Let a=a_i i+a_2 j+a_3 k and b=b_1 i+b_2 j+b_3 k be two vectors su… - Vidyadip

MCQ

Let $\vec{a}=a_i \hat{i}+a_2 \hat{j}+a_3 \hat{k}$ and $\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}$ be two vectors such that $|\vec{a}|=1 ; \quad \vec{a} \cdot \vec{b}=2$ and $|\vec{b}|=4$. If $\vec{c}=2(\vec{a} \times \vec{b})-3 \vec{b}$, then the angle between $\vec{b}$ and $\vec{c}$ is equal to :

A
$\cos ^{-1}\left(\frac{2}{\sqrt{3}}\right)$
B
$\cos ^{-1}\left(-\frac{1}{\sqrt{3}}\right)$
✓
$\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$
D
$\cos ^{-1}\left(\frac{2}{3}\right)$

✓

Answer

Correct option: C.

$\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$

c
Given $|\vec{a}|=1,|\vec{b}|=4, \vec{a} \cdot \vec{b}=2$

$\vec{c}=2(\vec{a} \times \vec{b})-3 \vec{b}$

Dot product with $\overrightarrow{\mathrm{a}}$ on both sides

$\overrightarrow{\mathrm{c}} . \overrightarrow{\mathrm{a}}=-6$

Dot product with $\vec{b}$ on both sides

$ \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}=-48 $

$\overrightarrow{\mathrm{c}} \overrightarrow{\mathrm{c}}=4|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|^2+9|\overrightarrow{\mathrm{b}}|^2 $

$ |\overrightarrow{\mathrm{c}}|^2=4\left[|\mathrm{a}|^2|\mathrm{~b}|^2-(\mathrm{a} \cdot \overrightarrow{\mathrm{b}})^2\right]+9|\overrightarrow{\mathrm{b}}|^2 $

$ |\overrightarrow{\mathrm{c}}|^2=4\left[(1)(4)^2-(4)\right]+9(16) $

$ |\overrightarrow{\mathrm{c}}|^2=4[12]+144 $

$ |\overrightarrow{\mathrm{c}}|^2=48+144 $

$ |\overrightarrow{\mathrm{c}}|^2=192 $

$ \therefore \cos \theta=\frac{\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}}{|\overrightarrow{\mathrm{b}}||\overrightarrow{\mathrm{c}}|} $

$ \therefore \cos \theta=\frac{-48}{\sqrt{192} \cdot 4} $

$ \therefore \cos \theta=\frac{-48}{8 \sqrt{3} \cdot 4} $

$ \therefore \cos \theta=\frac{-3}{2 \sqrt{3}} $

$ \therefore \cos \theta=\frac{-\sqrt{3}}{2} \Rightarrow \theta=\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$

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