Evaluate the Magnitude of Vector y

To determine the value of $|\mathbf{y}|$, we proceed step-by-step using vector algebra.

Step 1: Express Vector $\mathbf{y}$ in Terms of Given Vectors

Let the given vectors be: $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ $\vec{c} = \hat{i} - \hat{j} - \hat{k}$

Since vector $\mathbf{y}$ lies in the plane of $\vec{a}$ and $\vec{b}$, it can be represented as a linear combination of $\vec{a}$ and $\vec{b}$: $\mathbf{y} = x\vec{a} + y\vec{b} \quad \text{for } x, y \in \mathbb{R}$

---

Step 2: Use the Projection Condition

We are given that the projection of $\mathbf{y}$ on $\vec{c}$ is $\frac{1}{\sqrt{3}}$: $\text{Projection of } \mathbf{y} \text{ on } \vec{c} = \frac{\mathbf{y} \cdot \vec{c}}{|\vec{c}|} = \frac{1}{\sqrt{3}}$

First, evaluate the magnitude of $\vec{c}$: $|\vec{c}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3}$

Substituting $|\vec{c}| = \sqrt{3}$ into the projection equation: $\frac{\mathbf{y} \cdot \vec{c}}{\sqrt{3}} = \frac{1}{\sqrt{3}} \implies \mathbf{y} \cdot \vec{c} = 1$

---

Step 3: Relate $x$ and $y$

Substitute $\mathbf{y} = x\vec{a} + y\vec{b}$ into $\mathbf{y} \cdot \vec{c} = 1$: $(x\vec{a} + y\vec{b}) \cdot \vec{c} = 1 \implies x(\vec{a} \cdot \vec{c}) + y(\vec{b} \cdot \vec{c}) = 1$

Compute the required dot products: $\vec{a} \cdot \vec{c} = (1)(1) + (1)(-1) + (1)(-1) = 1 - 1 - 1 = -1$ $\vec{b} \cdot \vec{c} = (1)(1) + (-1)(-1) + (1)(-1) = 1 + 1 - 1 = 1$

Substitute these values back: $-x + y = 1 \implies y = x + 1$

Thus, vector $\mathbf{y}$ can be written in terms of $x$ as: $\mathbf{y} = x\vec{a} + (x + 1)\vec{b}$

---

Step 4: Express $|\mathbf{y}|^2$ in Terms of $x$

Now, calculate the magnitude squared of $\mathbf{y}$: $|\mathbf{y}|^2 = |x\vec{a} + (x + 1)\vec{b}|^2 = x^2 |\vec{a}|^2 + (x + 1)^2 |\vec{b}|^2 + 2x(x + 1)(\vec{a} \cdot \vec{b})$

Compute the required norms and dot product: $|\vec{a}|^2 = 1^2 + 1^2 + 1^2 = 3$ $|\vec{b}|^2 = 1^2 + (-1)^2 + 1^2 = 3$ $\vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) + (1)(1) = 1$

Substituting these values: $|\mathbf{y}|^2 = 3x^2 + 3(x + 1)^2 + 2x(x + 1)(1)$ $|\mathbf{y}|^2 = 3x^2 + 3(x^2 + 2x + 1) + 2(x^2 + x)$ $|\mathbf{y}|^2 = 3x^2 + 3x^2 + 6x + 3 + 2x^2 + 2x$ $|\mathbf{y}|^2 = 8x^2 + 8x + 3$

---

Step 5: Minimize $|\mathbf{y}|^2$ to Find the Value of $|\mathbf{y}|$

To find the minimum value of $|\mathbf{y}|^2$, complete the square: $|\mathbf{y}|^2 = 8\left(x^2 + x\right) + 3$ $|\mathbf{y}|^2 = 8\left(x + \frac{1}{2}\right)^2 - 8\left(\frac{1}{4}\right) + 3$ $|\mathbf{y}|^2 = 8\left(x + \frac{1}{2}\right)^2 + 1$

Since $8\left(x + \frac{1}{2}\right)^2 \ge 0$, the minimum occurs when $x = -\frac{1}{2}$: $|\mathbf{y}|^2_{\text{min}} = 1$

Taking the square root: $|\mathbf{y}| = 1$