# Math Help - Vector Question

1. ## Vector Question

Hi
i need help on the following questions:

1)if a=4i + 3j - k, b= 2i - 4j +3k, c = 2i +k find 1.5c - b +0.5a:
$=1.5(2i+k)-2i-4j+3k+0.5(4i+3j-k)$
$=3i+1.5k-2i-4j+3k+2i+1.5j-0.5k$
$=3i-2.5j-2k$

Now the answer says its 3i+(11/2)j-2k, however i don't understand how they got this.

2)In moving a crate of electrical fittings, a force F of magnitude 84 newton is applied in the direction of b, where b=3i-2j+6k
If the create is moved point A(-2, -1, 1) to the point B(1,1,2) measured in meters, find the work done.

$W= F . \vec{AB}$

$\vec{AB} = \vec{OB} - \vec{OA}$
$=i+j+2k--2i-j+k$
$=3i+2j+k$

i problem i have is what is the force? is it 84 or is it b=3i-2j+6k?

3) The angle between x and y is arcos (4/21). Find the scalar t given that:
$x=6i+3j-2k$ and $y=-2i+tj-4k$

Lastly just a side question from vectors, how would covert 1.11m/s into minutes and seconds?

P.S

2. Originally Posted by Paymemoney
1)if a=4i + 3j - k, b= 2i - 4j +3k, c = 2i +k find 1.5c - b +0.5a:
$=1.5(2i+k)-2i-4j+3k+0.5(4i+3j-k)$
$=3i+1.5k-2i-4j+3k+2i+1.5j-0.5k$
$=3i-2.5j-2k$

Now the answer says its 3i+(11/2)j-2k, however i don't understand how they got this.
1. It's actually

$1.5(2\mathbf{i} + \mathbf{k}) - (2\mathbf{i}- 4\mathbf{j} + 3\mathbf{k}) + 0.5(4\mathbf{i} + 3\mathbf{j} - \mathbf{k})$

$= 3\mathbf{i} + 1.5\mathbf{k} -2\mathbf{i} \color{red}+ \color{black}4\mathbf{j} -3\mathbf{k} + 2\mathbf{i} + 1.5\mathbf{j} - 0.5\mathbf{k}$

$= 3\mathbf{i} + 5.5\mathbf{j} -2\mathbf{k}$

3. Originally Posted by Paymemoney
2)In moving a crate of electrical fittings, a force F of magnitude 84 newton is applied in the direction of b, where b=3i-2j+6k
If the create is moved point A(-2, -1, 1) to the point B(1,1,2) measured in meters, find the work done.

$W= F . \vec{AB}$

$\vec{AB} = \vec{OB} - \vec{OA}$
$=i+j+2k--2i-j+k$
$=3i+2j+k$

i problem i have is what is the force? is it 84 or is it b=3i-2j+6k?
The force is the vector that has magnitude $84$ and is in the direction of $\mathbf{b} = 3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}$.

So $F = k\mathbf{b}$.

We also know that $|F| = 84$.

$|F| = |k\mathbf{b}| = k|\mathbf{b}|$.

$|\mathbf{b}| = \sqrt{3^2 + (-2)^2 + 6^2}$

$= \sqrt{9 + 4 + 36}$

$= \sqrt{49}$

$= 7$.

Since $|F| = k|\mathbf{b}|$

$84 = 7k$

$k = 12$.

So that means

$F = 12\mathbf{b}$

$= 12(3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k})$

$= 36\mathbf{i} - 24\mathbf{j} + 72\mathbf{k}$.

4. Originally Posted by Paymemoney
3) The angle between x and y is arcos (4/21). Find the scalar t given that:
$x=6i+3j-2k$ and $y=-2i+tj-4k$

Lastly just a side question from vectors, how would covert 1.11m/s into minutes and seconds?

P.S
3. You should know that

$\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos{\theta}$, where $\theta$ is the angle between them.

If $\mathbf{x} = 6\mathbf{i} + 3\mathbf{j} -2\mathbf{k}$

then $|\mathbf{x}| = \sqrt{6^2 + 3^2 + (-2)^2}$

$= \sqrt{36 + 9+ 4}$

$= \sqrt{49}$

$= 7$.

If $\mathbf{y} = -2\mathbf{i} + t\mathbf{j} - 4\mathbf{k}$

then $|\mathbf{y}| = \sqrt{(-2)^2 + t^2 + (-4)^2}$

$= \sqrt{4 + t^2 + 16}$

$= \sqrt{t^2 + 20}$.

$\mathbf{x}\cdot \mathbf{y} = (6\mathbf{i} + 3\mathbf{j} -2\mathbf{k})\cdot (-2\mathbf{i} + t\mathbf{j} - 4\mathbf{k})$

$= -12 + 3t + 8$

$= 3t - 4$.

Since $\mathbf{x} \cdot \mathbf{y} = |\mathbf{x}||\mathbf{y}|\cos{\theta}$

$3t - 4 = 7\sqrt{t^2 + 20}\cos{\left(\arccos{\frac{4}{21}}\right)}$

$3t - 4 = 7\sqrt{t^2 + 20}\cdot \frac{4}{21}$

$3t - 4 = \frac{4}{3}\sqrt{t^2 + 20}$

$\frac{3}{4}(3t - 4) = \sqrt{t^2 + 20}$

$\frac{9}{4}t - 3 = \sqrt{t^2 + 20}$

$\left(\frac{9}{4}t - 3\right)^2 = t^2 + 20$

$\frac{81}{16}t^2 - \frac{27}{2}t + 9 = t^2 + 20$

$\frac{65}{16}t^2 - \frac{27}{2}t - 11 = 0$.

Solve for $t$.

5. Originally Posted by Paymemoney
Lastly just a side question from vectors, how would covert 1.11m/s into minutes and seconds?

P.S
Seeing as what you have posted is a unit of speed and not a unit of time, you can't convert it into minutes OR seconds.

6. Originally Posted by Prove It
The force is the vector that has magnitude $84$ and is in the direction of $\mathbf{b} = 3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}$.

So $F = k\mathbf{b}$.

We also know that $|F| = 84$.

$|F| = |k\mathbf{b}| = k|\mathbf{b}|$.

$|\mathbf{b}| = \sqrt{3^2 + (-2)^2 + 6^2}$

$= \sqrt{9 + 4 + 36}$

$= \sqrt{49}$

$= 7$.

Since $|F| = k|\mathbf{b}|$

$84 = 7k$

$k = 12$.

So that means

$F = 12\mathbf{b}$

$= 12(3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k})$

$= 36\mathbf{i} - 24\mathbf{j} + 72\mathbf{k}$.
i don't understand why wouldn't, 84 be the magnitude of $\mathbf{b} = 3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}$

7. Originally Posted by Paymemoney
i don't understand why wouldn't, 84 be the magnitude of $\mathbf{b} = 3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}$
I've shown you why. Use Pythagoras on it and work out the magnitude of $\mathbf{b}$. Its magnitude is $7$.

A vector has magnitude and direction. Like I said, $F$ is a vector with a magnitude of $84$ and in the DIRECTION of $\mathbf{b}$. $F$ DOES NOT EQUAL $\mathbf{b}$. It is a SCALAR MULTIPLE of $\mathbf{b}$. (In other words it goes in the same direction but has been magnified to a different length). You need to figure out the magnification factor so you can multiply it by $\mathbf{b}$ and thus FIND $F$.

8. ok, so what does the k mean in the forumla F=kb? And how did you get that forumla?

9. Originally Posted by Paymemoney
ok, so what does the k mean in the forumla F=kb? And how did you get that forumla?
Since $F$ is a scalar multiple of $\mathbf{b}$, this means that $F = k\mathbf{b}$, where $k$ is the scalar that you multiply $\mathbf{b}$ by...

10. Originally Posted by Prove It
3. You should know that

$\frac{81}{16}t^2 - \frac{27}{2}t + 9 = t^2 + 20$

$\frac{65}{16}t^2 - \frac{27}{2}t - 11 = 0$.

Solve for $t$.
when you do these lines wouldn't you square root both sides and the result would be:

$\sqrt{\frac{81}{16}t^2-\frac{27}{2}t-11} = t$
$\frac{9}{4}t^2-\sqrt{\frac{27}{2}t}-\sqrt{{11}}=0$

11. Originally Posted by Paymemoney
when you do these lines wouldn't you square root both sides and the result would be:

$\sqrt{\frac{81}{16}t^2-\frac{27}{2}t-11} = t$
$\frac{9}{4}t^2-\sqrt{\frac{27}{2}t}-\sqrt{{11}}=0$
No you definitely would not. What you end up with is a Quadratic Equation, so set it equal to 0 and use the Quadratic Formula.