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Mathematics: Post your doubts here!

unhellty

unhellty

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Referred to the origin O, the points A, B and C have position vectors given by
−−→ OA = i + 2j + 3k, −−→ OB = 2i + 4j + k and −−→ OC = 3i + 5j − 3k.
(i) Find the exact value of the cosine of angle BAC. [4]
(ii) Hence find the exact value of the area of triangle ABC. [3]
(iii) Find the equation of the plane which is parallel to the y-axis and contains the line through B and C. Give your answer in the form ax + by + cÏ = d. [5]

Please help me with (ii), how do I get the sin of the angle sqrt41/21 as referred to mark scheme? My friend told me to use A^2=B^2+C^2, but how can we know if triangle ABC is a right-angled triangle? Is there any other way to solve this besides my friends method and sine rule?

ASAP. Thanks!!
 
qwertypoiu

qwertypoiu

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unhellty said:
Referred to the origin O, the points A, B and C have position vectors given by
−−→ OA = i + 2j + 3k, −−→ OB = 2i + 4j + k and −−→ OC = 3i + 5j − 3k.
(i) Find the exact value of the cosine of angle BAC. [4]
(ii) Hence find the exact value of the area of triangle ABC. [3]
(iii) Find the equation of the plane which is parallel to the y-axis and contains the line through B and C. Give your answer in the form ax + by + cÏ = d. [5]

Please help me with (ii), how do I get the sin of the angle sqrt41/21 as referred to mark scheme? My friend told me to use A^2=B^2+C^2, but how can we know if triangle ABC is a right-angled triangle? Is there any other way to solve this besides my friends method and sine rule?

ASAP. Thanks!!
Click to expand...
If you know cosine of an angle you can easily find sine of it as well as follows:
sin^2(X) + cos^2(X) = 1
You probably know this identity already. Substitute what you know to find the other.
 
A

awesomaholic101

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areeba240 said:
yea the ans is correct but how can u change the figure....the force Q is horizontal and R make angle with Q.
Click to expand...
It's the same thing. I've just rotated the diagram. You could otherwise draw a line perpendicular to the resultant as it is, then resolve in the direction of the resultant and in the direction perpendicular to it.
Hope you get what I mean.
 
areeba240

areeba240

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awesomaholic101 said:
It's the same thing. I've just rotated the diagram. You could otherwise draw a line perpendicular to the resultant as it is, then resolve in the direction of the resultant and in the direction perpendicular to it.
Hope you get what I mean.
Click to expand...
no what i meant to say was that if the force Q is horizontal then it cannot be 6sin(20) it should be 6
 
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awesomaholic101

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areeba240 said:
no what i meant to say was that if the force Q is horizontal then it cannot be 6sin(20) it should be 6
Click to expand...
Yes, if you take Q as horizontal. But I have taken the resultant to be horizontal to make calculations simpler.
Anyway, let's do it that way. You're resolving the forces in the direction of Q and in the direction perpendicular to it.

Resolving horizontally:
Rcos20 = 6 + 3cos(x+20)

Resolving vertically:
Rsin20 = 3sin(x+20)

So basically we are taking the resultant of the forces acting in the horizontal and the vertical directions. Then you will have to use simultaneous eqns to solve. Long process. Try working out both ways anyway and you will get a better understanding.
There may be many ways to solve a problem in mechanics. Your job is to do it the easiest and fastest way.
 
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