If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ and $\mathbf{d}$ are vectors in which $|\mathbf{d}|=1$ and given…

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
  1. 7
  2. 8
  3. -1
  4. 4

Solution

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} $

Asked in: AP EAMCET 2021 (23 Aug Shift 1)

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