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Let $\mathbf{a}=-\hat{\mathbf{i}}-\hat{\mathbf{k}}, \mathbf{b}=-\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $\mathbf{c}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ be three given vectors. If $\mathbf{r}$ is a vector such that $\mathbf{r} \times \mathbf{b}=\mathbf{c} \times \mathbf{b}$ and $\mathbf{r} \cdot \mathbf{a}=0$, then the value of $\mathbf{r} \cdot \mathbf{b}$ is
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Verified Answer
The correct answer is:
9
$$
\begin{aligned}
& \mathbf{r} \times \mathbf{b}=\mathbf{c} \times \mathbf{b} \\
& \Rightarrow \quad(\mathbf{r}-\mathbf{c}) \times \mathbf{b}=0 \Rightarrow \mathbf{r}-\mathbf{c}+\lambda \mathbf{b} \\
& \text { or } \quad \mathbf{r}=\mathbf{c}+\lambda \mathbf{b} \\
&
\end{aligned}
$$
Given, $\mathbf{r} \cdot \mathbf{a}=0$, taking dot product with a for Eq. (i).
Now, $\mathbf{r} \cdot \mathbf{a}=\mathbf{a} \cdot \mathbf{c}+\lambda \mathbf{a} \cdot \mathbf{b}$
$\therefore \quad \lambda=\frac{-\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}} \quad[\because \vec{r} \cdot \vec{a}=0] \ldots$ (i)
From Eqs. (i) and (ii), we get
$$
\mathbf{r}=\mathbf{c}-\frac{\mathbf{a} \cdot \mathbf{c}}{\mathbf{a} \cdot \mathbf{b}} \mathbf{b}
$$
Taking dot with $\mathbf{b}$, we get
$$
\mathbf{r} \cdot \mathbf{b}=\mathbf{c} \cdot \mathbf{b}-\frac{\mathbf{a} \cdot \mathbf{c}}{\mathbf{a} \cdot \mathbf{b}}(\mathbf{b} \cdot \mathbf{b})
$$
where, $\left[\begin{array}{l}\mathbf{a}=-\hat{\mathbf{i}}-\hat{\mathbf{k}} \\ \mathbf{b}=-\hat{\mathbf{i}}+\hat{\mathbf{j}} \\ \mathbf{c}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\end{array}\right]$
$$
=(-1+2)-\frac{(-1-3)}{(1)}(1+1)=1+8=9
$$
\begin{aligned}
& \mathbf{r} \times \mathbf{b}=\mathbf{c} \times \mathbf{b} \\
& \Rightarrow \quad(\mathbf{r}-\mathbf{c}) \times \mathbf{b}=0 \Rightarrow \mathbf{r}-\mathbf{c}+\lambda \mathbf{b} \\
& \text { or } \quad \mathbf{r}=\mathbf{c}+\lambda \mathbf{b} \\
&
\end{aligned}
$$
Given, $\mathbf{r} \cdot \mathbf{a}=0$, taking dot product with a for Eq. (i).
Now, $\mathbf{r} \cdot \mathbf{a}=\mathbf{a} \cdot \mathbf{c}+\lambda \mathbf{a} \cdot \mathbf{b}$
$\therefore \quad \lambda=\frac{-\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}} \quad[\because \vec{r} \cdot \vec{a}=0] \ldots$ (i)
From Eqs. (i) and (ii), we get
$$
\mathbf{r}=\mathbf{c}-\frac{\mathbf{a} \cdot \mathbf{c}}{\mathbf{a} \cdot \mathbf{b}} \mathbf{b}
$$
Taking dot with $\mathbf{b}$, we get
$$
\mathbf{r} \cdot \mathbf{b}=\mathbf{c} \cdot \mathbf{b}-\frac{\mathbf{a} \cdot \mathbf{c}}{\mathbf{a} \cdot \mathbf{b}}(\mathbf{b} \cdot \mathbf{b})
$$
where, $\left[\begin{array}{l}\mathbf{a}=-\hat{\mathbf{i}}-\hat{\mathbf{k}} \\ \mathbf{b}=-\hat{\mathbf{i}}+\hat{\mathbf{j}} \\ \mathbf{c}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\end{array}\right]$
$$
=(-1+2)-\frac{(-1-3)}{(1)}(1+1)=1+8=9
$$
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