1. ## Vector problem

HI
Can someone tell me how would i complete the following question:

If $a=2i-3j+k, b=2i-4j+5k$and $c=-i-4j+2k$find the values of p and q such that a+pb+qc is parallel to the x-axis.

This is what i just started:
$2i-3j+k+p(2i-4j+5k)+q(-i-4j+2k)
$

P.S

2. Originally Posted by Paymemoney
HI
Can someone tell me how would i complete the following question:

If $a=2i-3j+k, b=2i-4j+5k$and $c=-i-4j+2k$find the values of p and q such that a+pb+qc is parallel to the x-axis.

This is what i just started:
$2i-3j+k+p(2i-4j+5k)+q(-i-4j+2k)
$

P.S
If the resultant vector is parallel to the $x$ axis, then its $\mathbf{j}$ and $\mathbf{k}$ components are 0.

3. the answers says $p=\frac{1}{6}$ and $q=\frac{-11}{12}$

4. Do what I said.

Collect the $\mathbf{i},\mathbf{j}$ and $\mathbf{k}$ terms.

Set the $\mathbf{j}$ and $\mathbf{k}$ terms to 0.

Solve the resulting equations.

5. ok i'll tried that

6. I still cannot get it can someone tell me the working to this.

7. continuing from where u left off

$
2\mathbf{i}-3\mathbf{j}+\mathbf{k}+p(2\mathbf{i}-4\mathbf{j}+5\mathbf{k})+q(-\mathbf{i}-4\mathbf{j}+2\mathbf{k})=(2+2p-2q)\mathbf{i}-(3+4p+4q)\mathbf{j}+(1+5p+2q)\mathbf{k}
$

as stated in the question, the vector a+pb+qc is parallel to the x-axis

therefore, the $\mathbf{j}$ and $\mathbf{k}$ component are null
using that result;

$3+4p+4q = 0$ --> (1)
$1+5p+2q = 0$ --> (2)

(2) multiply by 2

$2+10p+4q=0$ --> (2a)

(2a)-(1);

$-1+6p = 0$
$p = \frac{1}{6}$

substitute $p = \frac{1}{6}$ into (1)
$q = \frac{-11}{12}$