VECTORS-Which is the largest internal angle of the triangle?

Question

Consider the triangle ABC with vertices A(-6, -4, -2), B(-5, -5, -7), and C(-1, 4, -4). Which is the largest internal angle of the triangle?

ABC, BCA, or CAB?

WHAT is the size of this angle...?

HELP!

Answer 1

So, there are a few ways to do this problem.

The easiest is to decide which side of the triangle is the longest. Then of course the opposite angle to that side is the largest angle.

To finish the problem you have to actually calculate the angle. This is done using the dot product. You should know that for two vectors a and b

a.b = |a|*|b|*cos(p)

where p is the angle between a and b. Here I'm using . for the dot product and | | for the length of a vector. Thus in order to find p you can use the formula

acos(a.b/(|a|*|b|)) = p

Answer 2

(a million) enable 2 elements of a parallelogram be both vectors a and b. both diagonals are given by means of way of a+b and a-b. they're orthogonal if and on condition that the scalar product is 0: 0 = (a + b).(a - b)= |a|² - |b|². i.e., a and b have equivalent length. (2) Circle radius r. 3 elements on it with position vectors a, -a, b, with |a| = |b| = r. the elements of the triangle are the vectors 2a, b-a and b+a. |b-a|² + |b+a|² = |a|² - 2a.b + |b|² + |a|² + 2a.b + |b|² = 4r² = |2a|². for this reason, by means of way of Pythag, the triangle is a perfect thoughts-set triangle. enable the realm vectors of three elements A, B and X (from a puzzling and quickly starting up position) be a, b and x. The lines XA and XB are represented by means of way of the vectors a-x and b-x. the thoughts-set between them is given by means of way of the scalar product (a-x).(b-x) = |a-x|*|b-x|cos?. If ? is consistent the locus is (a-x).(b-x) = const.*|a-x|*|b-x|]. (this is going to grant a quartic curve (!) - an equation interior the component to x with 4th powers.) (3) position vectors a, b, c (wiith savour to a puzzling and quickly starting up position) of three vertices of a triangle A, B, C. elements given by means of way of b-c, c-a and a-b. Unit vectors (b-c)/|b-c| and so on. thoughts-set bisector at A is for this reason the line a + ?[(c-a)/|c-a| + (b-a)/|b-a|] (the placement ? is a variable parameter.)....and so on and so on and so on.... Sorry, i'm feeling lazy about this there is an marvelous style of lengthy algebra ahead... [hint: the line by means of way of a element A with with position vector a, and route given by means of way of a vector v has parametric sort a+?v.]

Answer 3

I’d better do like this:
Vector AB = (-5+6, -5+4, -7+2) = (1, -1, -5), |AB|^2 = 27;
Vector AC = (-1+6, 4+4, -4+2) = (5, 8, -2), |AC|^2= 25+64+4 = 93
Vector BC = (-1+5, 4+5, -4+7) = (4, 9, 3), |BC|^2 = 16+81+9 = 106
The longest side is opposite of biggest angle; thus angle A is the biggest!

Added:
(AB*AC) = 1*5 – 1*8 + 5*2 = 7 = |AB|*|AC|*cos(A) = 50.11*cos(A), hence cos(A)=0.14 >0 thus A=82°