stuck on a few questions any help will be much appreciated
1.The triangle PQR has vertices P(1,4,-1) Q(3,4,2) and R (-2,4,1). Show that PQR is:
(i)isosceles
(ii)right-angled at P
2.Two vectors u and v are such that magnitude of u=7, magnitude of v = 4 and u.v=14
The vector w is defined by w=2u+1/2v
Evaluate w.w and hence find the magnitude of w.
vectors
dfx
Posted 10 March 2007 - 10:48 PM
stuck on a few questions any help will be much appreciated
1.The triangle PQR has vertices P(1,4,-1) Q(3,4,2) and R (-2,4,1). Show that PQR is:
(i)isosceles
1.The triangle PQR has vertices P(1,4,-1) Q(3,4,2) and R (-2,4,1). Show that PQR is:
(i)isosceles
Isosceles implies 2 lengths are the same and 2 angles are equal if that helps.
(ii)right-angled at P
Dot product of 2 vectors extruding from the point P must be zero if they're right angled.
2.Two vectors u and v are such that magnitude of u=7, magnitude of v = 4 and u.v=14
The vector w is defined by w=2u+1/2v
Evaluate w.w and hence find the magnitude of w.
The vector w is defined by w=2u+1/2v
Evaluate w.w and hence find the magnitude of w.
w.w = (2u+1/2v).(2u+1/2v) . The dot product is distributive so you can multiply out like you would in regular multiplication (i.e. 2u.2u = 4u.u ... etc). Note that u.u = 1 (Since Cos of zero - the angle between u and itself - is 1). u.v = v.u (the dot product is also commutative) and the value of this is given to you.
xClairex
Posted 10 March 2007 - 11:01 PM
2.Two vectors u and v are such that magnitude of u=7, magnitude of v = 4 and u.v=14
The vector w is defined by w=2u+1/2v
Evaluate w.w and hence find the magnitude of w.
The vector w is defined by w=2u+1/2v
Evaluate w.w and hence find the magnitude of w.
w.w = (2u+1/2v).(2u+1/2v) . The dot product is distributive so you can multiply out like you would in regular multiplication (i.e. 2u.2u = 4u.u ... etc). Note that u.u = 1 (Since Cos of zero - the angle between u and itself - is 1). u.v = v.u (the dot product is also commutative) and the value of this is given to you.
i dont know what you are meaning for this bit :s
thnx 4 the help
Steve
Posted 11 March 2007 - 12:24 PM
w.w = (2u+1/2v).(2u+1/2v) . The dot product is distributive so you can multiply out like you would in regular multiplication (i.e. 2u.2u = 4u.u ... etc). Note that u.u = 1 (Since Cos of zero - the angle between u and itself - is 1). u.v = v.u (the dot product is also commutative) and the value of this is given to you.
That's not quite right.
For any vector u,

There is a similar example on page 124 of the Vectors notes. Does this help?
xClairex
Posted 11 March 2007 - 05:29 PM
i still dont really understand.heres what i think im supposed to do but im not sure
if u.u=u²
then w.w=w² = (2u+1/2v)² = (2u+1/2v)(2u+1/2v) = 4u² + 2u1/2v + 2u1/2v + 1/4v²
= 4u² + 4uv + 1/4v²
i dont know where to go from here
is this right or completly wrong?
if u.u=u²
then w.w=w² = (2u+1/2v)² = (2u+1/2v)(2u+1/2v) = 4u² + 2u1/2v + 2u1/2v + 1/4v²
= 4u² + 4uv + 1/4v²
i dont know where to go from here
is this right or completly wrong?
The Wedge Effect
Posted 11 March 2007 - 09:37 PM
Completely wrong. These lines on either side means the magnitude, or length of w, which is
, where x, y and z are the coefficients (number the variable is multiplied by) of the vector.

dfx
Posted 11 March 2007 - 10:12 PM
w.w = (2u+1/2v).(2u+1/2v) . The dot product is distributive so you can multiply out like you would in regular multiplication (i.e. 2u.2u = 4u.u ... etc). Note that u.u = 1 (Since Cos of zero - the angle between u and itself - is 1). u.v = v.u (the dot product is also commutative) and the value of this is given to you.
That's not quite right.
Hmm got abit carried away with the us and vs as is and js. Sorry about that.
xClairex
Posted 13 March 2007 - 05:55 PM
i dont think i have got w.w right i thought i would need to find u and then v so tht i get an actual number instead of equation for the magnitude of w.
me and my friend were working together and we got w.w to be 4u² + 4uv + 1/4v² i think someone said before that this was wrong
me and my friend were working together and we got w.w to be 4u² + 4uv + 1/4v² i think someone said before that this was wrong
George
Posted 14 March 2007 - 03:26 PM
i dont think i have got w.w right i thought i would need to find u and then v so tht i get an actual number instead of equation for the magnitude of w.
me and my friend were working together and we got w.w to be 4u² + 4uv + 1/4v² i think someone said before that this was wrong
me and my friend were working together and we got w.w to be 4u² + 4uv + 1/4v² i think someone said before that this was wrong
Yes, your answer isn't quite right, but you have the right idea.
2.Two vectors u and v are such that magnitude of u=7, magnitude of v = 4 and u.v=14
The vector w is defined by w=2u+1/2v
Evaluate w.w and hence find the magnitude of w.
The vector w is defined by w=2u+1/2v
Evaluate w.w and hence find the magnitude of w.
The way to do this is to use the properties of the scalar product, so:

So now we can get a value for
