Which is correct about vectors
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Which is correct about vectors

[From: ] [author: ] [Date: 12-10-17] [Hit: ]
a doesnt equal 0, b doesnt equal 0,i think b an c, cause not sure.-ALL THE ANSWERS ARE HIGHLIGHTED, THE REST IS EXPLANATION.......
Which of the following statements are true for
all vectors a, b?
A. |a + b|^2= |a|^2+ 2 a · b + |b|^2
B. |a · b| ≤ |a| |b|,
C. |a·b| = |a||b|, a doesn't equal 0, b doesn't equal 0, ⇒ a ⊥ b

i think b an c, cause not sure.

-
ALL THE ANSWERS ARE HIGHLIGHTED, THE REST IS EXPLANATION.

********************************
A and B are true, C is false.
********************************
For two vectors:

p = p_x i + p_y j + p_z k (1)
q = q_x i + q_y j + q_z k (2)

The dot product is given by:

p.q = q.p = p_xq_x + p_yq_y + p_zq_z = |p||q|cosθ (3)

where θ is the interior angle between p and q. So, from this it is clear that:

|p|^2 = p.p and |q|^2 = q.q.

Also, from the middle expression in (3) it should be clear that the dot product is distributive:

(p + q).r = p.r + q.r

for another vector r = r_x i + r_y j + r_z k.

So now:
************************************
A.
***|a + b|^2 = (a + b).(a + b) = a.a + b.a + a.b + b.b = |a|^2 + 2a.b + |b|^2***

B.
|a.b| = |a||b| |cosθ|

and |cosθ| ≤ 1

=> ***|a.b| ≤ |a||b|***

C.
***|a.b| = |a||b| => |cosθ| = 1 => θ = 0 => a || b***
***************************************…

-
Indeed. Thank you.

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