Key Concepts and Formulas

• Projection of a Vector: The projection vector of $\vec b$ onto $\vec a$ is given by $\text{proj}_{\vec a} \vec b = \left( \frac{\vec b \cdot \vec a}{|\vec a|^2} \right) \vec a$.
• Condition for Perpendicular Vectors: Two non-zero vectors $\vec u$ and $\vec v$ are perpendicular if and only if their dot product is zero, i.e., $\vec u \cdot \vec v = 0$.
• Dot Product: For vectors $\vec u = u_1\widehat i + u_2\widehat j + u_3\widehat k$ and $\vec v = v_1\widehat i + v_2\widehat j + v_3\widehat k$, their dot product is $\vec u \cdot \vec v = u_1v_1 + u_2v_2 + u_3v_3$.
• Magnitude of a Vector: For a vector $\vec v = x\widehat i + y\widehat j + z\widehat k$, its magnitude is $|\vec v| = \sqrt{x^2 + y^2 + z^2}$.

Step-by-Step Solution

Step 1: Utilize the Projection Condition The problem states that the projection vector of $\overrightarrow b$ on $\overrightarrow a$ is $\overrightarrow a$. Applying the formula for the projection vector: $\text{proj}_{\vec a} \vec b = \left( \frac{\vec b \cdot \vec a}{|\vec a|^2} \right) \vec a$ Given that this is equal to $\overrightarrow a$: $\left( \frac{\vec b \cdot \vec a}{|\vec a|^2} \right) \vec a = \vec a$ Since $\overrightarrow a$ is a non-zero vector, we can equate the scalar coefficients: $\frac{\vec b \cdot \vec a}{|\vec a|^2} = 1$ This implies: $\vec b \cdot \vec a = |\vec a|^2$ Let's calculate $|\vec a|^2$: $|\vec a|^2 = (1)^2 + (1)^2 + (\sqrt 2)^2 = 1 + 1 + 2 = 4$ Now, let's calculate the dot product $\vec b \cdot \vec a$: $\vec b \cdot \vec a = (b_1)(1) + (b_2)(1) + (\sqrt 2)(\sqrt 2) = b_1 + b_2 + 2$ Equating the dot product with $|\vec a|^2$: $b_1 + b_2 + 2 = 4$ $b_1 + b_2 = 2 \quad \ldots (1)$ This equation provides the first relationship between the components $b_1$ and $b_2$.

Step 2: Utilize the Perpendicularity Condition The problem states that $\overrightarrow a + \overrightarrow b$ is perpendicular to $\overrightarrow c$. This means their dot product is zero: $(\overrightarrow a + \overrightarrow b) \cdot \overrightarrow c = 0$ First, let's find the vector $\overrightarrow a + \overrightarrow b$: $\overrightarrow a + \overrightarrow b = (\widehat i + \widehat j + \sqrt 2 \widehat k) + (b_1\widehat i + b_2\widehat j + \sqrt 2 \widehat k)$ $\overrightarrow a + \overrightarrow b = (1+b_1)\widehat i + (1+b_2)\widehat j + 2\sqrt 2 \widehat k$ Now, we compute the dot product with $\overrightarrow c = 5\widehat i + \widehat j + \sqrt 2 \widehat k$: $(\overrightarrow a + \overrightarrow b) \cdot \overrightarrow c = ((1+b_1)\widehat i + (1+b_2)\widehat j + 2\sqrt 2 \widehat k) \cdot (5\widehat i + \widehat j + \sqrt 2 \widehat k)$ $= (1+b_1)(5) + (1+b_2)(1) + (2\sqrt 2)(\sqrt 2)$ $= 5 + 5b_1 + 1 + b_2 + 4$ $= 5b_1 + b_2 + 10$ Since this dot product is zero: $5b_1 + b_2 + 10 = 0$ $5b_1 + b_2 = -10 \quad \ldots (2)$ This equation provides the second relationship between $b_1$ and $b_2$.

Step 3: Solve the System of Equations for $b_1$ and $b_2$ We have a system of two linear equations:

1. $b_1 + b_2 = 2$
2. $5b_1 + b_2 = -10$ Subtract equation (1) from equation (2): $(5b_1 + b_2) - (b_1 + b_2) = -10 - 2$ $4b_1 = -12$ $b_1 = -3$ Substitute $b_1 = -3$ into equation (1): $-3 + b_2 = 2$ $b_2 = 5$ Thus, the vector $\overrightarrow b$ is $\overrightarrow b = -3\widehat i + 5\widehat j + \sqrt 2 \widehat k$.

Step 4: Calculate the Magnitude of $\overrightarrow b$ We need to find $|\overrightarrow b|$: $|\overrightarrow b| = \sqrt{(b_1)^2 + (b_2)^2 + (\sqrt 2)^2}$ $|\overrightarrow b| = \sqrt{(-3)^2 + (5)^2 + (\sqrt 2)^2}$ $|\overrightarrow b| = \sqrt{9 + 25 + 2}$ $|\overrightarrow b| = \sqrt{36}$ $|\overrightarrow b| = 6$

Common Mistakes & Tips

• Misinterpreting Projection: Ensure you are using the correct formula for the projection vector, not just the scalar projection.
• Algebraic Errors: Carefully check your calculations when solving the system of linear equations for $b_1$ and $b_2$.
• Dot Product Calculation: Remember to multiply corresponding components and sum them up for the dot product. For the magnitude, square all components, sum them, and then take the square root.

Summary The problem was solved by first using the condition on the projection of $\overrightarrow b$ onto $\overrightarrow a$ to establish a linear relation between $b_1$ and $b_2$. Subsequently, the condition of perpendicularity between $\overrightarrow a + \overrightarrow b$ and $\overrightarrow c$ provided a second linear relation. Solving these two equations simultaneously yielded the values of $b_1$ and $b_2$. Finally, the magnitude of $\overrightarrow b$ was calculated using its components.

The final answer is $\boxed{6}$.