Let a = 2 i + j - 2 k and b = i + j . Let c be vector such that |… - Vidyadip

Question

Let $\vec a = 2\hat i + \hat j - 2\hat k$ and $\vec b = \hat i + \hat j$ . Let $\vec c$ be vector such that $\left| {\vec c - \vec a} \right| = 3,\;\left| {\left( {\vec a \times \vec b} \right) \times \vec c} \right| = 3$ and the angle between $\vec c$ and $\vec a \times \vec b$ be $30^\circ $ . Then $\vec a \cdot \vec c$ is equal to :

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Answer

c
$\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}, \quad \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}$

$\Rightarrow|\overrightarrow{\mathrm{a}}|=3$

$\therefore \vec{a} \times \vec{b}=2 \hat{i}-2 \hat{j}+\hat{k}$

$|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=\sqrt{2^{2}+2^{2}+1^{2}}=3$

We have $|(\vec{a} \times \vec{b}) \times \vec{c}|=|\vec{a} \times \vec{b} \| \vec{c}| \sin 30^{\circ}$

$\Rightarrow|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}|=3|\overrightarrow{\mathrm{c}}| \cdot \frac{1}{2}$

$\Rightarrow 3=3|\overrightarrow{\mathrm{c}}| \cdot \frac{1}{2}$

$\therefore|\overrightarrow{\mathrm{c}}|=2$

Now $|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|=3$

On squaring, we get

$\Rightarrow c^{2}+a^{2}-2 \vec{c} \cdot \vec{a}=9$

$\Rightarrow 4+9-2 \vec{a} \cdot \vec{c}=9$

$\Rightarrow \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=2$

[$\because $ $\vec c.\vec a = \vec a.\vec c$]

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