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If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ and $\mathbf{d}$ are vectors in which $|\mathbf{d}|=1$ and given $\mathbf{a}+\mathbf{b}+\mathbf{c}=s \mathbf{d}, \mathbf{b}+\mathbf{c}+\mathbf{d}=\mathbf{a}$, $\mathbf{a} \cdot \mathbf{d}=4$, then $s$ is equal to
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2444 Upvotes
Verified Answer
The correct answer is:
7
Given, $\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}$ are vector and
$$
\begin{aligned}
|\mathbf{d}| & =1, \mathbf{a}+\mathbf{b}+\mathbf{c}=s \mathbf{d} \\
\mathbf{b}+\mathbf{c}+\mathbf{d} & =\mathbf{a}, \cdot \mathbf{a} \cdot \mathbf{d}=4
\end{aligned}
$$
Using $\mathbf{b}+\mathbf{c}=\mathbf{a}-\mathbf{d}$ in $\mathbf{a}+\mathbf{b}+\mathbf{c}=s \mathbf{d}$, we obtain
$$
\begin{aligned}
\mathbf{a}+\mathbf{a}-\mathbf{d} & =s \mathbf{d} \Rightarrow 2 \mathbf{a}=(s+1) \mathbf{d} \\
\Rightarrow \quad \mathbf{a} & =\frac{(s+1)}{2} \mathbf{d}
\end{aligned}
$$
Given, $\mathbf{a} \cdot \mathbf{d}=4 \Rightarrow \frac{(s+1)}{2} \mathbf{d} \cdot \mathbf{d}=4$
$$
\begin{array}{ll}
\Rightarrow & \frac{(s+1)}{2}=4 \\
\Rightarrow & s=8-1=7
\end{array}
$$
$$
\begin{aligned}
|\mathbf{d}| & =1, \mathbf{a}+\mathbf{b}+\mathbf{c}=s \mathbf{d} \\
\mathbf{b}+\mathbf{c}+\mathbf{d} & =\mathbf{a}, \cdot \mathbf{a} \cdot \mathbf{d}=4
\end{aligned}
$$
Using $\mathbf{b}+\mathbf{c}=\mathbf{a}-\mathbf{d}$ in $\mathbf{a}+\mathbf{b}+\mathbf{c}=s \mathbf{d}$, we obtain
$$
\begin{aligned}
\mathbf{a}+\mathbf{a}-\mathbf{d} & =s \mathbf{d} \Rightarrow 2 \mathbf{a}=(s+1) \mathbf{d} \\
\Rightarrow \quad \mathbf{a} & =\frac{(s+1)}{2} \mathbf{d}
\end{aligned}
$$
Given, $\mathbf{a} \cdot \mathbf{d}=4 \Rightarrow \frac{(s+1)}{2} \mathbf{d} \cdot \mathbf{d}=4$
$$
\begin{array}{ll}
\Rightarrow & \frac{(s+1)}{2}=4 \\
\Rightarrow & s=8-1=7
\end{array}
$$
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