find the cosines of the interior angles of the triangle with vertices A(3,-1,1), B(4,1,0),C(2,-3,0), find the vector component of AC orthogonal to AB. Solution We will find the position vectors AB, BC and CA which are: (4-3,-3--1,0-1) = (1, -2, -1) =vectAB (2-4,-3-1,0-0)=(-2, -4, 0) =vectBC (3-2,-1--3,1-0)=(1, 2, 1) =vect CA. |AB|^2= 6 |BC|^2=20 and |CA|^2 = 6, Threfore, ABC is Isosceles with AB=AC and BC greatest and there fore, A is greatest. VectAB.VectCB = |vectAB|*|VectBC| cosB or cosB = vectAB.vectBC/{|vectAB|*|VectBC|}^(1/2) = (1*-2+ 2*-4+-1*0)/{1^2+2^2+(-1)^2)((- 2)^2+(-4)^2+(0)^2} =-10/ {- or+sqrt(120)] or Similarly, cosC = (-2-8+0)/[- or+sqrt(120) =-10/{-or+sqrt(120) } cos A = -10/sqrt(120) and cosA = 4/6 or A =131.82 degree.and B = 24.09 degree and 24.09 degree. The postion vector AC = (2-3, -3--1, 0-1) = (-1, -2, -1) Let CD be perpendicular to AB, then D = A+ k(vectAB) = (3+k, -1+2k, -k).Now, P vect CD = p vectD -P vectC = (1+k, 4+4k, k). p vect CD = (3+k, -1+2k, -k) - (2, -3, 0) = (1+k, 2+2k -k) Since AB and CD are given to be orthogonal, AB.CD = 0 gives: (1, 2, -1).(1+k, 2+2k, -k) = 0 or 1+k+ 4+4k+k = 0 , giving k = -5/6. Therefore, CD = (1+k, 2+2k, -k) = (1-5/6, 2-10/6, 5/6) = (1/6, 2/6, 5/6) Therefore, the component of AC on orthogonal to AB = AC.CD |AB| = (1, 2, 1)*(1/6, 2/6, 5/6)/|(1/6, 2/6, 5/6)| =(1/6+4/6+5/6){(1/6)^2+(2/6)^2+(5/6)^2}^(1/2) = (1/3)sqrt(30).

1. find the cosines of the interior angles of the triangle with vertices A(3,-1,1), B(4,1,0),C(2,-3,0),
find the vector component of AC orthogonal to AB.
Solution
We will find the position vectors AB, BC and CA which are:
(4-3,-3--1,0-1) = (1, -2, -1) =vectAB
(2-4,-3-1,0-0)=(-2, -4, 0) =vectBC
(3-2,-1--3,1-0)=(1, 2, 1) =vect CA.
|AB|^2= 6
|BC|^2=20 and
|CA|^2 = 6,
Threfore, ABC is Isosceles with AB=AC and BC greatest and there fore, A is greatest.
VectAB.VectCB = |vectAB|*|VectBC| cosB or
cosB = vectAB.vectBC/{|vectAB|*|VectBC|}^(1/2) = (1*-2+ 2*-4+-1*0)/{1^2+2^2+(-1)^2)((-
2)^2+(-4)^2+(0)^2}
=-10/ {- or+sqrt(120)] or
Similarly, cosC = (-2-8+0)/[- or+sqrt(120) =-10/{-or+sqrt(120) }
cos A = -10/sqrt(120) and cosA = 4/6
or A =131.82 degree.and B = 24.09 degree and 24.09 degree.
The postion vector AC = (2-3, -3--1, 0-1) = (-1, -2, -1)
Let CD be perpendicular to AB, then D = A+ k(vectAB) = (3+k, -1+2k, -k).Now, P vect CD = p
vectD -P vectC = (1+k, 4+4k, k).
p vect CD = (3+k, -1+2k, -k) - (2, -3, 0) = (1+k, 2+2k -k)
Since AB and CD are given to be orthogonal, AB.CD = 0 gives:
(1, 2, -1).(1+k, 2+2k, -k) = 0 or
1+k+ 4+4k+k = 0 , giving k = -5/6.
Therefore,
CD = (1+k, 2+2k, -k) = (1-5/6, 2-10/6, 5/6) = (1/6, 2/6, 5/6)
Therefore, the component of AC on orthogonal to AB = AC.CD |AB| = (1, 2, 1)*(1/6, 2/6,
5/6)/|(1/6, 2/6, 5/6)|
=(1/6+4/6+5/6){(1/6)^2+(2/6)^2+(5/6)^2}^(1/2)
= (1/3)sqrt(30)