Vectors 2 Question 8
8. If $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ are vectors such that $|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|=\sqrt{29}$ and $\overrightarrow{\mathbf{a}} \times(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})=(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \times \overrightarrow{\mathbf{b}}$, then a possible value of $(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot(-7 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})$ is
(2012)
(a) 0
(b) 3
(c) 4
(d) 8
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Answer:
Correct Answer: 8. (b)
Solution:
- Plan If $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}}$
$$ \begin{aligned} & \Rightarrow \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}}=0 \Rightarrow \overrightarrow{\mathbf{a}} \times(\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}})=0 \\ & \text { i.e. } \quad \overrightarrow{\mathbf{a}} |(\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}}) \text { or } \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}}=\lambda \overrightarrow{\mathbf{a}} \end{aligned} $$
Here, $\overrightarrow{\mathbf{a}} \times(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})=(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \times \overrightarrow{\mathbf{b}}$
$\Rightarrow \overrightarrow{\mathbf{a}} \times(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})-(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \times \overrightarrow{\mathbf{b}}=0$
$\Rightarrow(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \times(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})=0$
$\Rightarrow \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}=\lambda(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})$
Since, $\quad|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|=\sqrt{29}$
$\Rightarrow \pm \lambda \sqrt{4+9+16}=\sqrt{29}$
$\Rightarrow \quad \lambda= \pm 1$
$\therefore \quad \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}= \pm(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})$
Now, $(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot(-7 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})= \pm(-14+6+12)= \pm 4$
