Vector projections

[TsAmE]

Suppose that the vector a = (13, 1,2) is written as the sum of a vector u in the direction of b = (3, -1, 2) and a vector orthogonal to b. What is the length of v?

\(\displaystyle u = Proj_ba = \frac{b \cdot a}{|b|^2}b\)

\(\displaystyle =\frac{(3, -1, 2) \cdot (13, 1, 2)}{\sqrt{9 + 1 + 4}}(3, -1,2)\)

\(\displaystyle =\frac{42}{\sqrt{14}}(3, -1, 2)\)

\(\displaystyle = (\frac{126}{\sqrt{14}}, \frac{-42}{\sqrt{14}}, \frac{84}{\sqrt{14}})\)

I was gonna then use:

\(\displaystyle a = v + u\)

\(\displaystyle v = a - u\)

to find v, but u looks to messy and I dont think its right

 

[Deleted member 4993]

Suppose that the vector a = (13, 1,2) is written as the sum of a vector u in the direction of b = (3, -1, 2) and a vector orthogonal to b. What is the length of v?

\(\displaystyle u = Proj_ba = \frac{b \cdot a}{|b|^2}b\)

\(\displaystyle =\frac{(3, -1, 2) \cdot (13, 1, 2)}{\sqrt{9 + 1 + 4}}(3, -1,2)\)

\(\displaystyle =\frac{42}{\sqrt{14}}(3, -1, 2)\)

\(\displaystyle = (\frac{126}{\sqrt{14}}, \frac{-42}{\sqrt{14}}, \frac{84}{\sqrt{14}})\)

I was gonna then use:

\(\displaystyle a = v + u\)

\(\displaystyle v = a - u\)

to find v, but u looks to messy and I dont think its right

It is correct (and messy) - just continue and finish it.

 

[TsAmE]

From v = a -u:

\(\displaystyle = (13, 1, 2) - (\frac{126}{\sqrt{14}}, -\frac{42}{\sqrt{14}}, \frac{84}{\sqrt{14}})\)

\(\displaystyle \frac{1}{\sqrt{14}}(-113, -41, -82)\)

\(\displaystyle =\sqrt{\frac{12769}{14} + \frac{1681}{14} + \frac{6724}{14}}\)

\(\displaystyle = 38.89\)

but this isnt right

 

[Deleted member 4993]

From v = a -u:

\(\displaystyle = (13, 1, 2) - (\frac{126}{\sqrt{14}}, -\frac{42}{\sqrt{14}}, \frac{84}{\sqrt{14}})\)

\(\displaystyle \frac{1}{\sqrt{14}}(-113, -41, -82)\)

\(\displaystyle =\sqrt{\frac{12769}{14} + \frac{1681}{14} + \frac{6724}{14}}\)

\(\displaystyle = 38.89\)

but this isnt right

That's because your subtraction is incorrect!

13 - 126/?14 = (13*?14 - 126)/?14 = -20.67491648

 

[TsAmE]

Oh, sorry. Here is my correction:

v = a -u

=\(\displaystyle (\frac{13\sqrt{14} - 126}{\sqrt{14}}, \frac{\sqrt{14} + 42}{\sqrt{14}}, \frac{2\sqrt{14} - 84}{\sqrt{14}})\)

\(\displaystyle |v| = \sqrt{\frac{5984}{14} + \frac{2092}{14} + \frac{5854}{14}}\)
\(\displaystyle = 32 (rounded off)\)

It is a multiple choice question and has the possible answers:

A)\(\displaystyle \sqrt{5}\)
B) \(\displaystyle 3\)
C) \(\displaystyle 3\sqrt{2}\)
D) \(\displaystyle 4\sqrt{2}\)
E) \(\displaystyle 4\sqrt{3}\)

I checked but cant see any errors I have made

 

[Deleted member 4993]

Suppose that the vector a = (13, 1,2) is written as the sum of a vector u in the direction of b = (3, -1, 2) and a vector orthogonal to b. What is the length of v?

\(\displaystyle u = Proj_ba = \frac{b \cdot a}{|b|^2}b\)

\(\displaystyle =\frac{(3, -1, 2) \cdot (13, 1, 2)}{\sqrt{9 + 1 + 4}}(3, -1,2)\)

In here you should not have ? sign - since you are dividing by (|b|)[sup:1qzgq9qu]2[/sup:1qzgq9qu]

\(\displaystyle =\frac{42}{\sqrt{14}}(3, -1, 2)\)

\(\displaystyle = (\frac{126}{\sqrt{14}}, \frac{-42}{\sqrt{14}}, \frac{84}{\sqrt{14}})\)

I was gonna then use:

\(\displaystyle a = v + u\)

\(\displaystyle v = a - u\)

to find v, but u looks to messy and I dont think its right

It is correct (and messy) - just continue and finish it.

 

[TsAmE]

Oh ok thanks I got 4\sqrt{3}. I think thats right.