Vector Algebra

|a|=1 , |b|=1 because they are the unit vectors.

The given vectors are $(\overrightarrow{i}+3\overrightarrow{j}-2\overrightarrow{k})$ and $(-\overrightarrow{i}+3\overrightarrow{k})$.

$(\overrightarrow{i}+3\overrightarrow{j}-2\overrightarrow{k})\times (-\overrightarrow{i}+3\overrightarrow{k})$ represents the cross product of two vectors.

$$\begin{gathered} (\overrightarrow{i}+3\overrightarrow{j}-2\overrightarrow{k})\times (-\overrightarrow{i}+3\overrightarrow{k})=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 1& 3& -2\\ -1& 0& 3\end{array}\right| \\ (\overrightarrow{i}+3\overrightarrow{j}-2\overrightarrow{k})\times (-\overrightarrow{i}+3\overrightarrow{k})=\overrightarrow{i}\left|\begin{array}{cc}3& -2\\ 0& 3\end{array}\right|-\overrightarrow{j}\left|\begin{array}{cc}1& -2\\ -1& 3\end{array}\right|+\overrightarrow{k}\left|\begin{array}{cc}1& 3\\ -1& 0\end{array}\right| \\ (\overrightarrow{i}+3\overrightarrow{j}-2\overrightarrow{k})\times (-\overrightarrow{i}+3\overrightarrow{k})=\overrightarrow{i}(9-0)-\overrightarrow{j}(3-2)+\overrightarrow{k}(0+3) \\ (\overrightarrow{i}+3\overrightarrow{j}-2\overrightarrow{k})\times (-\overrightarrow{i}+3\overrightarrow{k})=9\overrightarrow{i}-\overrightarrow{j}+3\overrightarrow{k} \end{gathered}$$

Hence, the value of $(\overrightarrow{i}+3\overrightarrow{j}-2\overrightarrow{k})\times (-\overrightarrow{i}+3\overrightarrow{k})$ is $9\overrightarrow{i}-\overrightarrow{j}+3\overrightarrow{k}.$

Given, $\left|\overrightarrow{p}+\overrightarrow{q}\right|=\left|\overrightarrow{p}\right|=\left|\overrightarrow{q}\right|=1$

Now,

${\left|\overrightarrow{p}+\overrightarrow{q}\right|}^2={\left|\overrightarrow{p}\right|}^2+{\left|\overrightarrow{q}\right|}^2+2\overrightarrow{p}·\overrightarrow{q}$

$\Rightarrow 1^2=1+1+2\cos \alpha$

$\Rightarrow 1=2+2\cos \alpha$

$\Rightarrow \cos \alpha =-\frac{1}{2}$

$\Rightarrow \alpha =\frac{2\pi }{3}$

|a|=1 , |b|=1 because they are the unit vectors.

$\overrightarrow{a}=2\widehat{i}-\widehat{j}+\widehat{k}$

$\overrightarrow{b}=\lambda \widehat{j}+2\widehat{k};\lambda \in Z$

$\overrightarrow{c}=\overrightarrow{a}\times \overrightarrow{b}=\left(-2-\lambda \right)\widehat{i}-4\widehat{j}+2\lambda \widehat{k}$

$\left|\overrightarrow{c}\right|=\sqrt{53}$

$\Rightarrow 5{\lambda }^2+4\lambda -33=0$

$\lambda =2.2$ or \(–3\)

$\Rightarrow \lambda =-3$

$\overrightarrow{c}=\widehat{i}-4\widehat{j}-6\widehat{k}$

Let $\overrightarrow{d}=y\widehat{j}+z\widehat{k}$

$\left|\overrightarrow{d}\right|=2$

$\Rightarrow y^2+z^2=4$

\(\begin{aligned} & (\vec{c} \cdot \vec{d})^2=(4 y+6 z)^2 \\ & \leq\left(\sqrt{4^2+6^2} \times \sqrt{y^2+z^2}\right)^2 \leq 208\end{aligned}\)

The given vector is $\overrightarrow{i}-\overrightarrow{j}-\overrightarrow{k}.$

The term $\left|\overrightarrow{i}-\overrightarrow{j}-\overrightarrow{k}\right|$ represents

the magnitude of the vector.

The formula for magnitude of

a vector $x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}$ is

$|x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}|=\sqrt{x^2+y^2+z^2}.$

$$\begin{gathered} |\overrightarrow{i}-\overrightarrow{j}-\overrightarrow{k}| \\ =\sqrt{{\left(1\right)}^2+{(-1)}^2+{(-1)}^2} \\ =\sqrt{(1+1+1)} \\ =\sqrt{3} \end{gathered}$$

Hence,

the value of $\left|\overrightarrow{i}-\overrightarrow{j}-\overrightarrow{k}\right|$ is $\sqrt{3}.$

\(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta\)

We know that: \(-1 \leq \cos \theta \leq 1\)

\(\Rightarrow -|\vec{a}||\vec{b}| \leq |\vec{a}||\vec{b}| \cos \theta \leq |\vec{a}||\vec{b}|\)

\(\Rightarrow-|\vec{a}||\vec{b}| \leq \vec{a} \cdot \vec{b} \leq|\vec{a}||\vec{b}|\)

\(\Rightarrow \vec{a} \cdot \vec{b} \leq|\vec{a}||\vec{b}|\)

\(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta\)

\(-1 \leq \cos \theta \leq 1\)

\(\Rightarrow-|\vec{a}||\vec{b}| \leq \vec{a} \cdot \vec{b} \leq|\vec{a}||\vec{b}|\)

\(\Rightarrow \vec{a} \cdot \vec{b} \leq|\vec{a}||\vec{b}|\)

The projection of vector $\overrightarrow{a}$ on $\overrightarrow{b}$ is given by:

${\text{Proj}}_{\overrightarrow{b}}\overrightarrow{a}=\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{b}{|}^2}\overrightarrow{b}$

\(\overrightarrow{B A}=\langle 2-3,-3-(-4), 5-7\rangle\)

\(\overrightarrow{B A}=\langle-1,1,-2\rangle\)

\(\overrightarrow{B A}=-\hat{i}+\hat{j}-2 \hat{k}\)

Let \(\vec{a}=\hat{i}+\hat{k}, \vec{b}=\hat{i}-\hat{k}\)

Now perpendicular vector \(\vec{c} = \vec{a} \times \vec{b}=(\hat{i}+\hat{k}) \times(\hat{i}-\hat{k})\)

\(=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 1 \\ 1 & 0 & -1\end{array}\right|=-\hat{j}(-1-1)=2 \hat{j}\)

Unit vector \(\hat{c}=\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{2 \hat{j}}{2}=\hat{j}\)

The projection of $\overrightarrow{a}$ on $\overrightarrow{b}$ is given by:

$=\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{\left|\overrightarrow{b}\right|}$

Hence, Projection vector along $\overrightarrow{b}$ :

$=\left(\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{\left|\overrightarrow{b}\right|}\right)\hat{b}==\left(\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{{\left|\overrightarrow{b}\right|}^2}\right)\overrightarrow{b}$

The term $\overrightarrow{j}.\overrightarrow{j}$ represents the dot product of two vectors $\overrightarrow{j}$ and $\overrightarrow{j}.$

The dot product of two vectors gives a scalar value.

The vector $\overrightarrow{j}$ can be written as $0\overrightarrow{i}+\overrightarrow{j}+0\overrightarrow{k}$.

$$\begin{gathered} \overrightarrow{j}.\overrightarrow{j}=(0\overrightarrow{i}+\overrightarrow{j}+0\overrightarrow{k}).(0\overrightarrow{i}+\overrightarrow{j}+0\overrightarrow{k}) \\ \overrightarrow{j}.\overrightarrow{j}=0+1+0 \\ \overrightarrow{j}.\overrightarrow{j}=1 \end{gathered}$$

Hence, the value of $\overrightarrow{j}.\overrightarrow{j}$ is $1.$

There given two vectors $(4\overrightarrow{i}+3\overrightarrow{j}+3\overrightarrow{k})$ and $(6\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k})$.

$(4\overrightarrow{i}+3\overrightarrow{j}+3\overrightarrow{k}).(6\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k})$ represents the dot product of two vectors.

$$\begin{gathered} (4\overrightarrow{i}+3\overrightarrow{j}+3\overrightarrow{k}).(6\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k})=4\times 6+3\times (-4)+3\times 1 \\ (4\overrightarrow{i}+3\overrightarrow{j}+3\overrightarrow{k}).(6\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k})=24-12+3 \\ (4\overrightarrow{i}+3\overrightarrow{j}+3\overrightarrow{k}).(6\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k})=12+3 \\ (4\overrightarrow{i}+3\overrightarrow{j}+3\overrightarrow{k}).(6\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k})=15 \end{gathered}$$

Hence, the value of $(4\overrightarrow{i}+3\overrightarrow{j}+3\overrightarrow{k}).(6\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k})$ is $15.$

Using the section formula, the position vector of $\overrightarrow{r}$ is given by:

$\overrightarrow{r}=\frac{1·\overrightarrow{p}+3·\overrightarrow{q}}{3+1}=\frac{\overrightarrow{p}+3\overrightarrow{q}}{4}$

Point $S$ is the midpoint of $PR$.

$\begin{array}{rcl}\overrightarrow{s}& =& \frac{\overrightarrow{p}+\overrightarrow{r}}{2}\\ & =& \frac{\overrightarrow{p}+\frac{\overrightarrow{p}+3\overrightarrow{q}}{4}}{2}\\ & =& \frac{\frac{4\overrightarrow{p}+\overrightarrow{p}+3\overrightarrow{q}}{4}}{2}\\ & =& \frac{5\overrightarrow{p}+3\overrightarrow{q}}{8}\end{array}$

The given vectors are $(\overrightarrow{i}+3\overrightarrow{j}-2\overrightarrow{k})$ and $(-\overrightarrow{i}+3\overrightarrow{k})$.

$(\overrightarrow{i}+3\overrightarrow{j}-2\overrightarrow{k})\times (-\overrightarrow{i}+3\overrightarrow{k})$ represents the cross product of two vectors.

$$\begin{gathered} (\overrightarrow{i}+3\overrightarrow{j}-2\overrightarrow{k})\times (-\overrightarrow{i}+3\overrightarrow{k})=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 1& 3& -2\\ -1& 0& 3\end{array}\right| \\ (\overrightarrow{i}+3\overrightarrow{j}-2\overrightarrow{k})\times (-\overrightarrow{i}+3\overrightarrow{k})=\overrightarrow{i}\left|\begin{array}{cc}3& -2\\ 0& 3\end{array}\right|-\overrightarrow{j}\left|\begin{array}{cc}1& -2\\ -1& 3\end{array}\right|+\overrightarrow{k}\left|\begin{array}{cc}1& 3\\ -1& 0\end{array}\right| \\ (\overrightarrow{i}+3\overrightarrow{j}-2\overrightarrow{k})\times (-\overrightarrow{i}+3\overrightarrow{k})=\overrightarrow{i}(9-0)-\overrightarrow{j}(3-2)+\overrightarrow{k}(0+3) \\ (\overrightarrow{i}+3\overrightarrow{j}-2\overrightarrow{k})\times (-\overrightarrow{i}+3\overrightarrow{k})=9\overrightarrow{i}-\overrightarrow{j}+3\overrightarrow{k} \end{gathered}$$

Hence, the value of $(\overrightarrow{i}+3\overrightarrow{j}-2\overrightarrow{k})\times (-\overrightarrow{i}+3\overrightarrow{k})$ is $9\overrightarrow{i}-\overrightarrow{j}+3\overrightarrow{k}.$

\(\hat{a} \cdot \hat{b}=|\hat{a}||\hat{b}| \cos \theta=\cos \theta\)

\(\sin \theta=\frac{3}{5} \Rightarrow \sin ^2 \theta=\frac{9}{25}\)

\(\cos ^2 \theta=1-\sin ^2 \theta=\frac{16}{25}\)

\(\cos \theta= \pm \frac{4}{5}\)

\(\Rightarrow \hat{a} \cdot \hat{b}= \pm \frac{4}{5}\)

Given: \(\hat{\mathrm{a}}\) and \(\hat{\mathrm{b}}\) are two unit vectors.

So, \(|\hat{\mathrm{a}}|=1\), \(|\hat{\mathrm{b}}|=1\)

Now, \(\hat{a} \cdot \hat{b}=|\hat{a}||\hat{b}| \cos \theta=\cos \theta\)

\(\sin \theta=\frac{3}{5} \Rightarrow \sin ^2 \theta=\frac{9}{25}\)

\(\cos ^2 \theta=1-\sin ^2 \theta=\frac{16}{25}\)

\(\Rightarrow \cos \theta= \pm \frac{4}{5}\)

\(\therefore\hat{a} \cdot \hat{b}= \pm \frac{4}{5}\)

\(\vec{a}=\hat{i}+\hat{k}, \vec{b}=\hat{i}-\hat{k}\)

\(\vec{a} \times \vec{b}=(\hat{i}+\hat{k}) \times(\hat{i}-\hat{k})\)

\(=\hat{i} \times \hat{i}-\hat{i} \times \hat{k}+\hat{k} \times \hat{i}-\hat{k} \times \hat{k}\)

\(=0-(-\hat{j})+\hat{j}-0=2 \hat{j}\)

Unit vector \(=\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{2 \hat{j}}{2}=\hat{j}\)

Given: \(\mathrm{A}(2,-3,5)\) and \(\mathrm{B}(3,-4,7)\)

So, \(\overrightarrow{OA} = 2\hat{i}-3\hat{j}+5 \hat{k}\),

\(\overrightarrow{OB} = 3\hat{i}-4\hat{j}+7 \hat{k}\)

Now, \(\overrightarrow{B A}=\overrightarrow{OA} -\overrightarrow{OB} \)

\(\Rightarrow \overrightarrow{B A}=(2-3)\hat{i}+(-3+4)\hat{j}+(5-7) \hat{k}\)

\(\Rightarrow \overrightarrow{B A}=-\hat{i}+\hat{j}-2 \hat{k}\)

There given two vectors $(4\overrightarrow{i}+3\overrightarrow{j}+3\overrightarrow{k})$ and $(6\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k})$.

$(4\overrightarrow{i}+3\overrightarrow{j}+3\overrightarrow{k}).(6\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k})$ represents the dot product of two vectors.

$$\begin{gathered} (4\overrightarrow{i}+3\overrightarrow{j}+3\overrightarrow{k}).(6\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k})=4\times 6+3\times (-4)+3\times 1 \\ (4\overrightarrow{i}+3\overrightarrow{j}+3\overrightarrow{k}).(6\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k})=24-12+3 \\ (4\overrightarrow{i}+3\overrightarrow{j}+3\overrightarrow{k}).(6\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k})=12+3 \\ (4\overrightarrow{i}+3\overrightarrow{j}+3\overrightarrow{k}).(6\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k})=15 \end{gathered}$$

Hence, the value of $(4\overrightarrow{i}+3\overrightarrow{j}+3\overrightarrow{k}).(6\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k})$ is $15.$

The term $\overrightarrow{j}.\overrightarrow{j}$ represents the dot product of two vectors $\overrightarrow{j}$ and $\overrightarrow{j}.$

The dot product of two vectors gives a scalar value.

The vector $\overrightarrow{j}$ can be written as $0\overrightarrow{i}+\overrightarrow{j}+0\overrightarrow{k}$.

$$\begin{gathered} \overrightarrow{j}.\overrightarrow{j}=(0\overrightarrow{i}+\overrightarrow{j}+0\overrightarrow{k}).(0\overrightarrow{i}+\overrightarrow{j}+0\overrightarrow{k}) \\ \overrightarrow{j}.\overrightarrow{j}=0+1+0 \\ \overrightarrow{j}.\overrightarrow{j}=1 \end{gathered}$$

Hence, the value of $\overrightarrow{j}.\overrightarrow{j}$ is $1.$

$|\overrightarrow{p}+\overrightarrow{q}{|}^2=|\overrightarrow{p}{|}^2+|\overrightarrow{q}{|}^2+2\overrightarrow{p}·\overrightarrow{q}$

$1^2=1+1+2\cos \alpha$

$1=2+2\cos \alpha$

$\cos \alpha =-\frac{1}{2}$

$\alpha =\frac{2\pi }{3}$

Given: Point \(R\) divides line segment \(PQ\) in the ratio \(3:1\)

Using the section formula, the position vector of $\overrightarrow{r}$ is given by:

$\overrightarrow{r}=\frac{1·\overrightarrow{p}+3·\overrightarrow{q}}{3+1}=\frac{\overrightarrow{p}+3\overrightarrow{q}}{4}$

And point $S$ is the midpoint of $PR$.

$\begin{array}{rcl}\overrightarrow{s}& =& \frac{\overrightarrow{p}+\overrightarrow{r}}{2}\\ & =& \frac{\overrightarrow{p}+\frac{\overrightarrow{p}+3\overrightarrow{q}}{4}}{2}\\ & =& \frac{\frac{4\overrightarrow{p}+\overrightarrow{p}+3\overrightarrow{q}}{4}}{2}\\ & =& \frac{5\overrightarrow{p}+3\overrightarrow{q}}{8}\end{array}$

The given vector is $\overrightarrow{i}-\overrightarrow{j}-\overrightarrow{k}.$

The term $\left|\overrightarrow{i}-\overrightarrow{j}-\overrightarrow{k}\right|$ represents

the magnitude of the vector.

The formula for magnitude of

a vector $x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}$ is

$|x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}|=\sqrt{x^2+y^2+z^2}.$

$$\begin{gathered} |\overrightarrow{i}-\overrightarrow{j}-\overrightarrow{k}| \\ =\sqrt{{\left(1\right)}^2+{(-1)}^2+{(-1)}^2} \\ =\sqrt{(1+1+1)} \\ =\sqrt{3} \end{gathered}$$

Hence,

the value of $\left|\overrightarrow{i}-\overrightarrow{j}-\overrightarrow{k}\right|$ is $\sqrt{3}.$