1. ## More Vector Questions

Hi
I need help on the following questions:
1)Given $a=i+2j+3k, b=2i-3j+2k$ and $c=i+j-k$, find the angle between b and c.
This is what i have done: $cos^{-1}\frac{2}{\sqrt17\sqrt1}$
=60'58'

2)Given $x=2i-j+k$ and $y=i+j+tk$ find the value of t if the angle between the two vectors is 60 degree

P.S

2. Originally Posted by Paymemoney
Hi
I need help on the following questions:
1)Given $a=i+2j+3k, b=2i-3j+2k$ and $c=i+j-k$, find the angle between b and c.
This is what i have done: $cos^{-1}\frac{2}{\sqrt17\sqrt1}$
=60'58'

2)Given $x=2i-j+k$ and $y=i+j+tk$ find the value of t if the angle between the two vectors is 60 degree

P.S
You have the wrong value for $\mid c \mid$

$\mid c \mid =\sqrt{1^2+1^2+(-1)^2}=\sqrt{3}$

Secondly, the dot product of $b\cdot c$ should be -3. So you need to find the inverse cosine $cos^{-1}\frac{-3}{\sqrt17\sqrt3}$ which was around 114.84 degrees on my calculator.

The dot product should have worked out like this:

$b \cdot c = (2)(1)+(-3)(1)+(2)(-1) =-3$

3. oops i c now

4. could someone answer my second question?

5. For the second problem, just start by writing what you know. The main equation you will need is:

$x \cdot y =\mid x\mid \mid y\mid cos(60^{\circ})$

The dot product is $x \cdot y = 2-1+t=1+t$

$cos(60^{\circ})=\frac{1}{2}$

So now the main equation becomes:

$1+t=\frac{\mid x\mid \mid y\mid}{2}$

You need the absolute values:

$\mid x \mid =2$

$\mid y \mid =\sqrt{2+t^2}$

Substituting into the main equation results in:

$1+t=\sqrt{2+t^2}$

Square both sides to obtain:

$1+2t+t^2=2+t^2$

$t=\frac{1}{2}$

6. ok thanks i'll tried that

For the second problem, just start by writing what you know. The main equation you will need is:

$x \cdot y =\mid x\mid \mid y\mid cos(60^{\circ})$

The dot product is $x \cdot y = 2-1+t=1+t$

$cos(60^{\circ})=\frac{1}{2}$

So now the main equation becomes:

$1+t=\frac{\mid x\mid \mid y\mid}{2}$

You need the absolute values:

$\mid x \mid =2$

$\mid y \mid =\sqrt{2+t^2}$

Substituting into the main equation results in:

$1+t=\sqrt{2+t^2}$

Square both sides to obtain:

$1+2t+t^2=2+t^2$

$t=\frac{1}{2}$
the book's answers says t=2 not a $\frac{1}{2}$

8. Originally Posted by Paymemoney
the book's answers says t=2 not a $\frac{1}{2}$
Looking back, I see that I calculated $\mid x \mid$ incorrectly. Sorry about that:

$\mid x\mid =\sqrt{2^2+(-1)^2+1^2}=\sqrt{6}$

I originally had this as 2. So now we have to solve:

$1+t=\frac{\sqrt{6}\sqrt{2+t^2}}{2}$

The solution t=2 works for this. Sorry about that error.

9. that's alright i get it now