00:01
Hi, in this question, given that a is 2, -3 ,4 and b is 0 ,1 ,2 and c is –1 ,2 ,0.
00:17
In part a, we need to find the angle a.
00:22
Here we know that the formula angle between the two vector a and b is cos inverse of a vector dot b vector divided by modulus a vector into modulus b vector.
00:44
Here oa vector can be written as 2 i cap – 3 j cap plus 4 k cap and ob vector equals 0 i cap plus j cap plus 2 k cap and oc vector equals – i cap plus 2 j cap plus 0 k cap and here ab vector can be found by ob vector minus oc vector.
01:17
On subtracting these two vector, then we get – 2 i cap plus 4 j cap minus 2 k cap.
01:25
Next we need to find ac vector which is equal to oc vector minus oa vector.
01:32
On subtracting this and this, then we get – 3 i cap plus 5 j cap minus 4 k cap.
01:45
So that angle a equals cos inverse of ab vector into ac vector divided by modulus ab into modulus ac.
02:02
Here cos inverse of ab vector into ac vector can be written as 6 plus 20 plus 8 and modulus ab vector is square root of 24, modulus ac vector is square root of 50 which is equal to cos inverse of 34 divided by 20 root 3.
02:18
Therefore the required angle a equals 11 .039 degree.
02:26
Next we need to find the area of triangle abc...