mathematics-class-12-unit-10-chapter-07-vectors-n98z7fis9ma

### <Success style="font-size:25px"> Vectors L - 7 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ \overrightarrow{A B} \cdot \overrightarrow{A D}=\overrightarrow{A B}|| \overrightarrow{A D} \mid \cos (\eta)$
- $\cos \eta=\frac{\overrightarrow{A B} \cdot \overrightarrow{A D}}{|\overrightarrow{A B}||\overrightarrow{A D}|}=\frac{40}{\sqrt{225} \sqrt{9}}=\frac{40}{45} $
- $ =\frac{8}{9} $

- $\boxed{\cos \eta=\frac{8}{9}} , \quad \cos ^2 \eta+\sin ^2 \eta=1\ $

- $ \eta+\theta=90^{\circ} $
- $\cos (\theta)=\cos \left(90^{\circ}-\eta\right)=\sin (\eta)=\frac{\sqrt{17}}{9} $

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<footer style = "font-size: 18px">Vectors lecture-7 $\rightarrow$ Problem-1 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-2 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Vectors L - 7 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- If $A,B \text{and} C$ are collinear then $\overrightarrow{A B} = k \overrightarrow{A C}$   
- $k=1 \quad B ~ \text{and} ~ C$ are same. (Coincident)
- $ k \neq 1, \quad \overrightarrow{A B}=k \overrightarrow{A C} $
- $ (1-x)=-k(1+x) $
- $ (z-y)=-k(1+y) $
- $ +z=+k z \Rightarrow z=0 $

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- $ +y=+k(1+y) \Rightarrow y-k y=k $
- $\Rightarrow y=\frac{k}{1-k} \text {. }$
- $ \Rightarrow \boxed{k=\frac{y}{1+y}} $
- $ \boxed{k=\frac{x-1}{1+x}} $
- $ \boxed{\frac{x-1}{x+1}=\frac{y}{1+y}} $
- $ \boxed{x=2y+1} \quad \text{and} \boxed{z=0} \$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-2 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-3 $\rightarrow$ Problem-4 </footer>

### <Success style="font-size:25px"> Vectors L - 7 </Success>
### <primary style="font-size:24px"> Problem-3 </primary>
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- Find the cosine of the angle between vectors $\vec{p} \\& \vec{q}$ such that

- $ 2 \vec{p}+\vec{q}=i+j \quad \\& \vec{p}+2 \vec{q}=i-j . $

- Solution:
- $ 3 \vec{p}+3 \vec{q}=2 i \quad \vec{p}-\vec{q}=2 j $

- $ \vec{p}+\vec{q}=\frac{2}{3} i$ and $ 2 \vec{p}=\frac{2}{3} i+2 j \quad 2 \vec{q}=\frac{2}{3} i-2 j $

- $ 4 \vec{p} \cdot \vec{q}=\frac{4}{9}-4=\frac{-32}{9} \Rightarrow \vec{p} \cdot \vec{q}=\frac{-8}{9}$

- $|\vec{p}|^2=\frac{4}{36} \times 10  |\vec{p}|=\frac{\sqrt{10}}{3}$
- $|\vec{q}|^2=\frac{10}{q}  |\vec{q}|=\frac{\sqrt{10}}{3} $
- $\cos \theta =\frac{\vec{p} \cdot \vec{q}}{|\vec{p}||\vec{q}|}=\frac{-81 d}{10 / x}=\frac{-4}{5} .$
- $\theta  =\cos ^{-1}\left(\frac{-4}{5}\right)$

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<footer style = "font-size: 18px">Problem-2 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-3 </span> $\rightarrow$ Problem-4 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Vectors L - 7 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ \cos \theta<0, \quad|\cos \theta| \leqslant 1$

- $\frac{c x^2-12+6 c x}{\sqrt{\left(1+4 c^2\right) x^2+4} \sqrt{c^2 x^2+45}}<0, \forall x \in \mathbb{R}$ .

- $c x^2+6 c x-12<0 $

- $ c < 0 ~ \text{and} ~ (6 c)^2-4 c(-12) < 0 $
- $\Leftrightarrow  c < 0 ~ \text{and} ~ 36 c^2+48 c <0 $
- $\Leftrightarrow  c < 0 ~ \text{and} ~  12 c(3 c+4) <0$

- $\Leftrightarrow  c < 0 ~ \text{and} ~  3 c+4>0 $
- $\Leftrightarrow  c < 0 ~ \text{and} ~ c > -\frac{4}{3} $
- $\Leftrightarrow  \frac{4}{3} < c < 0 .$

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<footer style = "font-size: 18px">Problem-3 $\rightarrow$ Problem-4 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-5 $\rightarrow$ Problem-6 </footer>

### <Success style="font-size:25px"> Vectors L - 7 </Success>
### <primary style="font-size:24px"> Problem-5 </primary>
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- If $\vec{a} ~\\& ~ \vec{b}$ are two unit vectors such that $\vec{a}+2 \vec{b}  ~\\& ~ 5 \vec{a}-4 \vec{b}$ are perpendicular to each other then find the angle between $\vec{a} ~ \\& ~ \vec{b}$.

- Solution:

- $ (\vec{a}+2 \vec{b}) \cdot(5 \vec{a}-4 \vec{b})=0 .  |\vec{a}|=1-|\vec{b}| $
- $\Leftrightarrow  5 \vec{a} \cdot \vec{a}-8 \vec{b} \cdot \vec{b}+(10-4) \vec{a} \cdot \vec{b}=0  \vec{a} \cdot \vec{a}=1 $
- $\Leftrightarrow  5-8+6 \vec{a} \cdot \vec{b}=0 $
- $\Leftrightarrow  \vec{a} \cdot \vec{b}=\frac{3}{6}=\frac{1}{2} $
- $\Leftrightarrow  \frac{1}{\left|\vec{a}\right|}\frac{1}{\left|\vec{b}\right|} \cos \theta=\frac{1}{2} $
- $\Leftrightarrow  \theta=\cos ^{-1}(1 / 2)=60^{\circ} $

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<footer style = "font-size: 18px">Problem-4 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-5 </span> $\rightarrow$ Problem-6 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Vectors L - 7 </Success>
### <primary style="font-size:24px"> Problem-6 </primary>
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- If $ \vec{a}, \vec{b} ~ \\& ~ \vec{c}$ are unit vectors, then find $ |\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2 \leqslant$ ?

- Solution:

- $ 0 \leqslant|\vec{a}-\vec{b}|^2=(\vec{a}-\vec{b}) \cdot(\vec{a}-\vec{b}) $
- $ =|\vec{a}|^2+|\vec{b}|^2-2 \vec{a} \cdot \vec{b} $
- $ =2(1-\vec{a} \cdot \vec{b}) $
- $ 1-\vec{a} \cdot \vec{b} \geq 0 \quad-1 \leqslant \vec{a} \cdot \vec{b} \leqslant 1 $
- $ |\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2 $
- $ =2\left(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2\right)$ \\
- $ -2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) $
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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-5 $\rightarrow$ <span style = "color:blue"> Problem-6 </span> $\rightarrow$ Solution $\rightarrow$ Problem-7 </footer>

### <Success style="font-size:25px"> Vectors L - 7 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ =2(3)+2(3)=12 \text {. } $
- $ 0 \leq(\vec{a}+\vec{b}+\vec{c})^2=|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2 \vec{a} \cdot \vec{b}+2 \vec{b} \vec{c}+2 \vec{c} \cdot \vec{a}$
- $ =3+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) $

- $ 2(\vec{a} \cdot  \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \geqslant-3 $
- $ -2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \leqslant 3 $
- $ |\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2 \leqslant 2(3)+3=9 \text {. } $

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<footer style = "font-size: 18px">Problem-5 $\rightarrow$ Problem-6 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-7 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Vectors L - 7 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\vec{b}$ & $\vec{c}$ angle is $\pi / 6$

- $|\vec{b} \times \vec{c}|=|\vec{b}||\vec{c}| \sin \frac{\pi}{6} =\sin \frac{\pi}{6} = \frac{1}{2} $

- $ 1=|k| \frac{1}{2} $

- $ |k|=2 $

- $ k= \pm 2 $
 
- $\vec{a}= \pm 2(\vec{b} \times \vec{c}) .$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-7 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Thank you $\rightarrow$  </footer>