Find the three angles of the triangle formed using the position vectors i hat − j 4k and i hat 2j 3k and the line segment connecting their endpoints. Give your answers in degrees to two decimal places. (Enter your answers as a comma-separated list with no degree symbols.)

Answer:(75.31  ,  58.55,   46.14)Step-by-step explanation:The known two vectors have three components, i.e, they are vectors at [tex]R^{3}[/tex]. An easy strategy to deal with, it is focused on a product between vectors. Specially, the dot product and with its answer, it is easy to find the requested anglesFirstly, the dot product consists in an overlaping of vector onto the other one. Normaly, the vectors at [tex]R^{2}[/tex] and [tex]R^3[/tex] are easily operated, but if the dimension is increased, let say [tex]R^{4} or R^{5}[/tex], it is a must to use another strategy. For now,  the dot product can be understood as a multiplication of every single component  and then it sums the components in order to have the value of the overlapping. Some properties of this product are beneficial to algebraic purposes, then bear in mind: [tex]i\cdot i=1, j\cdot j=1, k\cdot k=1[/tex] and  [tex]i\cdot j=0, j\cdot k=0, k\cdot i=0[/tex]Therein, there are two formal definitions of dot product, either[tex]a\cdot b= |a||b|cos(\theta_{ab})[/tex]or[tex]a\cdot b= ai\cdot bi + aj\cdot bj +ak\cdot bk[/tex]When the product is equal to zero means the vectors are perpendicular between each other, otherwise they are not , With above definitions, it is convenient to use the second definition and then it will be used the first one. as followsPart a. Calculation of the length on vector "a"[tex]|a|=\sqrt{1^{2}+(-1)^{2}+4^2}=\sqrt(18)[/tex]Part b. Calculation of the length on vector "a"[tex]|b|=\sqrt{1^{2}+(2)^{2}+3^2}=\sqrt(14)[/tex]This procedures are related to the equation of the line segment. Now, using the second definition of dot product, it is calculated the overlapping between the two vectos, thus, [tex]a\cdot b=(i, -j, 4k) \cdot (i, 2j, 3k)= (i\cdot i,  -j\cdot2j , 4k\cdot3k)= (1-2+12)=11[/tex]These results are replaced in the first definion of the dot product, but the cosine function must be separated, thus, [tex]cos^-1(\theta_{ab})=cos^-1(\frac{a\cdot b}{|a||b|} )=cos^-1(\frac{11}{\sqrt{14}\sqrt{18}} ) \\\theta_{ab}=46.14[/tex]This is the first angle required by the exercise!It will be convenient to recall the calculated angle as [tex]C=46.14[/tex] and it is related between the vector a and b. To calculate the angles between the line segment connecting the their endpoints and the implied vectors, the law of cosines would be ideal technique to determine it, thus[tex]a^{2}=b^2+c^2-2bc *cos(\theta_{ab})\\b^{2}=a^2+c^2-2ac *cos(\theta_{ca})\\c^{2}=a^2+b^2-2ab *cos(\theta_{ab})[/tex]  The above equations allows to calculate the requested angles and the length if it is required. The first and the second equations provides the angles between the line segment and vector a. Also provides the angle between the line segment and vector b. The third one equation will be useful to calculate the length of the line segment. Use the following expression as follows For the length, [tex]c=\sqrt{(\sqrt{18})^2+(\sqrt{14})^2-2\sqrt{14}\sqrt{18}*cos(46.14)}=3.1624[/tex]For the angles [tex]cos(\theta_{ac})=\frac{a ^2-b^2-c^2}{-2bc} \\cos^-1(\theta_{ac})=\frac{(\sqrt{18}) ^2-(\sqrt{14})^2-(3.1624)^2}{-2*\sqrt{14}*3.1624} \\\\\theta_{ac}=75.31[/tex]Using exactly the same procedure, the angle between b and c is calculated, but keep in mind the values, thus[tex]cos(\theta_{bc})=\frac{b ^2-a^2-c^2}{-2ac} \\cos^-1(\theta_{ac})=\frac{(\sqrt{14}) ^2-(\sqrt{18})^2-(3.1624)^2}{-2*\sqrt{18}*3.1624} \\\\\theta_{ac}=58.55[/tex]To prove if thecalculated values were right, then it is convenient to sum all of them and if its sum is equal to 180 degrees means the calculations are well done, thus[tex]58.55+46.14+75.31=180[/tex]Prove!In addition of this calculation an attachment is included in which geometrical view is introduced.