Given two vectors $\vec{a} = \vec{i} - \vec{j}$ and $\vec{b} = -\vec{i} + p\vec{j} + 2\vec{k}$ in $R^3$. The length of the projection of vector $\vec{a}$ onto $\vec{b}$ is equal to 1. Find the value of $p$.

1. Problem Description

Given two vectors $\vec{a} = \vec{i} - \vec{j}$ and $\vec{b} = -\vec{i} + p\vec{j} + 2\vec{k}$ in $R^3$. The length of the projection of vector $\vec{a}$ onto $\vec{b}$ is equal to

1. Find the value of $p$.

2. Solution Steps

The formula for the projection of vector $\vec{a}$ onto vector $\vec{b}$ is given by:

$proj_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b}$

The length of the projection is given by:

$|proj_{\vec{b}} \vec{a}| = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{b}|}$

Given that the length of the projection of $\vec{a}$ onto $\vec{b}$ is

1. $|\vec{a} \cdot \vec{b}| = |\vec{b}|$

We have $\vec{a} = \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}$ and $\vec{b} = \begin{pmatrix} -1 \\ p \\ 2 \end{pmatrix}$.

The dot product $\vec{a} \cdot \vec{b}$ is:

$\vec{a} \cdot \vec{b} = (1)(-1) + (-1)(p) + (0)(2) = -1 - p$

The magnitude of $\vec{b}$ is:

$|\vec{b}| = \sqrt{(-1)^2 + p^2 + 2^2} = \sqrt{1 + p^2 + 4} = \sqrt{p^2 + 5}$

Since $|proj_{\vec{b}} \vec{a}| = 1$, we have:

$\frac{|\vec{a} \cdot \vec{b}|}{|\vec{b}|} = 1$

$|-1 - p| = \sqrt{p^2 + 5}$

$(-1-p)^2 = p^2 + 5$

$(1+p)^2 = p^2 + 5$

$1 + 2p + p^2 = p^2 + 5$

$2p = 4$

$p = 2$

3. Final Answer

The value of $p$ is

2.