### <Success style="font-size:25px"> Vectors L - 8 </Success> ### <primary style="font-size:24px"> Vectors <br> lecture-8 </primary> <hr> <main> <div style='float: left; width:50%; font-size:30px; text-align:left;'> </div> <div style='float: right; width:0%;'> <!-- (https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-12-unit-10-chapter-08-vectors-d0zjavv9f5k-01. --> </main> <hr> <footer style = "font-size: 18px"> $\rightarrow$ $\rightarrow$ <span style = "color:blue"> Vectors lecture-8 </span> $\rightarrow$ Problem-1 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Vectors L - 8 </Success> ### <primary style="font-size:24px"> Problem-1 </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px; text-align:left;'> - If $\vec{a}$, $\vec{b}$, $\vec{c}$ and $\vec{d}$ are unit vectors such that $(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d})=1$ \& $\vec{a} \cdot \vec{c}=1 / 2$ then - (P) $\vec{a}$, $\vec{b}$ and $\vec{c}$ are non-coplanar - (Q) $\vec{a}$, $\vec{b}$ and $\vec{d}$ are non-coplanar - (R) $\vec{b}$ and $\vec{d}$ are non-parallel - (S) $\vec{a}$ and $\vec{d}$ are parallel and $ \vec{b}$ and $\vec{c}$ are parallel. </div> </main> <hr> <footer style = "font-size: 18px"> $\rightarrow$ Vectors lecture-8 $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Vectors L - 8 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px; text-align:left;'> - $|\vec{a}|=|\vec{b}|=|\vec{c}|=|\vec{d}|=1$ - Let $\theta$ be the angle bet $(\vec{a} \times \vec{b})$ and $(\vec{c} \times \vec{d})$ - $ \theta_1$ be the angle between $\vec{a}$ and $\vec{b} $ - $ \theta_2$ be the angle between $\vec{c}$ and $\vec{d} $ - $ I=(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d})=|\vec{a} \times \vec{b}||\vec{c} \times \vec{d}| \cos \theta=|\vec{a}||\vec{b}| \sin \theta_1|\vec{c}||\vec{d}| \sin \theta_2 \cos \theta $ - $=\sin \theta_1 \sin \theta_2 \cos \theta $ </div> </main> <hr> <footer style = "font-size: 18px">Vectors lecture-8 $\rightarrow$ Problem-1 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Vectors L - 8 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px; text-align:left;'> - $\theta_1=\theta_2=\frac{\pi}{2}$ - $ \theta=0 $ $ \Rightarrow \vec{a} \perp \vec{b}, \vec{c} \perp \vec{d},(\vec{a} \times \vec{b}) \|(\vec{c} \times \vec{d})$ - $(\vec{a} \times \vec{b})=k(\vec{c} \times \vec{d})-\cdots * $ - $ (\vec{a} \times \vec{b}) \cdot \vec{c}=k(\vec{c} \times \vec{d}) \cdot \vec{c}=0 $ - $\Rightarrow[\vec{a} \vec{b} \vec{c}]=0 \Rightarrow$ $\vec{a}$, $\vec{b}$ and $\vec{c}$ one couple - $(\vec{a} \times \vec{b}) \cdot \vec{d}=k(\vec{c} \times \vec{d}) \cdot \vec{d}=0$ - $\Rightarrow[\vec{a} \vec{b} \vec{d}]=0 \Rightarrow$ $\vec{a}$, $\vec{b}$ and $\vec{d}$ one </div> </main> <hr> <footer style = "font-size: 18px">Problem-1 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Vectors L - 8 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px; text-align:left;'> - Suppose - $ \vec{b} \|| \vec{d} $ - $\Rightarrow \vec{b} =k \vec{d} $ - $ = \pm \vec{d}$ $(\vec{d}$ and $\vec{b}$ are unit vectors $)$ - $ 1=(\vec{a} \times \vec{b}) \cdot\left(\vec{c} \times d^2\right) \Rightarrow \pm 1=(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{b}) $ - $ \Rightarrow[\vec{a} \times \vec{b} \vec{c}, \vec{b}]= \pm 1=[\vec{c} \vec{b} \vec{a} \times \vec{b}] $ - $ \Rightarrow \vec{c} \cdot(\vec{b} \times(\vec{a} \times \vec{b}))= \pm 1 \Rightarrow \vec{c}(\vec{a}-(\vec{b} \cdot \vec{a}) \vec{b})= \pm 1 $ - $ \Rightarrow \vec{c} \cdot \vec{a}= \pm 1 \Rightarrow \leftarrow $ - $ \vec{a}\|\vec{a}$ and $\vec{b}\| \vec{c} $ </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-2 </footer>

### <Success style="font-size:25px"> Vectors L - 8 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px; text-align:left;'> - $ \vec{a} \cdot \vec{d}=1=\vec{b} \cdot \vec{c} $ - $ \vec{a}= \pm \vec{d}$ and $\vec{b}= \pm \vec{c} $ - $(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d})=(\vec{a} \times \vec{b})(\vec{b} \times \vec{a})= \pm 1 $ - $(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d})=(\vec{a} \times \vec{b})(\vec{b} \times \vec{a})= \pm 1$ - $(R)$ is correct </div> <div style='float: right; width:1%;'> <!-- (https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-12-unit-10-chapter-08-vectors-d0zjavv9f5k-02-A. --> <!-- (https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-12-unit-10-chapter-08-vectors-d0zjavv9f5k-02-B. --> <!-- (https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-12-unit-10-chapter-08-vectors-d0zjavv9f5k-02-C. --> <!-- (https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-12-unit-10-chapter-08-vectors-d0zjavv9f5k-02-D. --> <!-- (https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-12-unit-10-chapter-08-vectors-d0zjavv9f5k-02-E. --> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-2 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Vectors L - 8 </Success> ### <primary style="font-size:24px"> Problem-2 </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px; text-align:left;'> - Let $\vec{a}$, $\vec{b}$ and $\vec{c}$ are three non-coplanar vectors. - $(\vec{a}+\vec{b}+\vec{c}) \cdot[(\vec{a}+\vec{b}) \times(\vec{a}+\vec{c})]$ equals - (P) 0 - (Q) $[\vec{c} \quad \vec{b} \vec{c}]$ - (R) $2[\vec{c} \vec{b} \vec{c}]$ - (S) $-\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]$ </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-2 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Vectors L - 8 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px; text-align:left;'> - $ (\vec{a}+\vec{b}+\vec{c}) \cdot[(\vec{a}+\vec{b}) \times(\vec{a}+\vec{c})] $ - $ =(\vec{a}+\vec{b}+\vec{c}) \cdot[\vec{a} \times \vec{a}+\vec{b} \times \vec{a}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}] $ - $ =\vec{a} [ \vec{a} \times \vec{a}+\vec{b} \times \vec{a}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c} ] +\vec{b}[ \vec{a} \times \vec{a}+\vec{b} \times \vec{a}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c} ] +\vec{c}[ \vec{a} \times \vec{a}+\vec{b} \times \vec{a}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c} ]$ - $ =\left[\begin{array}{lll}\vec{a} & \vec{a} & \vec{a}\end{array}\right]+\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{a}\end{array}\right]+\left[\begin{array}{lll}\vec{a} & \vec{a} & \vec{c}\end{array}\right]$ </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Problem-2 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Vectors L - 8 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px; text-align:left;'> - $+\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]+\left[\begin{array}{lll}\vec{b} & \vec{a} & \vec{a}\end{array}\right]+\left[\begin{array}{lll}\vec{b} & \vec{b} & \vec{a}\end{array}\right]$ - $ +\left[\begin{array}{lll}\vec{b} & \vec{a} & \vec{c}\end{array}\right]+\left[\begin{array}{lll}\vec{b} & \vec{b} & \vec{c}\end{array}\right]+\left[\begin{array}{lll}\vec{c} & \vec{a} & \vec{a}\end{array}\right]$ - $+\left[\begin{array}{lll}\vec{c} & \vec{b} & \vec{a}\end{array}\right]+\left[\begin{array}{lll}\vec{c} & \vec{a} & \vec{c}\end{array}\right]+\left[\begin{array}{lll}\vec{c} & \vec{b} & \vec{c}\end{array}\right] $ </div> </main> <hr> <footer style = "font-size: 18px">Problem-2 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-3 </footer>

### <Success style="font-size:25px"> Vectors L - 8 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px; text-align:left;'> - $ =\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]+\left[\begin{array}{lll}\vec{b} & \vec{a} & \vec{c}\end{array}\right]+\left[\begin{array}{lll}\vec{c} & \vec{b} & \vec{a}\end{array}\right] $ - $ =\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]-\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]-\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right] $ - (S) is correct. </div> <div style='float: right; width:1%;'> <!-- (https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-12-unit-10-chapter-08-vectors-d0zjavv9f5k-03-A. --> <!-- (https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-12-unit-10-chapter-08-vectors-d0zjavv9f5k-03-B. --> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-3 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Vectors L - 8 </Success> ### <primary style="font-size:24px"> Problem-3 </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px; text-align:left;'> - Let $\vec{a}, \vec{b} \\& \vec{c}$ be three non-zero vectors such that no two of them are collinear \& $(\vec{a} \times \vec{b}) \times \vec{c}=\frac{1}{3}|\vec{b}||\vec{c}| \vec{a}$. If $\theta$ is the angle between ${ }^n$ $\vec{b} \\& \vec{c}$ then find the value of $\theta$. - Solution: - $ (\vec{a} \times \vec{b}) \times \vec{c}=\frac{1}{3}|\vec{b}||\vec{c}| \vec{a} $ - $ -\vec{c} \times(\vec{a} \times \vec{b})=\frac{1}{3}|\vec{b}||\vec{c}| \vec{a} $ - $ -(\vec{c} \cdot \vec{b}) \vec{a}+(\vec{c} \cdot \vec{a}) \vec{b}=\frac{1}{3}|\vec{b}||\vec{c}| \vec{a} .$ </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-3 </span> $\rightarrow$ Solution $\rightarrow$ Problem-4 </footer>

### <Success style="font-size:25px"> Vectors L - 8 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px; text-align:left;'> - $(\vec{c} \cdot \vec{a}) \vec{b}=\left(\frac{1}{3}|\vec{b}||\vec{c}|+\vec{c} \cdot \vec{b}\right) \vec{a}$ - $\vec{c} \cdot \vec{a}=0 $ - and $ \left(\frac{1}{3}|\vec{b}||\vec{c}|+|\vec{b}||\vec{c}| \cos \theta\right)=0 $ - $ \Rightarrow \quad \cos \theta+\frac{1}{3}=0$ - $\rightarrow \cos \theta=\frac{-1}{3} $ - $ \quad \sin \theta-\frac{\sqrt{8}}{3}=\frac{2 \sqrt{2}}{3}$ </div> <div style='float: right; width:1%;'> <!-- (https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-12-unit-10-chapter-08-vectors-d0zjavv9f5k-04-A. --> <!-- (https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-12-unit-10-chapter-08-vectors-d0zjavv9f5k-04-B. --> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Problem-3 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-4 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Vectors L - 8 </Success> ### <primary style="font-size:24px"> Problem-4 </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px; text-align:left;'> - $\vec{a}, \vec{b}$ , $\vec{c}$ are vector such that $|\vec{b}|=|\vec{c}|$ - Show that $[[(\vec{a}+\vec{b}) \times(\vec{a}+\vec{c})] \times(\vec{b} \times \vec{c})] \cdot(\vec{b}+\vec{c})=0$ - Solution: - $(\vec{a}+\vec{b}) \times(\vec{a}+\vec{c})= \vec{a} \times \vec{a}+\vec{b} \times \vec{a}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c} $ - $ {[(\vec{a}+\vec{b}) \times(\vec{a}+\vec{c})] \times(\vec{b} \times \vec{c})=[\vec{b} \times \vec{a}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}] \times(\vec{b} \times \vec{c}) } $ - $=(\vec{b} \times \vec{a}) \times(\vec{b} \times \vec{c})+(\vec{a} \times \vec{c}) \times(\vec{b} \times \vec{c}) $ - $= {[(\vec{b} \times \vec{a}) \cdot \vec{c}] \vec{b}-[(\vec{b} \times \vec{a}) \cdot \vec{b}] \vec{c} } +[(\vec{a} \times \vec{c}) \cdot \vec{c}] \vec{b}-[(\vec{a} \times \vec{c}) \cdot \vec{b}] \vec{c} $ - $= {[\vec{b} \vec{a} \vec{c}] \vec{b}-[\vec{b}\vec{a} \vec{b}] \vec{c} } +[\vec{a} \vec{c} \vec{c}] \vec{b}-[\vec{a} \vec{c} \vec{b}] \vec{c} $ - $=\[\vec{b} \vec{a} \vec{c} ](\vec{b}-\vec{c})$ </div> </main> <hr> <footer style = "font-size: 18px">Problem-3 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-4 </span> $\rightarrow$ Solution $\rightarrow$ Problem-5 </footer>

### <Success style="font-size:25px"> Vectors L - 8 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px; text-align:left;'> - $ {[[(\vec{a}+\vec{b}) \times(\vec{a}+\vec{c})] \times(\vec{b} \times \vec{c})] \cdot(\vec{b}+\vec{c})} $ - $ =\left[\begin{array}{lll}\vec{b} & \vec{a} & \vec{c}\end{array}\right] (\vec{b}-\vec{c}) \cdot(\vec{b}+\vec{c})$ - $ =\left[\begin{array}{lll}\vec{b} & \vec{a} & \vec{c}\end{array}\right](\vec{b} \cdot \vec{b}+\vec{c} \vec{b}-\vec{c} \vec{b}-\vec{c} \cdot \vec{c}) $ - $ =\left[\begin{array}{lll}\vec{b} & \vec{a} &\vec{c}\end{array}\right]\left(|\vec{b}|^2-|\vec{c}|^2\right)=\left[\begin{array}{lll}\vec{b} & \vec{a} & \vec{c}\end{array}\right] 0=0$ </div> <div style='float: right; width:1%;'> <!-- (https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-12-unit-10-chapter-08-vectors-d0zjavv9f5k-05-A. --> <!-- (https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-12-unit-10-chapter-08-vectors-d0zjavv9f5k-05-B. --> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Problem-4 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-5 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Vectors L - 8 </Success> ### <primary style="font-size:24px"> Problem-5 </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px; text-align:left;'> - $\vec{a}=i-j, \vec{b}=j-k$ \& $\vec{c}=k-i$. If $\vec{d}$ is unit vector such that $\vec{a} \cdot \vec{d}^2=0=\left[\vec{b} \vec{c} \vec{d}\right]$ Find $\vec{d}$. - Solution: - $ \vec{d}=x i+y j+z k $ - $\vec{a} \cdot \vec{d}=0 $ - $1 \cdot x-y. 1=0 \Rightarrow x=y $ </main> <hr> <footer style = "font-size: 18px">Problem-4 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-5 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Vectors L - 8 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px; text-align:left;'> - $\left[\begin{array}{lll}\vec{b} & \vec{c} & \vec{d}\end{array}\right]=0 $ - $\left|\begin{array}{ccc}0 & 1 & -1 \\\ -1 & 0 & 1 \\\ x & y & z\end{array}\right|=0$ - $\Rightarrow 2 x+z=0 \Rightarrow z=-2 x$ </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Problem-5 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Exercise </footer>

### <Success style="font-size:25px"> Vectors L - 8 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px; text-align:left;'> - $|\alpha| =1 =\sqrt{x^2+y^2+z^2}$ - $x^2+y^2+z^2=1 \Rightarrow x^2+x^2+4x^2=1$ - $6x^2=1 \Rightarrow 1/6 $ - $x\pm \frac{1}{\sqrt{6}} \Rightarrow 1/6 $ - $\Rightarrow xi +yj +zk = xi +xj + (-2x)k$ $=\pm\frac{1}{\sqrt{6}}(i + j -2k) $ </div> <div style='float: right; width:1%;'> <!-- (https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-12-unit-10-chapter-08-vectors-d0zjavv9f5k-06-A. --> <!-- (https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-12-unit-10-chapter-08-vectors-d0zjavv9f5k-06-B. --> <!-- (https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-12-unit-10-chapter-08-vectors-d0zjavv9f5k-06-C. --> </main> <hr> <footer style = "font-size: 18px">Problem-5 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Exercise $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Vectors L - 8 </Success> ### <primary style="font-size:24px"> Exercise </primary> <hr> <main> <div style='float: left; width:100%;font-size:30px; text-align:left;'> - Which of the following statements are meaningful? - (P) ${\vec{u}} \cdot(\vec{v} \times \vec{w})$ - (Q) $(\vec{u} \cdot \vec{v}) \cdot \vec{\omega}$ - (R) $(\vec{u} \cdot \vec{v}) \vec{\omega}$ - (s) $\vec{u} \times(\vec{v} \cdot \vec{w})$ - Answer: - $(P)$ and $(R)$ are meaningful. </div> <div style='float: right; width:1%;'> <!-- (https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-12-unit-10-chapter-08-vectors-d0zjavv9f5k-07. --> </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Exercise </span> $\rightarrow$ Thank you $\rightarrow$ </footer>

### <Success style="font-size:25px"> Vectors L - 8 </Success> ### <primary style="font-size:24px"> Thank you </primary> <hr> <main> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Exercise $\rightarrow$ <span style = "color:blue"> Thank you </span> $\rightarrow$ $\rightarrow$ </footer>