Let $\vec{c}$ be the projection vector of $\vec{b}=\lambda \hat{i}+4 \hat{k}, \lambda \gt 0$, 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 ________

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Aicharya · Accepted Answer

$\vec{c}=\left(\frac{\vec{b} \cdot \vec{a}}{|\vec{b}|}\right) \frac{\vec{a}}{|\vec{a}|}$
$\begin{aligned} & \quad=\left(\frac{\lambda+8}{9}\right)(\hat{i}+2 \hat{j}+2 \hat{k}) \\ & |\vec{a}+\vec{c}|=7 \\ & \left.\Rightarrow\left(\frac{\lambda+8}{9}+1\right) \hat{i}+\left(\frac{2(\lambda+8)}{9}+2\right) \hat{j}+\left(\frac{2(\lambda+8)}{9}+2\right) \hat{k} \right\rvert\,=7 \\ & \left(\frac{\lambda+8}{9}+1\right)^2+\left(\frac{2(\lambda+8)}{9}+2\right)^2+\left(\frac{2(\lambda+8)}{9}+2\right)^2=49\end{aligned}$
$\Rightarrow \lambda=4 \Rightarrow \vec{c}=\frac{4}{3} \hat{i}+\frac{8}{3} \hat{j}+\frac{8}{3} \hat{k}$
Area of parallelogram $=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ \frac{4}{3} & \frac{8}{3} & \frac{8}{3} \\ 4 & 0 & 4\end{array}\right|=16$

Let $\vec{c}$ be the projection vector of $\vec{b}=\lambda \hat{i}+4 \hat{k}, \lambda \gt 0$, on the vector…

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