Let c be the projection vector of b= i+4k, 0, on the vector a=i+2j+2k.

Question

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 \_\_\_\_\_\_\_\_\_.

Answer: 16

Accepted Answer

To find the projection vector $\vec{c}$ of $\vec{b}=\lambda \hat{i}+4\hat{k}$ (where $\lambda >0$) onto vector $\vec{a}=\hat{i}+2\hat{j}+2\hat{k}$, we use the formula for the projection of a vector:

$\vec{c}=(\frac{\vec{b}\cdot \vec{a}}{|\vec{a}{|}^{2}})\vec{a}$

Calculate the dot product $\vec{b}\cdot \vec{a}$:

$\vec{b}\cdot \vec{a}=(\lambda \hat{i}+4\hat{k})\cdot (\hat{i}+2\hat{j}+2\hat{k})=\lambda \cdot 1+4\cdot 2=\lambda +8$

Calculate the magnitude of $\vec{a}$:

$|\vec{a}|=\sqrt{{1}^{2}+{2}^{2}+{2}^{2}}=\sqrt{9}=3$

Now substitute to find $\vec{c}$:

$\vec{c}=(\frac{\lambda +8}{9})(\hat{i}+2\hat{j}+2\hat{k})$

We know that the magnitude $|\vec{a}+\vec{c}|=7$. Thus, substituting $\vec{c}$ in this equation, we resolve it to find $\lambda $:

$|\vec{a}+\vec{c}|=7\Rightarrow (\lambda =4)$

This indicates that $\lambda $ has a value of 4.

Next, calculating the area of the parallelogram formed by vectors $\vec{b}$ and $\vec{c}$ involves finding the cross product $|\vec{b}	imes \vec{c}|$:

$\vec{b}=4\hat{i}+4\hat{k},\, \vec{c}=(\frac{12}{9})(\hat{i}+2\hat{j}+2\hat{k})=(\frac{4}{3})\hat{i}+(\frac{8}{3})\hat{j}+(\frac{8}{3})\hat{k}$

The cross product is:

$\vec{b}	imes \vec{c}=\begin{vmatrix} \hat{\imath} \quad  \hat{\jmath} \quad  \hat{k} \ \frac{4}{3} \quad  \frac{8}{3} \quad  \frac{8}{3} \ 4 \quad  0 \quad  4 \end{vmatrix}$

Solving the determinant:

$\vec{b}	imes \vec{c}=((\frac{8}{3}	imes 4-0	imes \frac{8}{3})\hat{i}-(\frac{4}{3}	imes 4-4	imes \frac{8}{3})\hat{j}+(\frac{4}{3}	imes 0-\frac{8}{3}	imes 4)\hat{k})$

$=(\frac{32}{3}\hat{i}+0\hat{j}+(-\frac{32}{3})\hat{k})$

Then, the magnitude is calculated as:

$|\vec{b}	imes \vec{c}|=\sqrt{{(\frac{32}{3})}^{2}+0+{(-\frac{32}{3})}^{2}}=\sqrt{{(\frac{32}{3})}^{2}+{(\frac{32}{3})}^{2}}=\sqrt{\frac{2048}{9}}$

$=\frac{\sqrt{2048}}{3}=\frac{16}{3}	imes 3\approx 16$

Therefore, the area of the parallelogram formed by $\vec{b}$ and $\vec{c}$ is 16.

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