Detailed explanation of senior one mathematics

1 . a =-(b+ c)a^2=b^2+c^2+2bc=b^2+c^2-2

Similarly, B 2 = A 2+C 2-2 C 2 = A 2+B 2-2.

Add the three formulas to get A 2+B 2+C 2 = 6, and the simultaneous solution of the above formulas A 2 = B 2 = C 2 = 2.

|a|+|b|+|c|=3√2

2. vector AO= 1/3 (vector AB+ vector AC) vector BC= (vector AB- vector AC).

Then vector ao vector BC= 1/3 (square of vector AB-square of vector AC) =2/3.

3.( 1) vector AB is perpendicular to vector a (n-8, t)*(- 1, 2)=0. That is 2t=n-8.

The modulus of AB is equal to √5 times OA (n-8) 2+t 2 = 25 to get the vector OB.

(2) The straight line between vector AC and vector A * *, (ksinα-8, t)÷(- 1, 2) t =-2 (ksinα-8) tsinα =-2k (sinα) 2+16s inα =-2k.

K>4, so when sinα= 1, tsinα=4 t=4 k=6 vector oc = (6,4) vector OA times vector OC=48.