Find p so a vector expression is horizontal

Key takeaway: A vector parallel to the x-axis must have no vertical component. That means the coefficient of $\mathbf j$ has to vanish after combining the vectors.

We have

$\mathbf a=2\mathbf i-3\mathbf j=(2,-3),$ $\mathbf b=2\mathbf i-6\mathbf j=(2,-6),$ $\mathbf c=-\mathbf i+4\mathbf j=(-1,4).$

We want $\mathbf a+p\mathbf b+\mathbf c$ to be parallel to the x-axis. That means its $y$-component must be $0$.

Step 1: Write the y-component

The $y$-component is

$-3+p(-6)+4.$

Step 2: Set it equal to zero

$-3-6p+4=0$

$1-6p=0$

$6p=1$

$p=\frac16.$

So the required value is

$\boxed{\frac16}.$

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Pitfalls the pros know 👇 People often check the x-component first, but for parallel to the x-axis the x-component can be anything. The only thing that matters is that the y-component becomes zero.

What if the problem changes? If the vector had to be parallel to the y-axis instead, you would set the x-component equal to zero. Using the same vectors, that would produce a different value of $p$.

Tags: x-axis, components, vector equation