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:
Solution:
First, calculate the projection vector $\vec{c}$.
$$
\begin{aligned}
|\vec{a}| &= \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3 \\
\vec{c} &= \left( \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \right) \vec{a} \\
\therefore \vec{c} &= \left( \frac{(\lambda)(1) + (0)(2) + (4)(2) }{9} \right) \vec{a}
\end{aligned}
$$
Given that $|\vec{a} + \vec{c}| = 7$:
$$
\begin{aligned}
\left| \vec{a} + \frac{\lambda + 8}{9} \vec{a} \right| &= 7 \\
\left| \left( 1 + \frac{\lambda + 8}{9} \right) \vec{a} \right| &= 7 \\
\left( \frac{9 + \lambda + 8}{9} \right) |\vec{a}| &= 7 \\
\left( \frac{\lambda + 17}{9} \right) (3) &= 7 \\
\frac{\lambda + 17}{3} &= 7 \\
\lambda + 17 &= 21 \\
\therefore \lambda &= 4
\end{aligned}
$$
Now calculate the Area of the parallelogram formed by $\vec{b}$ and $\vec{c}$.
Substitute $\lambda = 4$ in vectors:
$$
\begin{aligned}
\vec{b} &= 4\hat{i} + 4\hat{k} \\
\vec{c} &= \frac{4+8}{9}\vec{a} = \frac{12}{9}\vec{a} = \frac{4}{3}\vec{a}
\end{aligned}
$$
Area $= |\vec{b} \times \vec{c}|$
$$
\begin{aligned}
\text{Area} &= \left| \vec{b} \times \left( \frac{4}{3}\vec{a} \right) \right| \\
&= \frac{4}{3} |\vec{b} \times \vec{a}|
\end{aligned}
$$
Calculating $\vec{b} \times \vec{a}$:
$$
\begin{aligned}
\vec{b} \times \vec{a} &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 0 & 4 \\ 1 & 2 & 2 \end{vmatrix} \\
&= \hat{i}(0 – 8) – \hat{j}(8 – 4) + \hat{k}(8 – 0) \\
&= -8\hat{i} – 4\hat{j} + 8\hat{k}
\end{aligned}
$$
Magnitude:
$$
\begin{aligned}
|\vec{b} \times \vec{a}| &= \sqrt{(-8)^2 + (-4)^2 + (8)^2} \\
&= \sqrt{64 + 16 + 64} \\
&= \sqrt{144} \\
&= 12
\end{aligned}
$$
Final Area:
$$
\begin{aligned}
\text{Area} &= \frac{4}{3} (12) \\
&= 16
\end{aligned}
$$
Ans. (16)