**25.** Let \( \vec{c} \) be the projection vector of \( \vec{b} = \lambda \hat{i} + 4 \hat{k} \), \( \lambda > 0 \), on the vector \( \vec{a} = \hat{i} + 2 \hat{j} + 2 \hat{k} \). If \( |\vec{a} + \vec{c}| = 7 \), then the area of the parallelogram formed by the vectors \( \vec{b} \) and \( \vec{c} \) is _____.

Key Concepts

1. Projection of a Vector
   The projection of b on a is the part of b that lies along a:

   $$\vec c = \text{proj}_{\vec a}\,\vec b = \frac{\vec b \cdot \vec a}{|\vec a|^2}\;\vec a$$

2. Adding Vectors and Using the Magnitude Condition
   We know $|\vec a + \vec c| = 7$. Because $\vec c$ is already along $\vec a$, the sum $\vec a + \vec c$ will also be along $\vec a$. That makes the magnitude calculation very direct.

3. Area of a Parallelogram in 3-D
   If two adjacent sides are $\vec p$ and $\vec q$, the area is

   $$|\vec p \times \vec q|$$

   The cross product gives a vector whose length measures exactly that area.

Logical Chain to Solve

1. Write b and a in component form.
2. Compute $\vec b \cdot \vec a$ and $|\vec a|^2$.
3. Obtain $\vec c$ using the projection formula.
4. Write $\vec a + \vec c$, take its magnitude, and equate it to 7 to solve for $\lambda$.
5. Substitute $\lambda$ back to get explicit vectors b and c.
6. Find $\vec b \times \vec c$ and its magnitude; that magnitude is the required area.

🤔 What’s going on?

Imagine two arrows:

1. Arrow b points a bit forward (x-direction) and straight up (z-direction).
2. Arrow a points forward, a little to the right, and up.

We first drop b onto a like a shadow. That shadow is another arrow called c (the projection).
Then we glue a and that shadow c together tip-to-tail. Their total length must be exactly 7 units. Using this clue we find how long b really points forward (that missing number λ).
Finally, we stretch a rubber sheet with sides b and c. Its area is just the size of the 3-D parallelogram. We calculate that area = 16 square units.

Step-by-Step Solution

1. Given vectors
   $\vec a = \hat i + 2\hat j + 2\hat k$ $\vec b = \lambda \hat i + 4\hat k \quad (\lambda > 0)$

2. Projection of (\vec b) on (\vec a)
   $\vec b \cdot \vec a = \lambda(1) + 0(2) + 4(2) = \lambda + 8$ $|\vec a|^2 = 1^2 + 2^2 + 2^2 = 9$ $\boxed{\vec c = \dfrac{\lambda + 8}{9}\,\vec a}$

3. Vector (\vec a + \vec c)

   = \left(1 + \dfrac{\lambda + 8}{9}\right)\vec a = \dfrac{\lambda + 17}{9}\,\vec a$$

4. Use the magnitude condition
   $|\vec a + \vec c| = \left|\dfrac{\lambda + 17}{9}\right||\vec a| = \dfrac{\lambda + 17}{9}\times 3 = \dfrac{\lambda + 17}{3}$ Given that this equals 7: $\dfrac{\lambda + 17}{3} = 7 \;\;\Longrightarrow\;\; \lambda + 17 = 21 \;\;\Longrightarrow\;\; \boxed{\lambda = 4}$

5. Explicit vectors with (\lambda = 4)
   $\vec b = 4\hat i + 4\hat k$

   = \dfrac{4}{3}\hat i + \dfrac{8}{3}\hat j + \dfrac{8}{3}\hat k$$

6. Area of the parallelogram
   Compute the cross product:

   \hat i & \hat j & \hat k\\ 4 & 0 & 4\\ 4/3 & 8/3 & 8/3 \end{vmatrix} = \left(-\dfrac{32}{3}\right)\hat i - \left(\dfrac{16}{3}\right)\hat j + \left(\dfrac{32}{3}\right)\hat k$$ Magnitude: $$|\vec b \times \vec c| = \sqrt{\left(-\dfrac{32}{3}\right)^2 + \left(-\dfrac{16}{3}\right)^2 + \left(\dfrac{32}{3}\right)^2} = \dfrac{1}{3}\sqrt{1024 + 256 + 1024} = \dfrac{1}{3}\sqrt{2304} = \dfrac{1}{3}\times 48 = 16$$ $$\boxed{\text{Area} = 16}$$