21. $$ \boldsymbol{u}=\boldsymbol{i}-\mathbf{2}+\boldsymbol{j} $$, and $$ \boldsymbol{v}=2 \boldsymbol{i}+\boldsymbol{j}-\boldsymbol{k} $$ are vectors. Given that $$ \boldsymbol{w}=\lambda \boldsymbol{u}+\boldsymbol{\mu} \boldsymbol{v} $$ and that $$ w $$ is perpendicular to $$ \boldsymbol{u} $$, find the ratio of $$ \lambda $$ to $$ \mu $$. 22. Find the sum of the vectors $$ 2 \boldsymbol{i}+\boldsymbol{j}-\boldsymbol{k}, \boldsymbol{i}+3 \boldsymbol{j}+\boldsymbol{k}, 3 \boldsymbol{i}+2 \boldsymbol{j} $$. 23. If $$ (a+2) \boldsymbol{i}+(b-1) \boldsymbol{j} $$ and $$ (b-1) \boldsymbol{i}-a \boldsymbol{j} $$ are equal vectors, find the values of $$ a $$ and $$ b $$.

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UpStudy AI · Accepted Answer

**21. Finding the ratio $$\lambda:\mu$$ when $$\boldsymbol{w}=\lambda \boldsymbol{u}+\mu \boldsymbol{v}$$ is perpendicular to $$\boldsymbol{u}$$.**

We are given  
$$
\boldsymbol{u}=\boldsymbol{i}+\boldsymbol{j}-2\boldsymbol{k}, \quad \boldsymbol{v}=2\boldsymbol{i}+\boldsymbol{j}-\boldsymbol{k}.
$$

Since $$\boldsymbol{w}$$ is perpendicular to $$\boldsymbol{u}$$, we have
$$
\boldsymbol{w}\cdot \boldsymbol{u} = 0.
$$
Substitute $$\boldsymbol{w}=\lambda \boldsymbol{u}+\mu \boldsymbol{v}$$ to obtain:
$$
(\lambda \boldsymbol{u}+\mu \boldsymbol{v})\cdot \boldsymbol{u} = \lambda (\boldsymbol{u}\cdot \boldsymbol{u})+\mu (\boldsymbol{v}\cdot \boldsymbol{u})=0.
$$

First, calculate $$\boldsymbol{u}\cdot \boldsymbol{u}$$:
$$
\boldsymbol{u}\cdot \boldsymbol{u} = 1^2+1^2+(-2)^2 = 1+1+4=6.
$$

Next, calculate $$\boldsymbol{v}\cdot \boldsymbol{u}$$:
$$
\boldsymbol{v}\cdot \boldsymbol{u} = (2)(1) + (1)(1) + (-1)(-2) = 2+1+2=5.
$$

So the perpendicularity condition becomes:
$$
6\lambda+5\mu=0.
$$

Solve for $$\lambda$$ in terms of $$\mu$$:
$$
\lambda = -\frac{5}{6}\mu.
$$
Thus, the ratio of $$\lambda$$ to $$\mu$$ is
$$
\frac{\lambda}{\mu}=-\frac{5}{6}.
$$

---

**22. Finding the sum of the vectors $$2\boldsymbol{i}+\boldsymbol{j}-\boldsymbol{k}$$, $$\boldsymbol{i}+3\boldsymbol{j}+\boldsymbol{k}$$, and $$3\boldsymbol{i}+2\boldsymbol{j}$$.**

Write each vector in component form:
$$
\text{First vector: } (2,\,1,\,-1), \quad
\text{Second vector: } (1,\,3,\,1), \quad
\text{Third vector: } (3,\,2,\,0).
$$

Add the corresponding components:
$$
i\text{-component: } 2+1+3 = 6,
$$
$$
j\text{-component: } 1+3+2 = 6,
$$
$$
k\text{-component: } -1+1+0 = 0.
$$

Thus, the sum is:
$$
6\boldsymbol{i}+6\boldsymbol{j}.
$$

---

**23. Finding $$a$$ and $$b$$ such that $$(a+2)\boldsymbol{i}+(b-1)\boldsymbol{j}$$ equals $$(b-1)\boldsymbol{i}-a\boldsymbol{j}$$.**

Since the vectors are equal, their corresponding components must be equal. Equate the $$i$$-components and $$j$$-components:

For the $$i$$-components:
$$
a+2 = b-1 \quad \Rightarrow \quad b = a+3.
$$

For the $$j$$-components:
$$
b-1 = -a \quad \Rightarrow \quad b = 1-a.
$$

Setting the two expressions for $$b$$ equal:
$$
a+3 = 1-a.
$$
Solve for $$a$$:
$$
a+3 = 1-a \quad \Rightarrow \quad 2a = -2 \quad \Rightarrow \quad a = -1.
$$

Now substitute back to find $$b$$:
$$
b = 1 - (-1) = 2.
$$

---

$$\boxed{
\begin{aligned}
&\textbf{21.}\quad \frac{\lambda}{\mu}=-\frac{5}{6},\[1mm]
&\textbf{22.}\quad 6\boldsymbol{i}+6\boldsymbol{j},\[1mm]
&\textbf{23.}\quad a=-1,\quad b=2.
\end{aligned}
}$$
**21.** The ratio of $$\lambda$$ to $$\mu$$ is $$-\frac{5}{6}$$. **22.** The sum of the vectors is $$6\boldsymbol{i} + 6\boldsymbol{j}$$. **23.** The values are $$a = -1$$ and $$b = 2$$.

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