4\overrightarrow{j}+\overrightarrow{k}$ .

a)find the unit vector along the direction of nA, for n<0

$\overrightarrow{e}=-\frac{\overrightarrow{A}}{|\overrightarrow{A}|}=-\frac{2\overrightarrow{i} + 3\overrightarrow{j} - \overrightarrow{k}}{\sqrt{2^2+3^2+(-1)^2}}=$$-\frac{2}{\sqrt{14}}\overrightarrow{i} -\frac{3}{\sqrt{14}}\overrightarrow{j} +\frac{1}{\sqrt{14}} \overrightarrow{k}$

b) find the length of the projection of vector A on vector C

$proj_{\overrightarrow{C}}\overrightarrow{A}=\frac{\overrightarrow{A}\cdot\overrightarrow{C}}{|\overrightarrow{C}|}=$$\frac{2\cdot3+3\cdot4+(-1)\cdot1}{\sqrt{3^2+4^2+1^2}}=\frac{17}{\sqrt{26}}$

c)find the length of projection of vector C and vector A

$proj_{\overrightarrow{A}}\overrightarrow{C}=\frac{\overrightarrow{A}\cdot\overrightarrow{C}}{|\overrightarrow{A}|}=$$\frac{2\cdot3+3\cdot4+(-1)\cdot1}{\sqrt{2^2+3^2+(-1)^2}}=\frac{17}{\sqrt{14}}$

d) find the sine of the angle between vector A and vector C

$|\overrightarrow{A}\times\overrightarrow{C}|=|\overrightarrow{A}|\cdot|\overrightarrow{C}|\cdot\sin{\alpha}$ , where $\alpha$ is the angle between $\overrightarrow{A}$ and $\overrightarrow{C}$

$\sin{\alpha}=\frac{|\overrightarrow{A}\times\overrightarrow{C}|}{|\overrightarrow{A}|\cdot|\overrightarrow{C}|}$

$\overrightarrow{A}\times\overrightarrow{C}=\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ 2 & 3 & -1 \\ 3 & 4 & 1 \end{vmatrix}=$$\overrightarrow{i}(3\cdot1-4\cdot(-1))-\overrightarrow{j}(2\cdot1-3\cdot(-1))+$$\overrightarrow{k}(2\cdot4-3\cdot3)=$$7\overrightarrow{i}-5\overrightarrow{j}-\overrightarrow{k}$

$\sin{\alpha}=\frac{\sqrt{7^2+(-5)^2+(-1)^2}}{\sqrt{2^2+3^2+(-1)^2}\cdot\sqrt{3^2+4^2+1^2}}=\frac{\sqrt{75}}{\sqrt{14}\cdot\sqrt{26}}=\frac{\sqrt{75}}{\sqrt{364}}=$$\frac 52\sqrt{\frac{3}{91}}$