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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 you
Answer 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
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