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

Show Answer

Answer:

Correct Answer: 8. (c)

Solution:

  1. 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, $(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot(-7 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})= \pm(-14+6+12)= \pm 4$



NCERT Chapter Video Solution

Dual Pane