# SOLUTION: Hi Triangle ABC has a right angle at A where AB= [-3] [ 3] [-2] and where CA= [-1] [ 3] [ 6]

Algebra ->  Algebra  -> Vectors -> SOLUTION: Hi Triangle ABC has a right angle at A where AB= [-3] [ 3] [-2] and where CA= [-1] [ 3] [ 6]       Log On

 Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

 Algebra: Introduction to vectors, addition and scaling Solvers Lessons Answers archive Quiz In Depth

 Question 27347: Hi Triangle ABC has a right angle at A where AB= [-3] [ 3] [-2] and where CA= [-1] [ 3] [ 6] {the above vectors should each be in a long square bracket} a) Calculate the acute angle at B b) Find the length at AB I have a total blank on this question. I think the acute angle at B is 45 degrees,as it's a right angle triangle, but I don't know how to prove it using the vector information given. I think i have to find BC but I am stumped on how to do that, let alone find the length of AB. Please can you help? Thank youAnswer by venugopalramana(3286)   (Show Source): You can put this solution on YOUR website!Triangle ABC has a right angle at A where AB=....LET US REPRESENT VECTOR AB BY THE SYMBOL {AB}AND MODULUS OF VECTOR AB BY |AB| [-3] [ 3] [-2]...SO {AB}= -3I+3J-2K and where CA= [-1] [ 3] [ 6] ...SO {CA}= -1I+3J+6K...HENCE {BC}={AC}-{AB}=-{CA}-{AB}=-(-1I+3J+6K)- (-3I+3J-2K}=4I-6J-4K {BC}=4I-6J-4K {the above vectors should each be in a long square bracket} a) Calculate the acute angle at B.. LET US FIND THIS BY USING {AB}.{BC}=|AB|*|BC|COS(B) (-3I+3J-2K).(4I-6J-4K)=|-3I+3J-2K|.|4I-6J-4K|COS(B) -12-18+8=SQRT(9+9+4)*SQRT(16+36+16)COS(B) COS(B)=-22/SQRT(22*68)...THERE IS SOME THING WRONG WITH YOUR NUMBERS ..IF ANGLE B IS ACUTE WE SHOULD GET THIS AS A POSITIVE NUMBER OR LET US TAKE SQRT OF 22*68 AS -38.678.....SO THAT COS(B)=0.5688.... AND ANGLE B =0.9658 RADIANS OR 55.36 DEGREES.... IF ANGLE A IS RIGHT ANGLE THEN WE SHOULD HAVE {CA}.{AB} =0...LET US CHECK.. (-1I+3J+6K).(-3I+3J-2K)= +3+9-12=0...OK.... b) Find the length at AB WELL THIS IS |AB|= |-3I+3J-2K|=SQRT(9+9+4)=SQRT(22)=4.69 I have a total blank on this question. I think the acute angle at B is 45 degrees,as it's a right angle triangle, but I don't know how to prove it using the vector information given. I think i have to find BC but I am stumped on how to do that, let alone find the length of AB. Please can you help? Thank you